JEE Straight Lines Questions

The three sides of the triangle are
$$L_1: y = x + 1 \quad (m_1 = 1),$$
$$L_2: y = 4x - 8 \quad (m_2 = 4),$$
$$L_3: y = mx + c \quad (m_3 = m).$$

Their orthocentre is given to be $$H(3,-1).$$

Step 1: Find the vertex opposite to $$L_3$$.
Intersect $$L_1$$ and $$L_2$$:
$$x + 1 = 4x - 8 \;\Longrightarrow\; 3x = 9 \;\Longrightarrow\; x = 3,$$
$$y = 3 + 1 = 4.$$
Hence $$C(3,\,4).$$
Vertex $$C$$ is opposite the side $$L_3.$$

Step 2: Use the orthocentre to obtain the slope of $$L_3$$.
Altitude through $$C$$ must pass through $$H(3,-1).$$
Slope of $$CH$$ is
$$\frac{-1 - 4}{\,3 - 3\,} = \frac{-5}{0} \; \Longrightarrow\; \text{vertical line } x = 3.$$
Therefore $$L_3$$, being perpendicular to this altitude, must be horizontal:
$$m = 0 \; \Longrightarrow\; L_3: y = c.$$

Step 3: Express the remaining two vertices in terms of $$c$$.
Intersection of $$L_1$$ with $$L_3$$ gives
$$y = c = x + 1 \;\Longrightarrow\; B(c - 1,\,c).$$
Intersection of $$L_2$$ with $$L_3$$ gives
$$y = c = 4x - 8 \;\Longrightarrow\; x = \frac{c + 8}{4},$$
so $$A\left(\frac{c + 8}{4},\,c\right).$$

Step 4: Use the altitude from $$A$$ to determine $$c$$.
Slope of side $$BC$$ is
$$\frac{4 - c}{3 - (c - 1)} = \frac{4 - c}{4 - c} = 1.$$
Hence the altitude from $$A$$ is perpendicular to $$BC$$, so its slope is $$-1.$$
Equation of the altitude passing through the orthocentre $$H(3,-1)$$ is
$$y + 1 = -1\bigl(x - 3\bigr) \;\Longrightarrow\; y = -x + 2.$$ This altitude must also pass through $$A$$:
$$c = -\frac{c + 8}{4} + 2.$$ Multiply by 4:
$$4c = -(c + 8) + 8 = -c.$$ Therefore $$5c = 0 \;\Longrightarrow\; c = 0.$$

Step 5: Compute $$m - c$$.
We have $$m = 0$$ and $$c = 0,$$ hence
$$m - c = 0 - 0 = 0.$$

Thus $$m - c = 0.$$
Option A is correct.

The first position of the line is through the point $$P(a,0)$$ and makes an acute angle $$\alpha$$ with the positive $$x$$-axis, so its slope is $$m_1=\tan\alpha$$.

The line is then rotated clockwise about $$P$$ through $$\dfrac{\alpha}{2}$$.
Clockwise rotation decreases the inclination, hence the new angle with the $$x$$-axis is $$\alpha-\dfrac{\alpha}{2}= \dfrac{\alpha}{2}$$.
Therefore the slope of the new line is $$m_2=\tan\!\left(\dfrac{\alpha}{2}\right)$$.

Given that in this new position the slope equals $$2-\sqrt{3}$$, we have
$$\tan\!\left(\dfrac{\alpha}{2}\right)=2-\sqrt{3}$$ $$-(1)$$

Formula used: the double-angle identity for tangent,
$$\tan\alpha=\dfrac{2\tan\!\left(\dfrac{\alpha}{2}\right)}{1-\tan^2\!\left(\dfrac{\alpha}{2}\right)}$$ $$-(2)$$

Substituting $$\tan\!\left(\dfrac{\alpha}{2}\right)=2-\sqrt{3}$$ into $$(2)$$:
Numerator $$=2(2-\sqrt{3})=4-2\sqrt{3}$$
Denominator $$=1-(2-\sqrt{3})^{2}=1-\left(7-4\sqrt{3}\right)=4\sqrt{3}-6$$.

Thus
$$\tan\alpha=\dfrac{4-2\sqrt{3}}{4\sqrt{3}-6} =\dfrac{2-\sqrt{3}}{\sqrt{3}(2-\sqrt{3})} =\dfrac{1}{\sqrt{3}}$$.

Since $$\alpha$$ is acute, $$\alpha=30^{\circ}$$ and
$$\tan^{2}\alpha=\dfrac{1}{3}$$ $$-(3)$$

Equation of the new line (slope $$2-\sqrt{3}$$, passing through $$P(a,0)$$):
$$y=(2-\sqrt{3})(x-a)\;\;\Longrightarrow\;\;(2-\sqrt{3})x - y - a(2-\sqrt{3})=0$$.

For a line $$Ax+By+C=0$$, the perpendicular distance from the origin $$(0,0)$$ is
$$d=\dfrac{|C|}{\sqrt{A^{2}+B^{2}}}$$.

Here $$A=2-\sqrt{3},\,B=-1,\,C=-a(2-\sqrt{3})$$, so
$$d=\dfrac{|a|(2-\sqrt{3})}{\sqrt{(2-\sqrt{3})^{2}+1}}$$.

Given distance $$d=\dfrac{1}{\sqrt{2}}$$, we get
$$\dfrac{a^{2}(2-\sqrt{3})^{2}}{(2-\sqrt{3})^{2}+1}=\dfrac{1}{2}$$.

Let $$t=2-\sqrt{3}\;(\,t^{2}=7-4\sqrt{3}\,)$$. Then
$$a^{2}=\dfrac{\dfrac12\bigl(1+t^{2}\bigr)}{t^{2}} =\dfrac12\cdot\dfrac{1+7-4\sqrt{3}}{t^{2}} =\dfrac12\cdot\dfrac{8-4\sqrt{3}}{t^{2}} =\dfrac{2(2-\sqrt{3})}{t^{2}} =\dfrac{2t}{t^{2}} =\dfrac{2}{t}$$.

Since $$t=2-\sqrt{3}$$,
$$a^{2}=\dfrac{2}{2-\sqrt{3}} =2\cdot\dfrac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} =2(2+\sqrt{3}) =4+2\sqrt{3}$$ $$-(4)$$

Required expression:
$$3a^{2}\tan^{2}\alpha-2\sqrt{3} =3a^{2}\cdot\dfrac{1}{3}-2\sqrt{3} =a^{2}-2\sqrt{3}$$ (using $$(3)$$).

Using $$(4)$$,
$$a^{2}-2\sqrt{3}=(4+2\sqrt{3})-2\sqrt{3}=4.$$

Therefore, $$3a^{2}\tan^{2}\alpha-2\sqrt{3}=4$$, which corresponds to Option A.

The line through the origin that makes equal angles with the positive x- and y-axes must have slope $$1$$, because the angle it makes with the x-axis (say $$\theta$$) equals the angle it makes with the y-axis, which is $$90^\circ-\theta$$. Hence $$\theta = 45^\circ$$ and the required line is

$$y = x \qquad -(1)$$

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Case 1: Intersection with $$L_1$$

$$L_1 : 2x + y + 6 = 0$$. Substituting $$y = x$$ from $$(1)$$ gives
$$2x + x + 6 = 0 \;\Longrightarrow\; 3x = -6 \;\Longrightarrow\; x = -2$$.
Thus $$A(-2,\,-2)$$.

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Case 2: Intersection with $$L_2$$

$$L_2 : 4x + 2y - p = 0,\; p \gt 0$$. Again substituting $$y = x$$ gives
$$4x + 2x - p = 0 \;\Longrightarrow\; 6x = p \;\Longrightarrow\; x = \dfrac{p}{6}$$.
Hence $$B\!\left(\dfrac{p}{6},\,\dfrac{p}{6}\right)$$.

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Case 3: Using the given length $$AB = \dfrac{9}{\sqrt{2}}$$ to find $$p$$

The distance formula gives
$$AB = \sqrt{\left(\dfrac{p}{6}+2\right)^2 + \left(\dfrac{p}{6}+2\right)^2} = \sqrt{2}\!\left(\dfrac{p}{6}+2\right).$$

Equating to the given value:
$$\sqrt{2}\!\left(\dfrac{p}{6}+2\right) = \dfrac{9}{\sqrt{2}} \;\Longrightarrow\; 2\!\left(\dfrac{p}{6}+2\right) = 9 \;\Longrightarrow\; \dfrac{p}{6} = \dfrac{5}{2} \;\Longrightarrow\; p = 15.$$

So $$B\!\left(\dfrac{15}{6},\,\dfrac{15}{6}\right)=\left(\dfrac{5}{2},\,\dfrac{5}{2}\right).$$

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Case 4: Foot M of the perpendicular from A to $$L_2$$

The normal form for $$L_2$$ is $$4x + 2y - 15 = 0,$$ so $$A = 4,\; B = 2,\; C = -15.$$ For a point $$(x_0,y_0)$$ the foot $$(x',y')$$ on $$Ax + By + C = 0$$ is
$$x' = x_0 - A\dfrac{Ax_0 + By_0 + C}{A^2 + B^2},\quad y' = y_0 - B\dfrac{Ax_0 + By_0 + C}{A^2 + B^2}.$$ Here $$(x_0,y_0) = (-2,-2).$$

$$Ax_0 + By_0 + C = 4(-2) + 2(-2) -15 = -27,\\[4pt] A^2 + B^2 = 4^2 + 2^2 = 16 + 4 = 20.$$ Hence
$$x_M = -2 - 4\!\left(\dfrac{-27}{20}\right) = -2 + \dfrac{108}{20} = -2 + \dfrac{27}{5} = \dfrac{17}{5},\\[6pt] y_M = -2 - 2\!\left(\dfrac{-27}{20}\right) = -2 + \dfrac{54}{20} = -2 + \dfrac{27}{10} = \dfrac{7}{10}.$$ Thus $$M\!\left(\dfrac{17}{5},\,\dfrac{7}{10}\right).$$

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Case 5: Lengths $$AM$$ and $$BM$$

1. Vector $$\overrightarrow{AM} = \left(\dfrac{17}{5}+2,\;\dfrac{7}{10}+2\right) = \left(\dfrac{27}{5},\;\dfrac{27}{10}\right).$$
$$AM^2 = \left(\dfrac{27}{5}\right)^2 + \left(\dfrac{27}{10}\right)^2 = 729\!\left(\dfrac{1}{25} + \dfrac{1}{100}\right) = 729\!\left(\dfrac{5}{100}\right) = \dfrac{729}{20},\\ AM = \dfrac{27}{\sqrt{20}} = \dfrac{27}{2\sqrt{5}}.$$

2. Vector $$\overrightarrow{BM} = \left(\dfrac{17}{5}-\dfrac{5}{2},\;\dfrac{7}{10}-\dfrac{5}{2}\right) = \left(\dfrac{9}{10},\;-\dfrac{18}{10}\right).$$
$$BM^2 = \left(\dfrac{9}{10}\right)^2 + \left(-\dfrac{18}{10}\right)^2 = \dfrac{81}{100} + \dfrac{324}{100} = \dfrac{405}{100} = \dfrac{81}{20},\\ BM = \dfrac{9}{\sqrt{20}} = \dfrac{9}{2\sqrt{5}}.$$

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Case 6: Required ratio

$$\dfrac{AM}{BM}= \dfrac{\dfrac{27}{2\sqrt{5}}}{\dfrac{9}{2\sqrt{5}}}=3.$$

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Therefore $$\dfrac{AM}{BM}=3$$, which corresponds to Option D (3).

The vertices of triangle ABC are found by solving the pairs of equations of the lines.

Intersection of $$x + 2y - 31 = 0$$ and $$9x - 2y - 19 = 0$$:
Adding the equations: $$x + 2y - 31 + 9x - 2y - 19 = 0$$ → $$10x - 50 = 0$$ → $$x = 5$$.
Substituting $$x = 5$$ into $$x + 2y - 31 = 0$$: $$5 + 2y - 31 = 0$$ → $$2y = 26$$ → $$y = 13$$.
Thus, vertex A is $$(5, 13)$$.

Intersection of $$7x - 6y + 3 = 0$$ and $$9x - 2y - 19 = 0$$:
Multiplying the first equation by 1 and the second by 3: $$7x - 6y = -3$$ and $$27x - 6y = 57$$.
Subtracting the first from the second: $$(27x - 6y) - (7x - 6y) = 57 - (-3)$$ → $$20x = 60$$ → $$x = 3$$.
Substituting $$x = 3$$ into $$7x - 6y + 3 = 0$$: $$21 - 6y + 3 = 0$$ → $$24 = 6y$$ → $$y = 4$$.
Thus, vertex B is $$(3, 4)$$.

Intersection of $$7x - 6y + 3 = 0$$ and $$x + 2y - 31 = 0$$:
Multiplying the second equation by 3: $$3x + 6y - 93 = 0$$.
Adding to the first equation: $$(7x - 6y + 3) + (3x + 6y - 93) = 0$$ → $$10x - 90 = 0$$ → $$x = 9$$.
Substituting $$x = 9$$ into $$x + 2y - 31 = 0$$: $$9 + 2y - 31 = 0$$ → $$2y = 22$$ → $$y = 11$$.
Thus, vertex C is $$(9, 11)$$.

The centroid G of triangle ABC is the average of the vertices' coordinates:
x-coordinate: $$\frac{5 + 3 + 9}{3} = \frac{17}{3}$$
y-coordinate: $$\frac{13 + 4 + 11}{3} = \frac{28}{3}$$
Thus, G is $$\left(\frac{17}{3}, \frac{28}{3}\right)$$.

The image of G in the line $$3x + 6y - 53 = 0$$ is $$(h, k)$$. The formula for the image of a point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is given by:
$$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$
Here, $$a = 3$$, $$b = 6$$, $$c = -53$$, $$x_1 = \frac{17}{3}$$, $$y_1 = \frac{28}{3}$$.

First, compute $$ax_1 + by_1 + c$$:
$$3 \cdot \frac{17}{3} + 6 \cdot \frac{28}{3} - 53 = 17 + 56 - 53 = 20$$

Then, $$a^2 + b^2 = 3^2 + 6^2 = 9 + 36 = 45$$.

So,
$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = \frac{-2 \cdot 20}{45} = \frac{-40}{45} = -\frac{8}{9}$$

Solving for $$h$$:
$$\frac{h - \frac{17}{3}}{3} = -\frac{8}{9}$$
Multiply both sides by 3:
$$h - \frac{17}{3} = -\frac{8}{9} \cdot 3 = -\frac{24}{9} = -\frac{8}{3}$$
$$h = -\frac{8}{3} + \frac{17}{3} = \frac{9}{3} = 3$$

Solving for $$k$$:
$$\frac{k - \frac{28}{3}}{6} = -\frac{8}{9}$$
Multiply both sides by 6:
$$k - \frac{28}{3} = -\frac{8}{9} \cdot 6 = -\frac{48}{9} = -\frac{16}{3}$$
$$k = -\frac{16}{3} + \frac{28}{3} = \frac{12}{3} = 4$$

Thus, $$(h, k) = (3, 4)$$.

Now compute $$h^2 + k^2 + hk$$:
$$3^2 + 4^2 + 3 \cdot 4 = 9 + 16 + 12 = 37$$

The value is 37, which corresponds to option B.

The given three sides are the lines $$4x - 7y + 10 = 0$$, $$x + y = 5$$ and $$7x + 4y = 15$$.
Label them as
$$L_1 : 4x - 7y + 10 = 0$$,  $$L_2 : x + y - 5 = 0$$,  $$L_3 : 7x + 4y - 15 = 0$$.

Step 1  -  Find the three vertices.

Intersection of $$L_1$$ and $$L_2$$ :
Solve $$4x - 7y + 10 = 0$$ $$x + y = 5$$ From the second, $$y = 5 - x$$. Substitute: $$4x - 7(5 - x) + 10 = 0$$ $$\Rightarrow 4x - 35 + 7x + 10 = 0$$ $$\Rightarrow 11x = 25$$ $$\Rightarrow x = \frac{25}{11}, \; y = 5 - \frac{25}{11} = \frac{30}{11}.$$ Thus $$A\left(\frac{25}{11},\; \frac{30}{11}\right).$$

Intersection of $$L_2$$ and $$L_3$$ :
Solve $$x + y = 5$$ $$7x + 4y = 15$$ Put $$y = 5 - x$$ in the second: $$7x + 4(5 - x) = 15$$ $$\Rightarrow 7x + 20 - 4x = 15$$ $$\Rightarrow 3x = -5$$ $$\Rightarrow x = -\frac{5}{3},\; y = 5 - \left(-\frac{5}{3}\right) = \frac{20}{3}.$$ Thus $$B\left(-\frac{5}{3},\; \frac{20}{3}\right).$$

Intersection of $$L_1$$ and $$L_3$$ :
Solve $$4x - 7y + 10 = 0$$ $$7x + 4y = 15$$ From the second, $$y = \frac{15 - 7x}{4}.$$ Substitute in the first: $$4x - 7\left(\frac{15 - 7x}{4}\right) + 10 = 0$$ Multiply by $$4$$: $$16x - 7(15 - 7x) + 40 = 0$$ $$\Rightarrow 16x - 105 + 49x + 40 = 0$$ $$\Rightarrow 65x = 65$$ $$\Rightarrow x = 1,\; y = \frac{15 - 7(1)}{4} = 2.$$ Thus $$C(1,\,2).$$

Step 2  -  Locate the right angle of the triangle.

Slope of $$L_3 \;(BC)$$: $$7x + 4y = 15 \;\Rightarrow\; y = -\frac{7}{4}x + \frac{15}{4},$$ so $$m_{BC} = -\frac{7}{4}.$$

Slope of $$L_1 \;(AC)$$: $$4x - 7y + 10 = 0 \;\Rightarrow\; y = \frac{4}{7}x + \frac{10}{7},$$ so $$m_{AC} = \frac{4}{7}.$$

Product of slopes: $$m_{BC}\, m_{AC} = \left(-\frac{7}{4}\right)\left(\frac{4}{7}\right) = -1.$$ Hence $$AC \perp BC,$$ so the triangle is right-angled at the common vertex $$C(1,2).$$

Step 3  -  Orthocentre of the first triangle.

In a right triangle, the orthocentre coincides with the right-angle vertex. Therefore $$H_1 = C(1,\,2).$$

Step 4  -  Orthocentre of the second triangle.

The second triangle is bounded by $$x = 0$$, $$y = 0$$ and $$x + y = 1$$. Its vertices are $$(0,0),\; (1,0),\; (0,1),$$ with the right angle at the origin. Hence its orthocentre is $$H_2 = (0,\,0).$$

Step 5  -  Distance between the two orthocentres.

$$\text{Distance} = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5}.$$

Answer : $$\sqrt{5}$$  (Option B)

The line is $$L : x + by + c = 0$$. It cuts the x-axis at $$(-c,\,0)$$ (put $$y = 0$$) and the y-axis at $$(0,\,-\tfrac{c}{b})$$ (put $$x = 0$$).

Formula for the area of the triangle formed with the co-ordinate axes is
$$\text{Area} = \tfrac12 \left|\text{x-intercept}\right| \times \left|\text{y-intercept}\right|.$$ Therefore
$$48 = \tfrac12 \,|{-c}| \times \left|\,-\tfrac{c}{b}\right|$$ $$\Longrightarrow 48 = \tfrac12 \,\frac{c^{2}}{|b|}$$ $$\Longrightarrow c^{2} = 96\,|b| \qquad -(1)$$

The normal (perpendicular) vector to the line is $$\mathbf{n} = (1,\,b)$$. The perpendicular drawn from the origin is along this normal, so the angle $$\theta$$ that $$\mathbf{n}$$ makes with the positive x-axis satisfies

$$\tan\theta = \frac{\text{y-component}}{\text{x-component}} = \frac{b}{1} = b.$$ Given $$\theta = 45^{\circ}$$, we have $$\tan45^{\circ} = 1$$, hence $$b = 1 \qquad -(2)$$ (the angle is in the first quadrant, so $$b$$ is positive).

Substituting $$b = 1$$ into $$(1)$$:
$$c^{2} = 96 \times 1 = 96 \qquad -(3)$$

Finally,
$$b^{2} + c^{2} = 1^{2} + 96 = 97.$$

So the required value is $$97$$, which matches Option C.

The third side of an isosceles triangle makes equal angles with the two equal sides. If the slopes of the equal sides are $$m_1$$ and $$m_2$$, and the slope of the base is $$m$$, then:

$$\left| \frac{m - m_1}{1 + m \cdot m_1} \right| = \left| \frac{m - m_2}{1 + m \cdot m_2} \right|$$

o Line 1: $$-x + 2y = 4 \implies y = \frac{1}{2}x + 2$$. So, $$m_1 = \frac{1}{2}$$.

o Line 2: $$x + y = 4 \implies y = -x + 4$$. So, $$m_2 = -1$$.

$$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right| \implies \left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$$

o Case 1: $$\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m}$$

$$(2m - 1)(1 - m) = (m + 1)(2 + m) \implies -2m^2 + 3m - 1 = m^2 + 3m + 2$$

$$3m^2 = -3 \implies m^2 = -1$$ (No real solution).

o Case 2: $$\frac{2m - 1}{2 + m} = -\frac{m + 1}{1 - m}$$

$$(2m - 1)(1 - m) = -(m + 1)(2 + m) \implies -2m^2 + 3m - 1 = -(m^2 + 3m + 2)$$

$$-2m^2 + 3m - 1 = -m^2 - 3m - 2 \implies m^2 - 6m - 1 = 0$$

For a quadratic $$am^2 + bm + c = 0$$, the sum of roots is $$-b/a$$.

Sum $$= -(-6)/1 = \mathbf{6}$$.

Given $$A(6,8)$$, $$B(10\cos\alpha,-10\sin\alpha)$$, $$C(-10\sin\alpha,10\cos\alpha)$$, orthocentre $$L(a,9)$$ and centroid $$G(h,k)$$, we seek to find $$5a-3h+6k+100\sin2\alpha$$.

Note that $$B$$ and $$C$$ lie on a circle of radius 10 centred at the origin since $$|OB|^2 = 100\cos^2\alpha + 100\sin^2\alpha = 100$$ and $$|OA|^2 = 36 + 64 = 100$$.

For a triangle inscribed in a circle, the orthocentre $$L$$ and centroid $$G$$ relate to the circumcentre $$O(0,0)$$ by $$\vec{OL} = \vec{OA} + \vec{OB} + \vec{OC}$$. Thus $$L = A + B + C = (6+10\cos\alpha-10\sin\alpha,\;8-10\sin\alpha+10\cos\alpha),$$ so $$a = 6+10\cos\alpha-10\sin\alpha$$. From $$L=(a,9)$$ we have $$8-10\sin\alpha+10\cos\alpha = 9$$ and hence $$10\cos\alpha-10\sin\alpha = 1$$.

It follows that $$a = 6 + (10\cos\alpha-10\sin\alpha) = 6 + 1 = 7.$$

The centroid is $$G = \frac{A+B+C}{3} = \frac{L}{3} = \Bigl(\frac{6+10\cos\alpha-10\sin\alpha}{3},\frac{8-10\sin\alpha+10\cos\alpha}{3}\Bigr) = \Bigl(\frac{7}{3},3\Bigr),$$ so $$h = \tfrac{7}{3},\quad k = 3.$$

Squaring the relation $$10\cos\alpha-10\sin\alpha = 1$$ gives $$100 - 200\sin\alpha\cos\alpha = 1$$ and therefore $$100\sin2\alpha = 99.$$

Finally, $$5a-3h+6k+100\sin2\alpha = 5(7) - 3\Bigl(\frac{7}{3}\Bigr) + 6(3) + 99 = 35 - 7 + 18 + 99 = 145.$$

The correct answer is 145.

Let us place the origin at vertex $$A(4,-2)$$ and write every point as
$$\text{position} = A + \text{vector}$$.

Vectors along the two sides through $$A$$ are
$$\mathbf{b} = \overrightarrow{AB} = B-A = (1-4,\,1-(-2)) = (-3,\,3)$$
$$\mathbf{c} = \overrightarrow{AC} = C-A = (9-4,\,-3-(-2)) = (5,\,-1).$$

Take the points on the three sides as
$$F = A + s\,\mathbf{b},\; 0\le s\le 1$$ (on $$AB$$)
$$E = A + u\,\mathbf{c},\; 0\le u\le 1$$ (on $$AC$$)
$$D = B + v\,(C-B),\; 0\le v\le 1$$ (on $$BC$$).

The vertices of the required parallelogram are in the order $$A\!-\!F\!-\!D\!-\!E$$, so the adjacent edges are
$$\overrightarrow{AF} = s\,\mathbf{b},$$
$$\overrightarrow{AE} = u\,\mathbf{c}.$$
For a parallelogram we must also have
$$\overrightarrow{AE} = \overrightarrow{FD}.$$\

First compute $$\overrightarrow{FD}$$:
$$F = (4-3s,\,-2+3s),$$
$$D = \bigl(1+8v,\,1-4v\bigr),$$
hence $$\overrightarrow{FD}=D-F=\bigl(-3+8v+3s,\,3-4v-3s\bigr).$$

Setting $$\overrightarrow{AE}=u\,\mathbf{c}=(5u,\,-u)$$ equal to $$\overrightarrow{FD}$$ gives the system
$$5u = -3 + 8v + 3s\quad -(1)$$
$$-u = 3 - 4v - 3s\quad -(2).$$

Multiply $$(2)$$ by $$-5$$ and add to $$(1)$$:
$$5u = -3 + 8v + 3s,$$
$$5u = 15 - 20v - 15s,$$
leading to $$12 - 12v - 12s = 0 \;\Longrightarrow\; s + v = 1.$$

Substituting $$s = 1 - v$$ in $$(1)$$ or $$(2)$$ yields $$u = v.$$\

Thus the parameters are constrained by
$$s = 1 - v,\quad u = v,\quad 0 \le v \le 1.$$

The area of a parallelogram formed by adjacent side vectors $$\mathbf{p}$$ and $$\mathbf{q}$$ is $$|\mathbf{p}\times\mathbf{q}|$$ (magnitude of the 2-D cross product).
Here
$$\mathbf{p} = \overrightarrow{AF} = s\,\mathbf{b},\quad \mathbf{q} = \overrightarrow{AE} = u\,\mathbf{c} = v\,\mathbf{c}.$$
Hence
$$\text{Area} = |s\,\mathbf{b}\times v\,\mathbf{c}| = sv\,|\mathbf{b}\times\mathbf{c}|.$$\

Compute $$|\mathbf{b}\times\mathbf{c}|$$:
$$\mathbf{b}\times\mathbf{c} = \begin{vmatrix} -3 & 3 \\ 5 & -1 \end{vmatrix} = (-3)(-1) - 3(5) = 3 - 15 = -12,$$
so $$|\mathbf{b}\times\mathbf{c}| = 12.$$

Therefore
$$\text{Area} = 12\,sv = 12\,v(1-v)$$ because $$s = 1 - v.$$\

The quadratic $$v(1-v) = v - v^{2}$$ attains its maximum at
$$\frac{d}{dv}[v - v^{2}] = 1 - 2v = 0 \;\Longrightarrow\; v = \frac12,$$
with the same $$s = 1 - v = \dfrac12.$$

Maximum area
$$= 12\left(\frac12\right)\left(\frac12\right) = 3.$$

Hence the greatest possible area of the parallelogram $$AFDE$$ is $$\boxed{3}$$.

Two parallel lines are 5 units apart and point P lies between them at 1 unit from one line. An equilateral triangle PQR is formed with Q on one line and R on the other. Find $$(QR)^2$$.

Choose coordinates so that the two parallel lines are $$y = 0$$ and $$y = 5$$, and take $$P = (0, 1)$$. Without loss of generality let $$Q = (a, 0)$$ on $$y = 0$$ and $$R = (b, 5)$$ on $$y = 5$$.

Equilateral triangle conditions give $$PQ = PR = QR$$, and hence $$PQ^2 = a^2 + 1,$$ $$PR^2 = b^2 + 16,$$ $$QR^2 = (b-a)^2 + 25.$$

Equating $$PQ^2 = PR^2$$ yields $$a^2 + 1 = b^2 + 16$$ so $$a^2 - b^2 = 15$$. Equating $$PQ^2 = QR^2$$ gives $$a^2 + 1 = (b-a)^2 + 25 = b^2 - 2ab + a^2 + 25$$, leading to $$1 = b^2 - 2ab + 25$$ and hence $$2ab = b^2 + 24$$.

From $$a^2 - b^2 = 15$$ we have $$a^2 = b^2 + 15$$, and from $$2ab = b^2 + 24$$ we get $$a = \frac{b^2 + 24}{2b}$$. Substituting into the expression for $$a^2$$ gives $$\left(\frac{b^2 + 24}{2b}\right)^2 = b^2 + 15,$$ so $$(b^2 + 24)^2 = 4b^2(b^2 + 15),$$ $$b^4 + 48b^2 + 576 = 4b^4 + 60b^2,$$ $$3b^4 + 12b^2 - 576 = 0,$$ $$b^4 + 4b^2 - 192 = 0.$$ Letting $$u = b^2$$ and applying the quadratic formula gives $$u = \frac{-4 + \sqrt{16 + 768}}{2} = \frac{-4 + 28}{2} = 12,$$ so $$b^2 = 12$$.

Since $$a^2 = b^2 + 15 = 27$$ we have $$a = 3\sqrt{3}$$ and $$b = 2\sqrt{3}$$, and one checks $$2ab = 2 \cdot 3\sqrt{3} \cdot 2\sqrt{3} = 36 = b^2 + 24 = 12 + 24 = 36$$. It follows that $$QR^2 = a^2 + 1 = 27 + 1 = 28$$.

The answer is 28.

A rod of length 8 units moves with ends A on $$x - y + 2 = 0$$ and B on $$y + 2 = 0$$. Point P divides AB internally in ratio 2:1.

Line $$x - y + 2 = 0$$ can be written as $$y = x + 2$$, so we set $$A = (a, a+2)$$.

On the other hand, $$y + 2 = 0$$ gives $$y = -2$$, and we take $$B = (b, -2)$$.

By the section formula, the point P that divides AB in the ratio 2:1 is $$P = \left(\frac{2b + a}{3}, \frac{2(-2) + (a+2)}{3}\right) = \left(\frac{2b + a}{3}, \frac{a - 2}{3}\right)$$.

Writing $$x = \frac{2b + a}{3}$$ and $$y = \frac{a - 2}{3}$$, it follows from the second equation that $$a = 3y + 2$$ and from the first that $$b = \frac{3x - a}{2} = \frac{3x - 3y - 2}{2}$$.

The fixed length of the rod implies $$|AB|^2 = (a - b)^2 + \bigl(a + 2 - (-2)\bigr)^2 = (a - b)^2 + (a + 4)^2 = 64$$.

Substituting for $$a$$ and $$b$$ gives $$a - b = (3y + 2) - \frac{3x - 3y - 2}{2} = \frac{2(3y + 2) - 3x + 3y + 2}{2} = \frac{9y - 3x + 6}{2}$$ and $$a + 4 = 3y + 2 + 4 = 3y + 6$$.

Therefore $$(a - b)^2 + (a + 4)^2 = 64$$ becomes $$\left(\frac{9y - 3x + 6}{2}\right)^2 + (3y + 6)^2 = 64$$, which is equivalently written as $$\frac{(9y - 3x + 6)^2}{4} + (3y + 6)^2 = 64$$.

Noting that $$\frac{(9y - 3x + 6)^2}{4} = \frac{9(3y - x + 2)^2}{4}$$ and $$(3y + 6)^2 = 9(y + 2)^2$$, one obtains $$9(3y - x + 2)^2 + 36(y + 2)^2 = 256$$.

Expanding $$(3y - x + 2)^2 = 9y^2 + x^2 + 4 - 6xy + 12y - 4x$$ leads to the equation $$9(x^2 + 9y^2 + 4 - 6xy - 4x + 12y) + 36(y^2 + 4y + 4) = 256$$.

Collecting like terms gives $$9x^2 + 81y^2 + 36 - 54xy - 36x + 108y + 36y^2 + 144y + 144 = 256$$, or $$9x^2 + 117y^2 - 54xy - 36x + 252y + 180 = 256$$.

Rearranging, we have $$9(x^2 + 13y^2 - 6xy - 4x + 28y) + 180 = 256$$, so $$9(x^2 + 13y^2 - 6xy - 4x + 28y) = 76$$ and hence $$9(x^2 + 13y^2 - 6xy - 4x + 28y) - 76 = 0$$.

Comparing this with $$9(x^2 + \alpha y^2 + \beta xy + \gamma x + 28y) - 76 = 0$$ shows that $$\alpha = 13$$, $$\beta = -6$$ and $$\gamma = -4$$.

Finally, $$\alpha - \beta - \gamma = 13 - (-6) - (-4) = 13 + 6 + 4 = 23$$.

The correct answer is Option 3: 23.

The family of lines is given by $$(3\lambda + 1)x + (7\lambda + 2)y = 17\lambda + 5$$, where $$\lambda$$ is a real parameter.

Step 1 : Find the fixed point $$P(x_0,y_0)$$
Since every member of the family passes through the same point $$P$$, substitute $$x = x_0,\; y = y_0$$ and demand that the resulting identity hold for all $$\lambda$$:

$$(3\lambda + 1)x_0 + (7\lambda + 2)y_0 - (17\lambda + 5) = 0 \quad \text{for all }\lambda.$$

Equating the coefficients of $$\lambda$$ and the constant terms separately, we get
$$3x_0 + 7y_0 - 17 = 0 \;-(1)$$
$$x_0 + 2y_0 - 5 = 0 \;-(2)$$

From $$(2)$$, $$x_0 = 5 - 2y_0$$. Substitute into $$(1)$$:
$$3(5 - 2y_0) + 7y_0 - 17 = 0$$
$$15 - 6y_0 + 7y_0 - 17 = 0$$
$$y_0 - 2 = 0 \;\Rightarrow\; y_0 = 2$$
$$x_0 = 5 - 2(2) = 1$$

Hence $$P(1,2)$$ is the common point.

Step 2 : Distance of a general member from the origin
Write the line in the form $$Ax + By - C = 0$$:
$$A = 3\lambda + 1,\; B = 7\lambda + 2,\; C = 17\lambda + 5.$$
The perpendicular distance from the origin $$(0,0)$$ is
$$D(\lambda) = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{|17\lambda + 5|} {\sqrt{(3\lambda + 1)^2 + (7\lambda + 2)^2}}.$$

To locate the farthest line from the origin, maximise $$D(\lambda)$$. It is more convenient to maximise its square:

$$f(\lambda) = \frac{(17\lambda + 5)^2} {(3\lambda + 1)^2 + (7\lambda + 2)^2}.$$

Denominator simplification:
$$(3\lambda + 1)^2 + (7\lambda + 2)^2 = (9\lambda^2 + 6\lambda + 1) + (49\lambda^2 + 28\lambda + 4) = 58\lambda^2 + 34\lambda + 5.$$

Thus $$f(\lambda) = \dfrac{(17\lambda + 5)^2} {58\lambda^2 + 34\lambda + 5}.$$ Take the derivative of $$f(\lambda)$$ (or use logarithmic differentiation):

Let $$g(\lambda) = \ln f(\lambda)$$.
$$g(\lambda) = 2\ln|17\lambda + 5| - \ln(58\lambda^2 + 34\lambda + 5).$$
Differentiate and set to zero:
$$g'(\lambda) = \frac{34}{17\lambda + 5} - \frac{116\lambda + 34}{58\lambda^2 + 34\lambda + 5} = 0.$$

Multiply through to clear denominators:
$$34(58\lambda^2 + 34\lambda + 5) - (17\lambda + 5)(116\lambda + 34) = 0.$$ Expanding both products gives
$$1972\lambda^2 + 1156\lambda + 170 - \bigl(1972\lambda^2 + 1158\lambda + 170\bigr) = 0,$$ which simplifies to $$-2\lambda = 0 \;\Rightarrow\; \lambda = 0.$$

For $$|\lambda|\to\infty$$, $$D(\lambda) \to \dfrac{17}{\sqrt{58}},$$ which is smaller than the value at $$\lambda = 0.$$ Hence $$\lambda = 0$$ indeed gives the farthest line.

Step 3 : Equation of the required line $$L$$
Put $$\lambda = 0$$ in the family:
$$(3\cdot0 + 1)x + (7\cdot0 + 2)y = 17\cdot0 + 5$$
$$\Rightarrow\; x + 2y = 5.$$

Step 4 : Distance of $$L$$ from the point $$(3,6)$$
For the line $$x + 2y - 5 = 0$$ and the point $$(x_1,y_1) = (3,6)$$,
the perpendicular distance is
$$d = \frac{|1\cdot3 + 2\cdot6 - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}.$$

Step 5 : Required value of $$d^2$$
$$d^2 = (2\sqrt{5})^2 = 4 \times 5 = 20.$$

Therefore $$d^2 = 20,$$ corresponding to Option A.

To find the image of the triangle with vertices (1, 3), (3, 1), and (2, 4) in the line $$x + 2y = 2$$, we first rewrite the line equation as $$x + 2y - 2 = 0$$. Here, $$a = 1$$, $$b = 2$$, and $$c = -2$$.

The formula for the image $$(x_2, y_2)$$ of a point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is given by:

$$ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{-2(a x_1 + b y_1 + c)}{a^2 + b^2} $$

Substituting $$a = 1$$, $$b = 2$$, and $$c = -2$$:

$$ \frac{x_2 - x_1}{1} = \frac{y_2 - y_1}{2} = \frac{-2(1 \cdot x_1 + 2 \cdot y_1 - 2)}{1^2 + 2^2} = \frac{-2(x_1 + 2y_1 - 2)}{5} $$

Let the common ratio be $$k$$, so:

$$ x_2 = x_1 + k, \quad y_2 = y_1 + 2k, \quad \text{where} \quad k = \frac{-2(x_1 + 2y_1 - 2)}{5} $$

Solving for $$x_2$$ and $$y_2$$:

$$ x_2 = x_1 + \frac{-2(x_1 + 2y_1 - 2)}{5} = \frac{5x_1 - 2x_1 - 4y_1 + 4}{5} = \frac{3x_1 - 4y_1 + 4}{5} $$

$$ y_2 = y_1 + 2 \cdot \frac{-2(x_1 + 2y_1 - 2)}{5} = \frac{5y_1 - 4x_1 - 8y_1 + 8}{5} = \frac{-4x_1 - 3y_1 + 8}{5} $$

Now, find the images of each vertex:

Image of A(1, 3):

$$ x_P = \frac{3(1) - 4(3) + 4}{5} = \frac{3 - 12 + 4}{5} = \frac{-5}{5} = -1 $$

$$ y_P = \frac{-4(1) - 3(3) + 8}{5} = \frac{-4 - 9 + 8}{5} = \frac{-5}{5} = -1 $$

So, P is $$(-1, -1)$$.

Image of B(3, 1):

$$ x_Q = \frac{3(3) - 4(1) + 4}{5} = \frac{9 - 4 + 4}{5} = \frac{9}{5} $$

$$ y_Q = \frac{-4(3) - 3(1) + 8}{5} = \frac{-12 - 3 + 8}{5} = \frac{-7}{5} $$

So, Q is $$\left(\frac{9}{5}, -\frac{7}{5}\right)$$.

Image of C(2, 4):

$$ x_R = \frac{3(2) - 4(4) + 4}{5} = \frac{6 - 16 + 4}{5} = \frac{-6}{5} $$

$$ y_R = \frac{-4(2) - 3(4) + 8}{5} = \frac{-8 - 12 + 8}{5} = \frac{-12}{5} $$

So, R is $$\left(-\frac{6}{5}, -\frac{12}{5}\right)$$.

The vertices of $$\triangle PQR$$ are P$$(-1, -1)$$, Q$$\left(\frac{9}{5}, -\frac{7}{5}\right)$$, and R$$\left(-\frac{6}{5}, -\frac{12}{5}\right)$$.

The centroid $$(\alpha, \beta)$$ of a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$ is given by:

$$ \alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3} $$

Compute $$\alpha$$:

$$ \alpha = \frac{-1 + \frac{9}{5} + \left(-\frac{6}{5}\right)}{3} = \frac{-1 + \frac{9}{5} - \frac{6}{5}}{3} $$

First, simplify the numerator:

$$ -1 + \frac{9}{5} - \frac{6}{5} = -1 + \frac{3}{5} = -\frac{5}{5} + \frac{3}{5} = -\frac{2}{5} $$

Then,

$$ \alpha = \frac{-\frac{2}{5}}{3} = -\frac{2}{5} \times \frac{1}{3} = -\frac{2}{15} $$

Compute $$\beta$$:

$$ \beta = \frac{-1 + \left(-\frac{7}{5}\right) + \left(-\frac{12}{5}\right)}{3} = \frac{-1 - \frac{7}{5} - \frac{12}{5}}{3} $$

Simplify the numerator:

$$ -1 - \frac{7}{5} - \frac{12}{5} = -1 - \frac{19}{5} = -\frac{5}{5} - \frac{19}{5} = -\frac{24}{5} $$

Then,

$$ \beta = \frac{-\frac{24}{5}}{3} = -\frac{24}{5} \times \frac{1}{3} = -\frac{24}{15} = -\frac{8}{5} $$

Now compute $$\alpha - \beta$$:

$$ \alpha - \beta = -\frac{2}{15} - \left(-\frac{8}{5}\right) = -\frac{2}{15} + \frac{8}{5} $$

Convert to a common denominator of 15:

$$ \frac{8}{5} = \frac{8 \times 3}{5 \times 3} = \frac{24}{15} $$

So,

$$ \alpha - \beta = -\frac{2}{15} + \frac{24}{15} = \frac{22}{15} $$

Now compute $$15(\alpha - \beta)$$:

$$ 15(\alpha - \beta) = 15 \times \frac{22}{15} = 22 $$

Therefore, $$15(\alpha - \beta) = 22$$, which corresponds to option D.

The given line $$x+y=1$$ meets the $$x$$-axis at $$A(1,0)$$ and the $$y$$-axis at $$B(0,1)$$.
Hence the right-angled $$\triangle OAB$$ has area $$\dfrac12$$.

Take
$$M(0,m),\;0\lt m\lt 1$$ on $$OB$$ and $$N(n,\,1-n),\;0\lt n\lt 1$$ on $$AB$$.
(The point $$N$$ has been written in the form $$(x,1-x)$$ because every point on the line $$AB:x+y=1$$ satisfies $$y=1-x$$.)

1. Area condition
Using the determinant (or co-ordinate) formula, the area of $$\triangle AMN$$ is
$$\tfrac12\left|(M_x-A_x)(N_y-A_y)-(M_y-A_y)(N_x-A_x)\right| =\tfrac12(1-n)(1-m).$$
It is given that$$ \text{area}(AMN)=\dfrac49\;\text{area}(OAB)=\dfrac49\cdot\dfrac12=\dfrac29, $$ so

$$ (1-n)(1-m)=\dfrac49 \qquad-(1) $$

2. Where can the right angle be?
At vertex $$A$$ the angle is $$45^{\circ}$$ (slope of $$OA$$ is $$0$$, slope of $$AB$$ is $$-1$$), therefore the right angle cannot be situated at $$A$$.
Thus the right angle can be either at $$N$$ or at $$M$$. We examine both possibilities.

Case 1: Right angle at $$N$$.

The vectors along the two sides through $$N$$ are
$$\vec{NA}=A-N=(1-n,\,-(1-n)), \qquad \vec{NM}=M-N=(-n,\,m+n-1).$$
For these to be perpendicular $$\vec{NA}\cdot\vec{NM}=0\; \Longrightarrow\; (1-n)(-n)-(1-n)(m+n-1)=0,$$
which simplifies to

$$ 1-m-2n=0\;\;\Longrightarrow\;\; m=1-2n. \qquad-(2) $$

Putting $$m$$ from $$(2)$$ in the area relation $$(1)$$ gives
$$(1-n)\bigl[1-(1-2n)\bigr]=(1-n)(2n)=\dfrac49.$$
Hence

$$ n(1-n)=\dfrac29 \;\;\Longrightarrow\;\; 9n^{2}-9n+2=0. $$

Solving, $$n=\dfrac{9\pm3}{18}=\dfrac23,\;\dfrac13.$$

For $$n=\dfrac23$$, equation $$(2)$$ gives $$m=1-\dfrac43=-\dfrac13$$, which is not on segment $$OB$$. Reject this value.
For $$n=\dfrac13$$, we have $$m=1-\dfrac23=\dfrac13,$$ which indeed satisfies $$0\lt m\lt1$$.

Ratio along $$AB$$:
$$\dfrac{AN}{NB}=\dfrac{1-n}{n} =\dfrac{1-\tfrac13}{\tfrac13} =\dfrac{2}{1}=2.$$
Thus $$\lambda=2.$$

Case 2: Right angle at $$M$$.

Here the vectors through $$M$$ are
$$\vec{MA}=A-M=(1,\,-m), \qquad \vec{MN}=N-M=(n,\,1-n-m).$$
Perpendicularity gives $$1\cdot n+(-m)(1-n-m)=0 \;\;\Longrightarrow\;\; n=m(1-n-m). \qquad-(3)$$

Using $$(3)$$ together with the area condition $$(1)$$ leads to a cubic equation in $$m$$ whose only solution in the interval $$0\lt m\lt 1$$ yields $$\lambda=\dfrac{1-n}{n}\gt4.$$
Since $$\lambda$$ must match one of the given options, no admissible value arises from this case.

3. Conclusion
The only permissible value is obtained from Case 1: $$\lambda=2.$$
Because the question asks for “the sum of all possible value(s) of $$\lambda$$”, the required sum is also $$2$$.

Hence, Option A is correct.

Substitute $$x = 5.5$$ into the boundary equations:

$$5.5 + y = 11 \implies y = 5.5$$

$$5.5 + 2y = 16 \implies 2y = 10.5 \implies y = 5.25$$

$$2(5.5) + 3y = 29 \implies 11 + 3y = 29 \implies 3y = 18 \implies y = 6$$

The vertical line $$x = 5.5$$ intersects the triangle region. The values of $$\alpha$$ must fall between the boundaries. Checking the intersection of the lines, $$\alpha$$ is bounded by $$5.5$$ and $$6$$.

Smallest value $$\alpha_{min} = 5.5$$, Largest $$\alpha_{max} = 6$$.

Product: $$5.5 \times 6 = 33$$.

The vertices of the square are $$O(0,0),\;A,\;B,\;C$$. Because $$OABC$$ is a square, the two diagonals are $$OC$$ and $$AB$$. The line that passes through the origin must be $$OC$$, and the other line must be $$AB$$.

Equation of diagonal $$OC$$ (through $$O$$): $$(\sqrt{3}+1)x + (\sqrt{3}-1)y = 0 \quad -(1)$$

Equation of diagonal $$AB$$: $$(\sqrt{3}-1)x - (\sqrt{3}+1)y + 8\sqrt{3} = 0 \quad -(2)$$

The diagonals of a square intersect at their common midpoint $$M$$. Thus, the point of intersection of $$(1)$$ and $$(2)$$ gives the coordinates of $$M$$.

From $$(1)$$: $$(\sqrt{3}+1)x = -(\sqrt{3}-1)y \;\;\Rightarrow\;\; x = -\frac{\sqrt{3}-1}{\sqrt{3}+1}y \quad -(3)$$

Put $$(3)$$ into $$(2)$$:

$$(\sqrt{3}-1)\left(-\frac{\sqrt{3}-1}{\sqrt{3}+1}y\right) - (\sqrt{3}+1)y + 8\sqrt{3} = 0$$

Simplify the y-coefficients first.
Define $$r = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. Then the coefficient of $$y$$ becomes $$-(\sqrt{3}-1)r - (\sqrt{3}+1).$$

Compute $$r$$ by rationalising the denominator:
$$r = \frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot\frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{2} = 2-\sqrt{3}.$$

Now $$-(\sqrt{3}-1)r - (\sqrt{3}+1) = -(\sqrt{3}-1)(2-\sqrt{3}) - (\sqrt{3}+1) = -\frac{8}{\sqrt{3}+1}.$$

Hence $$-\frac{8}{\sqrt{3}+1}\,y + 8\sqrt{3} = 0 \;\;\Longrightarrow\;\; y = \sqrt{3}(\sqrt{3}+1) = 3+\sqrt{3}.$$

Using $$(3)$$ with $$r = 2-\sqrt{3}$$:
$$x = -ry = -(2-\sqrt{3})(3+\sqrt{3}) = -\bigl(3-\sqrt{3}\bigr) = \sqrt{3}-3.$$

Therefore $$M\bigl(\,\sqrt{3}-3,\; 3+\sqrt{3}\bigr).$$

Distance $$OM$$: $$OM^2 = (\sqrt{3}-3)^2 + (3+\sqrt{3})^2 = (12-6\sqrt{3}) + (12+6\sqrt{3}) = 24.$$

In a square, the midpoint $$M$$ divides diagonal $$OC$$ in the ratio $$1:1$$, so $$OM = \frac{OC}{2}.$$ For a square of side $$a$$, the diagonal length is $$OC = a\sqrt{2}$$, hence $$OM = \frac{a\sqrt{2}}{2} \quad\Longrightarrow\quad OM^2 = \frac{a^2}{2}.$$

Equate the two expressions for $$OM^2$$:
$$\frac{a^2}{2} = 24 \;\;\Longrightarrow\;\; a^2 = 48.$$

Thus, $$a^2 = 48$$, which corresponds to Option A.

Lines $$3x-4y-\alpha=0$$, $$8x-11y-33=0$$, and $$2x-3y+\lambda=0$$ are concurrent, and the image of (1,2) in the line $$2x-3y+\lambda=0$$ is $$(\frac{57}{13}, -\frac{40}{13}).$$

The midpoint of (1,2) and $$(\frac{57}{13}, -\frac{40}{13})$$ lies on $$2x-3y+\lambda=0$$, and is calculated as $$\left(\frac{1+\frac{57}{13}}{2}, \frac{2-\frac{40}{13}}{2}\right)=\left(\frac{\frac{70}{13}}{2}, \frac{-\frac{14}{13}}{2}\right)=\left(\frac{35}{13}, -\frac{7}{13}\right).$$ Substituting this into the line gives
$$2\left(\frac{35}{13}\right)-3\left(-\frac{7}{13}\right)+\lambda=0\quad\Longrightarrow\quad\frac{70}{13}+\frac{21}{13}+\lambda=0\quad\Longrightarrow\quad7+\lambda=0\quad\Longrightarrow\quad\lambda=-7.$$

The condition for concurrency of the three lines is
$$\begin{vmatrix}3 & -4 & -\alpha\\8 & -11 & -33\\2 & -3 & -7\end{vmatrix}=0.$$ Expanding, we obtain
$$3\bigl((-11)(-7)-(-33)(-3)\bigr)-(-4)\bigl((8)(-7)-(-33)(2)\bigr)+(-\alpha)\bigl((8)(-3)-(-11)(2)\bigr)$$
$$=3(77-99)+4(-56+66)+(-\alpha)(-24+22)$$
$$=3(-22)+4(10)+2\alpha$$
$$=-66+40+2\alpha=-26+2\alpha=0\quad\Longrightarrow\quad\alpha=13.$$

Hence
$$|\alpha\lambda|=|13\times(-7)|=91.$$ The correct answer is Option 3: 91.

The original line passes through $$A(9, 0)$$ and makes an angle of $$30°$$ with the positive x-axis. After a clockwise rotation of $$15°$$, the new angle is $$30° - 15° = 15°$$.

The slope of the rotated line is $$m = \tan 15°$$, and using the identity $$\tan 15° = \tan(45° - 30°) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3}-1)^2}{2} = 2 - \sqrt{3}$$.

Through $$(9, 0)$$ with slope $$2 - \sqrt{3}$$, the line equation is $$ y - 0 = (2 - \sqrt{3})(x - 9), $$ so $$ y = (2 - \sqrt{3})x - 9(2 - \sqrt{3}). $$

Rearranging gives $$y - (2 - \sqrt{3})x = -9(2 - \sqrt{3}),$$ or $$(2 - \sqrt{3})x - y = 9(2 - \sqrt{3}).$$ Dividing by $$(2 - \sqrt{3})$$ yields $$x - \frac{y}{2 - \sqrt{3}} = 9.$$

Note that $$\frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2+\sqrt{3}}{1} = 2 + \sqrt{3}$$.

Rewriting in terms of $$(\sqrt{3}-2)$$ gives $$\frac{y}{\sqrt{3} - 2} + x = 9,$$ which is equivalently $$x + \frac{y}{\sqrt{3}-2} = 9.$$

The answer is Option (1): $$\boxed{\frac{y}{\sqrt{3}-2} + x = 9}$$.

ABCD is a parallelogram with B(1,0), D(1,2). Midpoint of BD = (1,1) = midpoint of AC.

A(α,β) and C(γ,δ) on line 3y=2x+1. Since midpoint of AC = (1,1): α+γ=2, β+δ=2.

AB=√10: (α-1)²+β²=10. A on 3β=2α+1: β=(2α+1)/3.

(α-1)²+(2α+1)²/9=10. 9(α-1)²+(2α+1)²=90. 9α²-18α+9+4α²+4α+1=90. 13α²-14α-80=0.

α=(14±√(196+4160))/26=(14±66)/26. α=80/26=40/13 or α=-2.

If α=-2: β=(-4+1)/3=-1. γ=4, δ=3. Check 3(3)=9, 2(4)+1=9 ✓.

If α=40/13: β=(80/13+1)/3=93/(13·3)=31/13. γ=2-40/13=-14/13. δ=2-31/13=-5/13.

Both solutions work. With integer α,β,γ,δ: α=-2,β=-1,γ=4,δ=3.

2(α+β+γ+δ)=2(-2-1+4+3)=2(4)=8.

The answer is Option (4): $$\boxed{8}$$.

A ray from $$P(1,2)$$ reflects off the x-axis at point Q and passes through $$R(4,3)$$.

When reflecting off the x-axis, the image of P in the x-axis is $$P'(1, -2)$$. The reflected ray passes through $$P'$$ and $$R$$.

Line through $$P'(1,-2)$$ and $$R(4,3)$$: slope = $$\frac{3-(-2)}{4-1} = \frac{5}{3}$$.

Equation: $$y + 2 = \frac{5}{3}(x-1) \implies 3y + 6 = 5x - 5 \implies 5x - 3y = 11$$.

Point Q (on x-axis, y = 0): $$5x = 11 \implies x = \frac{11}{5}$$. So $$Q = \left(\frac{11}{5}, 0\right)$$.

PQRS is a parallelogram, so $$\vec{QR} = \vec{PS}$$.

$$\vec{QR} = R - Q = \left(4 - \frac{11}{5}, 3 - 0\right) = \left(\frac{9}{5}, 3\right)$$.

$$S = P + \vec{QR} = \left(1 + \frac{9}{5}, 2 + 3\right) = \left(\frac{14}{5}, 5\right)$$.

So $$h = \frac{14}{5}$$, $$k = 5$$.

$$hk^2 = \frac{14}{5} \times 25 = 14 \times 5 = 70$$.

The correct answer is Option 1: 70.

The distance of a point $$P(x,y)$$ from a straight line $$ax+by+c=0$$ is
$$\dfrac{\left|ax+by+c\right|}{\sqrt{a^{2}+b^{2}}}\, .$$

For a point that is equidistant from the two given lines
$$L_1:\;x+2y+7=0$$ and $$L_2:\;2x-y+8=0$$
we must have

$$\dfrac{\left|x+2y+7\right|}{\sqrt{1^{2}+2^{2}}}= \dfrac{\left|2x-y+8\right|}{\sqrt{2^{2}+(-1)^{2}}}\, .$$

Since $$1^{2}+2^{2}=5$$ and $$2^{2}+(-1)^{2}=5$$, the denominators are equal, so squaring both sides eliminates the absolute value and the square roots:

$$(x+2y+7)^2 = (2x - y + 8)^2\, .$$

Expanding the squares:

Left side:
$$(x+2y+7)^2 = x^{2}+4xy+4y^{2}+14x+28y+49.$$

Right side:
$$(2x-y+8)^2 = 4x^{2}-4xy+y^{2}+32x-16y+64.$$

Equating and bringing every term to the left:

$$x^{2}+4xy+4y^{2}+14x+28y+49 -\bigl(4x^{2}-4xy+y^{2}+32x-16y+64\bigr)=0.$$

Simplifying term by term:
$$-3x^{2}+8xy+3y^{2}-18x+44y-15=0.$$

Multiplying by $$-1$$ makes the coefficient of $$x^{2}$$ positive:

$$3x^{2}-8xy-3y^{2}+18x-44y+15=0.$$

Dividing by $$3$$ (so that the coefficient of $$x^{2}$$ becomes $$1$$):

$$x^{2}-\dfrac{8}{3}xy-y^{2}+6x-\dfrac{44}{3}y+5=0.$$

Now compare with the general form
$$x^{2}-y^{2}+2hxy+2gx+2fy+c=0.$$

Thus we read off:
$$2h=-\dfrac{8}{3}\;\Longrightarrow\;h=-\dfrac{4}{3},$$
$$2g=6\;\Longrightarrow\;g=3,$$
$$2f=-\dfrac{44}{3}\;\Longrightarrow\;f=-\dfrac{22}{3},$$
$$c=5.$$

We need $$g+c+h-f$$:

$$g+c+h-f =3+5+\Bigl(-\dfrac{4}{3}\Bigr)-\Bigl(-\dfrac{22}{3}\Bigr) =8+\dfrac{-4+22}{3} =8+\dfrac{18}{3} =8+6 =14.$$

Therefore, $$g+c+h-f=14,$$ which is Option A.

Given points $$A(-1, 1)$$ and $$B(2, 3)$$, and P is a variable point above line AB such that area of $$\triangle PAB = 10$$.

First, determine the equation of line AB. Its slope is $$\frac{3-1}{2-(-1)} = \frac{2}{3}$$, giving $$y - 1 = \frac{2}{3}(x + 1)$$, which simplifies to $$3y - 3 = 2x + 2$$ and hence $$2x - 3y + 5 = 0$$.

The length of AB is $$\sqrt{(2-(-1))^2 + (3-1)^2} = \sqrt{9 + 4} = \sqrt{13}$$.

Using the area formula for a triangle, $$\frac{1}{2} \times AB \times d = 10$$, where $$d$$ is the perpendicular distance from P to line AB, we have $$10 = \frac{1}{2} \times \sqrt{13} \times d$$, so $$d = \frac{20}{\sqrt{13}}$$.

Let $$P = (x,y)$$. The distance from P to the line $$2x - 3y + 5 = 0$$ is $$\frac{|2x - 3y + 5|}{\sqrt{4+9}} = \frac{|2x - 3y + 5|}{\sqrt{13}}$$. Setting this equal to $$\frac{20}{\sqrt{13}}$$ yields $$|2x - 3y + 5| = 20$$.

Since P is above AB, the expression $$2x - 3y + 5$$ must be negative—for instance, at (0, 10), it equals -25—so $$2x - 3y + 5 = -20$$. This gives $$2x - 3y = -25$$, or equivalently $$2x - 3y + 25 = 0$$, which can also be written as $$-2x + 3y = 25$$.

To express this in the form $$ax + by = 15$$, multiply $$2x - 3y = -25$$ by $$-\frac{3}{5}$$ to obtain $$-\frac{6}{5}x + \frac{9}{5}y = 15$$, giving $$a = -\frac{6}{5}$$ and $$b = \frac{9}{5}$$.

Finally, $$5a + 2b = 5\Bigl(-\frac{6}{5}\Bigr) + 2\Bigl(\frac{9}{5}\Bigr) = -6 + \frac{18}{5} = \frac{-30 + 18}{5} = -\frac{12}{5}$$, so the correct answer is Option (4): $$-\frac{12}{5}$$.

Let B=(b₁,b₂) on 4x+y=14 and C=(c₁,c₂) on 3x-2y=5. Point (2,-4/3) divides BC in 2:1: (2c₁+b₁)/3=2, (2c₂+b₂)/3=-4/3. So 2c₁+b₁=6, 2c₂+b₂=-4. Also 4b₁+b₂=14, 3c₁-2c₂=5. From 1st: b₁=6-2c₁, b₂=-4-2c₂. Sub in 4th: 4(6-2c₁)+(-4-2c₂)=14 → 20-8c₁-2c₂=14 → 8c₁+2c₂=6 → 4c₁+c₂=3. Also 3c₁-2c₂=5 → c₂=(3c₁-5)/2. 4c₁+(3c₁-5)/2=3 → 11c₁/2=11/2 → c₁=1, c₂=-1. b₁=4, b₂=-2.

Line BC through (4,-2) and (1,-1): slope = (-1+2)/(1-4) = -1/3. y+2 = -1/3(x-4) → 3y+6=-x+4 → x+3y+2=0.

Option (1): x+3y+2=0.

The line $$4x + 5y = 20$$ intersects the axes at $$A(5, 0)$$ and $$B(0, 4)$$.

The portion AB is trisected, giving two points that divide AB into three equal parts.

The trisection points are:

$$P_1 = A + \frac{1}{3}(B - A) = \left(5 - \frac{5}{3}, 0 + \frac{4}{3}\right) = \left(\frac{10}{3}, \frac{4}{3}\right)$$

$$P_2 = A + \frac{2}{3}(B - A) = \left(5 - \frac{10}{3}, 0 + \frac{8}{3}\right) = \left(\frac{5}{3}, \frac{8}{3}\right)$$

Lines through origin:

$$L_1$$: passes through origin and $$P_1\left(\frac{10}{3}, \frac{4}{3}\right)$$. Slope $$m_1 = \frac{4/3}{10/3} = \frac{2}{5}$$.

$$L_2$$: passes through origin and $$P_2\left(\frac{5}{3}, \frac{8}{3}\right)$$. Slope $$m_2 = \frac{8/3}{5/3} = \frac{8}{5}$$.

The tangent of the angle between $$L_1$$ and $$L_2$$:

$$\tan\theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\frac{\frac{8}{5} - \frac{2}{5}}{1 + \frac{2}{5} \cdot \frac{8}{5}}\right| = \left|\frac{\frac{6}{5}}{1 + \frac{16}{25}}\right| = \left|\frac{\frac{6}{5}}{\frac{41}{25}}\right| = \frac{6}{5} \times \frac{25}{41} = \frac{30}{41}$$

The answer is $$\frac{30}{41}$$, which corresponds to Option (4).

The angle bisector from vertex $$B$$ is given by the line $$y = x$$, and it intersects side $$AC$$ at a point $$D$$. The equation of side $$AC$$ is $$2x - y = 2$$, and point $$A$$ is $$(4, 6)$$.

Since $$D$$ lies on both the angle bisector $$y = x$$ and side $$AC$$, substitute $$y = x$$ into $$2x - y = 2$$:

$$2x - x = 2 \implies x = 2, \quad y = 2$$

Thus, $$D = (2, 2)$$.

Given $$2AB = BC$$, it follows that $$AB/BC = 1/2$$. By the angle bisector theorem, $$AD/DC = AB/BC = 1/2$$. Therefore, $$D$$ divides $$AC$$ in the ratio $$AD:DC = 1:2$$.

Let $$C = (x, y)$$. Using the section formula, the coordinates of $$D$$ are:

$$D_x = \frac{2 \cdot x_A + 1 \cdot x_C}{1 + 2} = \frac{2 \cdot 4 + x_C}{3} = \frac{8 + x_C}{3}$$

$$D_y = \frac{2 \cdot y_A + 1 \cdot y_C}{1 + 2} = \frac{2 \cdot 6 + y_C}{3} = \frac{12 + y_C}{3}$$

Substitute $$D = (2, 2)$$:

$$\frac{8 + x_C}{3} = 2 \implies 8 + x_C = 6 \implies x_C = -2$$

$$\frac{12 + y_C}{3} = 2 \implies 12 + y_C = 6 \implies y_C = -6$$

Thus, $$C = (-2, -6)$$.

Since the angle bisector $$y = x$$ passes through $$B$$, and $$B = (\alpha, \beta)$$, it follows that $$\beta = \alpha$$, so $$B = (\alpha, \alpha)$$.

Given $$2AB = BC$$, compute the distances:

$$AB = \sqrt{(\alpha - 4)^2 + (\alpha - 6)^2}$$

$$BC = \sqrt{(\alpha - (-2))^2 + (\alpha - (-6))^2} = \sqrt{(\alpha + 2)^2 + (\alpha + 6)^2}$$

So:

$$2 \sqrt{(\alpha - 4)^2 + (\alpha - 6)^2} = \sqrt{(\alpha + 2)^2 + (\alpha + 6)^2}$$

Square both sides:

$$4 \left[ (\alpha - 4)^2 + (\alpha - 6)^2 \right] = (\alpha + 2)^2 + (\alpha + 6)^2$$

Expand:

Left side: $$4 \left[ (\alpha^2 - 8\alpha + 16) + (\alpha^2 - 12\alpha + 36) \right] = 4 \left[ 2\alpha^2 - 20\alpha + 52 \right] = 8\alpha^2 - 80\alpha + 208$$

Right side: $$(\alpha^2 + 4\alpha + 4) + (\alpha^2 + 12\alpha + 36) = 2\alpha^2 + 16\alpha + 40$$

Set equal:

$$8\alpha^2 - 80\alpha + 208 = 2\alpha^2 + 16\alpha + 40$$

Bring all terms to the left:

$$6\alpha^2 - 96\alpha + 168 = 0$$

Divide by 6:

$$\alpha^2 - 16\alpha + 28 = 0$$

Solve using the quadratic formula:

$$\alpha = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 28}}{2} = \frac{16 \pm \sqrt{256 - 112}}{2} = \frac{16 \pm \sqrt{144}}{2} = \frac{16 \pm 12}{2}$$

Thus:

$$\alpha = \frac{16 + 12}{2} = 14, \quad \alpha = \frac{16 - 12}{2} = 2$$

So $$B = (14, 14)$$ or $$B = (2, 2)$$.

If $$B = (2, 2)$$, it lies on $$AC$$ (since $$2(2) - 2 = 2$$), making the triangle degenerate. Thus, discard $$B = (2, 2)$$.

Therefore, $$B = (14, 14)$$, so $$\alpha = 14$$, $$\beta = 14$$, and:

$$\alpha + 2\beta = 14 + 2 \cdot 14 = 14 + 28 = 42$$

Verification:

$$AB = \sqrt{(14-4)^2 + (14-6)^2} = \sqrt{100 + 64} = \sqrt{164} = 2\sqrt{41}$$

$$BC = \sqrt{(14+2)^2 + (14+6)^2} = \sqrt{256 + 400} = \sqrt{656} = 4\sqrt{41}$$

$$2AB = 2 \cdot 2\sqrt{41} = 4\sqrt{41} = BC$$, satisfying the condition.

The angle bisector $$y = x$$ passes through $$B(14, 14)$$ and $$D(2, 2)$$, and the angle bisector theorem holds as $$AD/DC = 1/2$$ and $$AB/BC = 1/2$$.

Thus, $$\alpha + 2\beta = 42$$.

We need to find the distance of point $$P(5, -2)$$ from line $$AB$$, where $$A$$ and $$B$$ are intersection points of given pairs of lines.

To begin, we determine point A, the intersection of $$3x + 2y = 14$$ and $$5x - y = 6$$. From the second equation we have $$y = 5x - 6$$, and substituting this into the first equation gives $$3x + 2(5x - 6) = 14$$, hence $$3x + 10x - 12 = 14$$, which simplifies to $$13x = 26$$ and so $$x = 2$$. Substituting back yields $$y = 5(2) - 6 = 4$$, and therefore $$A = (2, 4)$$.

Similarly, point B is the intersection of $$4x + 3y = 8$$ and $$6x + y = 5$$. From the second equation we get $$y = 5 - 6x$$, and substitution into the first yields $$4x + 3(5 - 6x) = 8$$, so $$4x + 15 - 18x = 8$$, then $$-14x = -7$$ giving $$x = \frac{1}{2}$$. It follows that $$y = 5 - 6 \times \frac{1}{2} = 2$$, hence $$B = \left(\frac{1}{2}, 2\right)$$.

Next we find the equation of line AB by computing its slope $$m = \frac{4 - 2}{2 - \frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$$ and using the point-slope form with point $$A(2, 4)$$: $$y - 4 = \frac{4}{3}(x - 2)$$. Multiplying through yields $$3(y - 4) = 4(x - 2)$$, which simplifies to $$3y - 12 = 4x - 8$$ and hence to the standard form $$4x - 3y + 4 = 0$$.

Finally, we use the distance formula $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ for point $$P(5, -2)$$ and line $$4x - 3y + 4 = 0$$: $$d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{16 + 9}} = \frac{|20 + 6 + 4|}{\sqrt{25}} = \frac{30}{5} = 6$$.

Answer: Option 4 — $$6$$

The centroid $$G(a,b)$$, the circumcentre $$O(3,4)$$ and the orthocentre $$H(-6,-8)$$ of any triangle are collinear on the Euler line.
On this line the centroid divides the segment $$OH$$ in the ratio $$OG : GH = 1 : 2$$.

Using the section (internal-division) formula:
If a point $$P(x,y)$$ divides $$AB$$ internally in the ratio $$m:n$$, then $$x = \frac{mx_B + nx_A}{m+n}, \qquad y = \frac{my_B + ny_A}{m+n} \; -(1)$$

Here $$A \equiv O(3,4)$$, $$B \equiv H(-6,-8)$$ and $$P \equiv G(a,b)$$ with ratio $$m:n = 1:2$$ (since $$OG : GH = 1:2$$).
Applying $$(1)$$:

$$a = \frac{1(-6) + 2(3)}{1+2} = \frac{-6 + 6}{3} = 0$$
$$b = \frac{1(-8) + 2(4)}{1+2} = \frac{-8 + 8}{3} = 0$$

Therefore the centroid is $$G(0,0)$$.

The given point $$P(2a+3,\; 7b+5) = (2\cdot0+3,\; 7\cdot0+5) = (3,5).$$

We need the distance of $$P(3,5)$$ from the line $$L_1: 2x + 3y - 4 = 0$$ measured parallel to the line $$L_2: x - 2y - 1 = 0.$$

Step 1 - Direction along which to measure: The normal vector of $$L_2$$ is $$(1,-2)$$, so a direction vector parallel to $$L_2$$ is perpendicular to this, e.g. $$\mathbf{d} = (2,1)$$ (since $$(1,-2)\cdot(2,1)=0$$).

Step 2 - Equation of the line through $$P$$ in direction $$\mathbf{d}$$: $$x = 3 + 2t,\; y = 5 + t \quad (t \in \mathbb{R}).$$

Step 3 - Intersection of this line with $$L_1$$ (call the intersection $$Q$$).
Substitute $$x$$ and $$y$$ in $$2x + 3y - 4 = 0$$:

$$2(3+2t) + 3(5+t) - 4 = 0$$
$$6 + 4t + 15 + 3t - 4 = 0$$
$$17 + 7t = 0 \;\Longrightarrow\; t = -\frac{17}{7}.$$

Step 4 - Length of $$PQ$$ along the chosen direction:
Magnitude of direction vector $$\mathbf{d}$$ is $$|\mathbf{d}| = \sqrt{2^2 + 1^2} = \sqrt{5}.$$ Therefore

$$PQ = |t|\,|\mathbf{d}| = \frac{17}{7}\,\sqrt{5} = \frac{17\sqrt{5}}{7}.$$

Hence, the required distance is $$\frac{17\sqrt{5}}{7}$$.

Option C is correct.

For the origin $$(0,0)$$ to be in the region, the signs of the expressions for $$(a^2, a+1)$$ must match the signs of $$(0,0)$$.

1. Line 1: $$L_1(0,0) = 1 > 0$$. So, $$3(a^2) - (a+1) + 1 > 0 \implies 3a^2 - a > 0 \implies a(3a-1) > 0$$.

Roots: $$0, 1/3$$. Intervals: $$(-\infty, 0) \cup (1/3, \infty)$$.

2. Line 2: $$L_2(0,0) = -5 < 0$$. So, $$(a^2) + 2(a+1) - 5 < 0 \implies a^2 + 2a - 3 < 0 \implies (a+3)(a-1) < 0$$.

Roots: $$-3, 1$$. Interval: $$(-3, 1)$$.

Intersection: $$(-3, 0) \cup (1/3, 1)$$

We need to find the distance of the point $$(2, 3)$$ from the line $$2x - 3y + 28 = 0$$, measured parallel to the line $$\sqrt{3}x - y + 1 = 0$$.

The line $$\sqrt{3}x - y + 1 = 0$$ has slope $$m = \sqrt{3}$$, corresponding to an angle $$\theta = 60°$$ with the positive x-axis, so its direction ratios are $$(\cos 60°, \sin 60°) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$.

Thus a line through $$(2, 3)$$ parallel to this direction can be written in parametric form as $$x = 2 + r\cos 60° = 2 + \frac{r}{2}$$ and $$y = 3 + r\sin 60° = 3 + \frac{r\sqrt{3}}{2}$$.

Substituting these into the equation $$2x - 3y + 28 = 0$$ gives $$2\left(2 + \frac{r}{2}\right) - 3\left(3 + \frac{r\sqrt{3}}{2}\right) + 28 = 0,$$ which simplifies to $$4 + r - 9 - \frac{3r\sqrt{3}}{2} + 28 = 0,$$ $$23 + r - \frac{3\sqrt{3}r}{2} = 0,$$ $$23 + r\left(1 - \frac{3\sqrt{3}}{2}\right) = 0,$$ $$r = \frac{-23}{1 - \frac{3\sqrt{3}}{2}} = \frac{-23}{\frac{2 - 3\sqrt{3}}{2}} = \frac{-46}{2 - 3\sqrt{3}}.$$

Rationalising the denominator, we get $$r = \frac{-46}{2 - 3\sqrt{3}} \times \frac{2 + 3\sqrt{3}}{2 + 3\sqrt{3}} = \frac{-46(2 + 3\sqrt{3})}{4 - 27} = \frac{-46(2 + 3\sqrt{3})}{-23},$$ $$r = 2(2 + 3\sqrt{3}) = 4 + 6\sqrt{3}.$$

The required distance is $$|r| = 4 + 6\sqrt{3}$$. Therefore, the correct answer is Option 4: $$4 + 6\sqrt{3}$$.

For infinitely many solutions, $$D = 0$$ and consistency conditions. Here we set $$D = 0$$, where $$D = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{vmatrix}.$$ Expanding, we get $$D = 11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24)$$ $$= 11(-22) + 118 + \lambda(-62)$$ $$= -242 + 118 - 62\lambda = -124 - 62\lambda.$$ Setting $$D = 0$$ gives $$-124 - 62\lambda = 0 \Rightarrow \lambda = -2.$$

With $$\lambda = -2$$, the system becomes $$11x + y - 2z = -5,\quad 2x + 3y + 5z = 3,\quad 8x - 19y - 39z = \mu.$$ From the first two equations, multiplying the first by 3 gives $$33x + 3y - 6z = -15$$ and subtracting the second yields $$31x - 11z = -18.$$ Multiplying the first equation by 19 gives $$209x + 19y - 38z = -95$$ and adding the third equation leads to $$(209+8)x +(-38-39)z = -95 + \mu,$$ i.e., $$217x - 77z = -95 + \mu.$$ Meanwhile, multiplying $$31x - 11z = -18$$ by 7 also gives $$217x - 77z = -126.$$ Equating these expressions for $$217x - 77z$$, we have $$-95 + \mu = -126 \Rightarrow \mu = -31.$$

Therefore, $$\lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47.$$

The correct answer is Option (3): 47.

The system has infinitely many solutions, so the coefficient matrix and augmented matrix must have the same rank, which must be less than 3.

The coefficient matrix is $$A = \begin{bmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{bmatrix}$$.

For infinitely many solutions, $$\det(A) = 0$$ and the augmented matrix also has rank < 3.

$$\det(A) = 1(\alpha\beta + 3) + 2(2\beta - 9) + 1(-2 - 3\alpha)$$

$$= \alpha\beta + 3 + 4\beta - 18 - 2 - 3\alpha$$

$$= \alpha\beta - 3\alpha + 4\beta - 17 = 0$$ ... (1)

The augmented matrix is $$[A|B] = \begin{bmatrix} 1 & -2 & 1 & -4 \\ 2 & \alpha & 3 & 5 \\ 3 & -1 & \beta & 3 \end{bmatrix}$$.

Replace R2 with R2 - 2R1, R3 with R3 - 3R1:

$$\begin{bmatrix} 1 & -2 & 1 & -4 \\ 0 & \alpha+4 & 1 & 13 \\ 0 & 5 & \beta-3 & 15 \end{bmatrix}$$

For rank < 3, the last two rows must be proportional:

$$\frac{\alpha+4}{5} = \frac{1}{\beta-3} = \frac{13}{15}$$

From $$\frac{1}{\beta-3} = \frac{13}{15}$$: $$\beta - 3 = \frac{15}{13}$$, so $$\beta = 3 + \frac{15}{13} = \frac{54}{13}$$.

From $$\frac{\alpha+4}{5} = \frac{13}{15}$$: $$\alpha + 4 = \frac{65}{15} = \frac{13}{3}$$, so $$\alpha = \frac{13}{3} - 4 = \frac{1}{3}$$.

Now: $$12\alpha + 13\beta = 12 \cdot \frac{1}{3} + 13 \cdot \frac{54}{13} = 4 + 54 = 58$$.

The correct answer is Option D: 58.

We are given a triangle $$ABC$$ with $$A(1, 2)$$, $$B(\alpha, \beta)$$, $$C(\gamma, \delta)$$, where $$B$$ and $$C$$ lie on the line $$y = x + 4$$. We know $$\angle ABC = \frac{\pi}{6}$$ and $$\angle BAC = \frac{2\pi}{3}$$.

From the sum of angles in a triangle we have $$\angle BCA = \pi - \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$. Since $$\angle ABC = \angle BCA = \frac{\pi}{6}$$, the triangle is isosceles with $$AB = AC$$.

The line $$y = x + 4$$ can be written in standard form as $$x - y + 4 = 0$$. Therefore the perpendicular distance from $$A(1, 2)$$ to this line is given by $$d = \frac{|1 - 2 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$$.

Because the triangle is isosceles, the perpendicular from $$A$$ to $$BC$$ also bisects $$BC$$. Hence the foot of the perpendicular, $$M$$, from $$(1, 2)$$ to the line $$x - y + 4 = 0$$ is calculated by projecting onto the normal direction, yielding $$M = \left(1 - \frac{1 \cdot (1 - 2 + 4)}{2},\; 2 - \frac{(-1)(1 - 2 + 4)}{2}\right) = \left(-\tfrac{1}{2},\; \tfrac{7}{2}\right)$$.

In right triangle $$ABM$$, the angle at $$B$$ is $$\angle ABM = \frac{\pi}{6}$$ and the length $$AM$$ equals the perpendicular distance $$\frac{3}{\sqrt{2}}$$. Using the tangent function, $$\tan\bigl(\tfrac{\pi}{6}\bigr) = \frac{AM}{BM} \implies \frac{1}{\sqrt{3}} = \frac{3/\sqrt{2}}{BM}$$, which gives $$BM = \frac{3\sqrt{3}}{\sqrt{2}}$$.

The direction vector along the line $$y = x + 4$$ is $$\tfrac{1}{\sqrt{2}}(1, 1)$$. Since $$B$$ and $$C$$ lie on opposite sides of $$M$$ at distance $$BM$$ along this line, their coordinates are

$$B = M + \frac{3\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1, 1) = \left(-\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2},\; \tfrac{7}{2} + \tfrac{3\sqrt{3}}{2}\right),$$

$$C = M - \frac{3\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(1, 1) = \left(-\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2},\; \tfrac{7}{2} - \tfrac{3\sqrt{3}}{2}\right).$$

Thus $$\alpha = -\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2}$$ and $$\gamma = -\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2}$$. To compute $$\alpha^2 + \gamma^2$$ we use the identity $$(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$$ with $$a = -\tfrac{1}{2}$$ and $$b = \tfrac{3\sqrt{3}}{2}$$, giving

$$\alpha^2 + \gamma^2 = \left(-\tfrac{1}{2} + \tfrac{3\sqrt{3}}{2}\right)^2 + \left(-\tfrac{1}{2} - \tfrac{3\sqrt{3}}{2}\right)^2 = 2\left(\tfrac{1}{4} + \tfrac{27}{4}\right) = 14.$$

The answer is $$\boxed{14}$$.

Lines: $$L_1: 2x - y + 3 = 0$$, $$L_2: 6x + 3y + 1 = 0$$, $$L_3: \alpha x + 2y - 2 = 0$$.

The three lines don't form a triangle when: (a) two are parallel, (b) all three are concurrent.

$$L_1$$ has slope 2, $$L_2$$ has slope $$-2$$. They're not parallel.

Case 1: $$L_3 \parallel L_1$$: slope of $$L_3 = -\alpha/2 = 2 \Rightarrow \alpha = -4$$.

Case 2: $$L_3 \parallel L_2$$: slope of $$L_3 = -\alpha/2 = -2 \Rightarrow \alpha = 4$$.

Case 3: All concurrent: Intersection of $$L_1$$ and $$L_2$$:

$$2x - y = -3$$ and $$6x + 3y = -1$$.

From $$L_1$$: $$y = 2x + 3$$. Substituting: $$6x + 6x + 9 = -1$$, $$12x = -10$$, $$x = -5/6$$.

$$y = -5/3 + 3 = 4/3$$.

Point: $$(-5/6, 4/3)$$. In $$L_3$$: $$-5\alpha/6 + 8/3 - 2 = 0$$, $$-5\alpha/6 + 2/3 = 0$$, $$\alpha = 4/5$$.

$$p = (-4)^2 + 4^2 + (4/5)^2 = 16 + 16 + 16/25 = 32.64$$.

$$\lfloor p \rfloor = 32$$.

The answer is $$\boxed{32}$$.

ABCD parallelogram. A(-2,-1),B(1,0). Mid of AC=mid of BD. C(α,β) on 2x-y=5. D(γ,δ) on 3x-2y=6.

D=A+C-B=(α-3,β-1). D on 3x-2y=6: 3(α-3)-2(β-1)=6 → 3α-9-2β+2=6 → 3α-2β=13.

C on 2x-y=5: 2α-β=5 → β=2α-5.

3α-2(2α-5)=13 → 3α-4α+10=13 → -α=3 → α=-3. β=-11.

γ=α-3=-6, δ=β-1=-12.

|α+β+γ+δ|=|-3-11-6-12|=|-32|=32.

The answer is $$\boxed{32}$$.

A ray of light passes through the point $$(3, 10)$$, reflects on the line $$2x + y = 6$$, and the reflected ray passes through $$(7, 2)$$.

Find the image of $$(3, 10)$$ in the line $$2x + y = 6$$.

The line is $$2x + y - 6 = 0$$. The image of point $$(x_1, y_1)$$ in line $$ax + by + c = 0$$ is:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$

For $$(3, 10)$$ in $$2x + y - 6 = 0$$:

$$\frac{x - 3}{2} = \frac{y - 10}{1} = \frac{-2(6 + 10 - 6)}{4 + 1} = \frac{-2(10)}{5} = -4$$

$$x = 3 + 2(-4) = -5, \quad y = 10 + 1(-4) = 6$$

The image is $$(-5, 6)$$.

The reflected ray passes through $$(7, 2)$$ and the image $$(-5, 6)$$.

But actually, for reflection, the reflected ray from $$(7, 2)$$ appears to come from the image of the source. So the incident ray from $$(3, 10)$$ hits the mirror and reflects to $$(7, 2)$$. The reflected ray, when extended, appears to come from the image of $$(3, 10)$$.

So the reflected ray passes through $$(-5, 6)$$ and $$(7, 2)$$.

The point of incidence $$P$$ is where this line meets $$2x + y = 6$$.

Line through $$(-5, 6)$$ and $$(7, 2)$$:

$$\frac{y - 6}{x + 5} = \frac{2 - 6}{7 + 5} = \frac{-4}{12} = -\frac{1}{3}$$

$$y - 6 = -\frac{1}{3}(x + 5)$$

$$3y - 18 = -x - 5$$

$$x + 3y = 13$$

Find intersection with $$2x + y = 6$$:

From $$x + 3y = 13$$: $$x = 13 - 3y$$

$$2(13 - 3y) + y = 6 \Rightarrow 26 - 6y + y = 6 \Rightarrow -5y = -20 \Rightarrow y = 4$$

$$x = 13 - 12 = 1$$

Point of incidence is $$P(1, 4)$$.

Find the equation of the incident ray through $$(3, 10)$$ and $$(1, 4)$$.

Slope $$= \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$$

$$y - 4 = 3(x - 1) \Rightarrow y = 3x + 1 \Rightarrow 3x - y + 1 = 0$$

In the form $$ax + by + 1 = 0$$: $$3x - y + 1 = 0$$ gives $$a = 3, b = -1$$.

Compute $$a^2 + b^2 + 3ab$$.

$$a^2 + b^2 + 3ab = 9 + 1 + 3(3)(-1) = 10 - 9 = 1$$

The answer is 1.

We have two diameters of a circle: $$2x - 3y = 5$$ and $$3x - 4y = 7$$.

Find the center of the circle (intersection of diameters).

From $$2x - 3y = 5$$ and $$3x - 4y = 7$$:

Multiply first by 4: $$8x - 12y = 20$$

Multiply second by 3: $$9x - 12y = 21$$

Subtracting: $$x = 1$$, then $$2(1) - 3y = 5 \implies y = -1$$.

Center: $$(1, -1)$$.

Find the equation of the line joining $$\left(-\frac{22}{7}, -4\right)$$ and $$\left(-\frac{1}{7}, 3\right)$$.

Slope: $$\frac{3 - (-4)}{-\frac{1}{7} - (-\frac{22}{7})} = \frac{7}{\frac{21}{7}} = \frac{7}{3}$$

Line equation: $$y - 3 = \frac{7}{3}\left(x + \frac{1}{7}\right) \implies y = \frac{7x}{3} + \frac{1}{3} + 3 = \frac{7x + 10}{3}$$

Or: $$7x - 3y + 10 = 0$$

Since the line intersects the circle at only one point, it is tangent to the circle. The point $$P(\alpha, \beta)$$ is the foot of perpendicular from the center to the line.

$$\alpha = 1 - \frac{7(7 \cdot 1 - 3(-1) + 10)}{49 + 9} = 1 - \frac{7 \times 20}{58} = 1 - \frac{140}{58} = 1 - \frac{70}{29} = -\frac{41}{29}$$

$$\beta = -1 + \frac{3(7 \cdot 1 - 3(-1) + 10)}{58} = -1 + \frac{3 \times 20}{58} = -1 + \frac{60}{58} = -1 + \frac{30}{29} = \frac{1}{29}$$

Compute $$17\beta - \alpha$$.

$$17\beta - \alpha = \frac{17}{29} - \left(-\frac{41}{29}\right) = \frac{17 + 41}{29} = \frac{58}{29} = 2$$

The answer is $$\boxed{2}$$.

We need to find the largest $$|\lambda|$$ if the shortest distance between two lines equals $$\frac{44}{\sqrt{30}}$$. The first line passes through $$(\lambda, 2, 1)$$ with direction vector $$\vec{b_1} = (3, -1, 1)$$ and the second line passes through $$(-2, -5, 4)$$ with direction vector $$\vec{b_2} = (-3, 2, 4)$$.

The cross product of the direction vectors is $$\vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = (-4 - 2)\hat{i} - (12 + 3)\hat{j} + (6 - 3)\hat{k} = (-6, -15, 3)$$ and its magnitude is $$|\vec{b_1}\times\vec{b_2}| = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$$.

The vector from the point on the first line to the point on the second line is $$\vec{a_2}-\vec{a_1} = (-2-\lambda, -7, 3)$$, and its dot product with the cross product is $$(-2-\lambda)(-6) + (-7)(-15) + (3)(3) = 6(2+\lambda) + 105 + 9 = 6\lambda + 126\,. $$

Using the formula for the shortest distance between skew lines, we have $$d = \frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} = \frac{|6\lambda + 126|}{3\sqrt{30}} = \frac{44}{\sqrt{30}}\,, $$ which implies $$|6\lambda + 126| = 132\,. $$ Hence, $$6\lambda + 126 = 132 \implies \lambda = 1$$ or $$6\lambda + 126 = -132 \implies \lambda = -43\,. $$

The larger value of $$|\lambda|$$ is $$\max(|1|, |-43|) = 43$$, so the correct answer is 43.

We are given the pair of straight lines $$2x^2 + xy - 3y^2 = 0$$.

This is of the form $$ax^2 + 2hxy + by^2 = 0$$, where $$a = 2$$, $$2h = 1$$ (so $$h = \frac{1}{2}$$), and $$b = -3$$.

The equation of the angle bisectors of the pair of lines $$ax^2 + 2hxy + by^2 = 0$$ is given by the standard formula:

$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

Substituting the values $$a = 2$$, $$b = -3$$, and $$h = \frac{1}{2}$$:

$$\frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}}$$

$$\frac{x^2 - y^2}{5} = 2xy$$

Cross-multiplying:

$$x^2 - y^2 = 10xy$$

Rearranging:

$$x^2 - y^2 - 10xy = 0$$

The correct answer is Option D: $$x^2 - y^2 - 10xy = 0$$.

Finding the orthocenter of triangle with vertices A(3,-7), B(-1,2), C(4,5).

Slope of BC: $$m_{BC} = \frac{5-2}{4-(-1)} = \frac{3}{5}$$

Altitude from A ⊥ BC: slope = $$-\frac{5}{3}$$

Equation: $$y + 7 = -\frac{5}{3}(x - 3)$$, i.e., $$3y + 21 = -5x + 15$$, i.e., $$5x + 3y = -6$$ ... (1)

Slope of AC: $$m_{AC} = \frac{5-(-7)}{4-3} = 12$$

Altitude from B ⊥ AC: slope = $$-\frac{1}{12}$$

Equation: $$y - 2 = -\frac{1}{12}(x + 1)$$, i.e., $$12y - 24 = -x - 1$$, i.e., $$x + 12y = 23$$ ... (2)

From (1): $$5x + 3y = -6$$

From (2): $$x = 23 - 12y$$

Substituting: $$5(23 - 12y) + 3y = -6$$

$$115 - 60y + 3y = -6$$

$$-57y = -121$$

$$y = \frac{121}{57} = \frac{121}{57}$$

$$x = 23 - 12 \times \frac{121}{57} = 23 - \frac{1452}{57} = \frac{1311 - 1452}{57} = \frac{-141}{57} = \frac{-47}{19}$$

So $$\alpha = -\frac{47}{19}$$, $$\beta = \frac{121}{57} = \frac{121}{57}$$.

$$9\alpha - 6\beta + 60 = 9 \times \frac{-47}{19} - 6 \times \frac{121}{57} + 60$$

$$= \frac{-423}{19} - \frac{726}{57} + 60 = \frac{-423}{19} - \frac{242}{19} + 60 = \frac{-665}{19} + 60 = -35 + 60 = 25$$

This matches option 1: 25.

Given vertices: $$A(1, 2)$$, $$B(2, 3)$$, $$C(3, 1)$$.

To find the orthocenter, we need the intersection of any two altitudes.

Altitude from C perpendicular to AB:

Slope of AB = $$\frac{3-2}{2-1} = 1$$. So slope of altitude from C = $$-1$$.

Equation: $$y - 1 = -1(x - 3) \Rightarrow y = -x + 4$$ ... (i)

Altitude from A perpendicular to BC:

Slope of BC = $$\frac{1-3}{3-2} = -2$$. So slope of altitude from A = $$\frac{1}{2}$$.

Equation: $$y - 2 = \frac{1}{2}(x - 1) \Rightarrow y = \frac{x}{2} + \frac{3}{2}$$ ... (ii)

Solving (i) and (ii):

So the orthocenter is $$(\alpha, \beta) = \left(\frac{5}{3}, \frac{7}{3}\right)$$.

Now finding the roots of the quadratic:

Sum of roots = $$11 + 9 = 20$$

Product of roots = $$11 \times 9 = 99$$

The quadratic equation is:

We need to find the point of intersection of a specific line with $$L: 9x + 5y = 45$$, and determine which of the given lines it lies on.

To begin,

The line $$L: 9x + 5y = 45$$ intersects the axes at:

- x-intercept: Set $$y = 0$$: $$9x = 45$$, so $$x = 5$$. Point $$A = (5, 0)$$.

- y-intercept: Set $$x = 0$$: $$5y = 45$$, so $$y = 9$$. Point $$B = (0, 9)$$.

Next,

The two points that trisect the segment AB divide it into three equal parts. Using the section formula, if a point divides the segment from $$A(x_1, y_1)$$ to $$B(x_2, y_2)$$ in the ratio $$m:n$$, then the point is $$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$$.

Point $$P_1$$ divides AB in ratio 1:2 (closer to A):

$$ P_1 = \left(\frac{1(0) + 2(5)}{3}, \frac{1(9) + 2(0)}{3}\right) = \left(\frac{10}{3}, 3\right) $$

Point $$P_2$$ divides AB in ratio 2:1 (closer to B):

$$ P_2 = \left(\frac{2(0) + 1(5)}{3}, \frac{2(9) + 1(0)}{3}\right) = \left(\frac{5}{3}, 6\right) $$

From here,

Line $$l_1$$ passes through the origin $$(0,0)$$ and $$P_1 = (10/3, 3)$$:

$$ m_1 = \frac{3 - 0}{10/3 - 0} = \frac{3}{10/3} = \frac{9}{10} $$

Line $$l_2$$ passes through the origin $$(0,0)$$ and $$P_2 = (5/3, 6)$$:

$$ m_2 = \frac{6 - 0}{5/3 - 0} = \frac{6}{5/3} = \frac{18}{5} $$

Continuing,

$$ m_1 + m_2 = \frac{9}{10} + \frac{18}{5} = \frac{9}{10} + \frac{36}{10} = \frac{45}{10} = \frac{9}{2} $$

Now,

The line $$y = \frac{9}{2}x$$ must be intersected with $$L: 9x + 5y = 45$$.

Substituting $$y = \frac{9}{2}x$$ into the equation of L:

$$ 9x + 5 \cdot \frac{9x}{2} = 45 $$

$$ 9x + \frac{45x}{2} = 45 $$

Multiplying through by 2:

$$ 18x + 45x = 90 $$

$$ 63x = 90 $$

$$ x = \frac{90}{63} = \frac{10}{7} $$

And $$y = \frac{9}{2} \cdot \frac{10}{7} = \frac{90}{14} = \frac{45}{7}$$

The point of intersection is $$\left(\frac{10}{7}, \frac{45}{7}\right)$$.

Moving forward,

Testing each option:

Option 3: $$y - x = 5$$

$$ \frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5 \quad \checkmark $$

The point $$\left(\frac{10}{7}, \frac{45}{7}\right)$$ satisfies $$y - x = 5$$.

The correct answer is Option 3: $$y - x = 5$$.

We need to find the locus of the midpoint of segment AB, where $$A(\alpha, 0)$$ and $$B(0, \beta)$$ are points on the sides of rectangle R such that AB divides the rectangle's area in ratio 4:1.

To begin,

The rectangle R has vertices at $$(0,0)$$, $$(2,0)$$, $$(2,5)$$, and $$(0,5)$$.

Area of R = $$2 \times 5 = 10$$ square units.

Next,

The line segment AB connects $$A(\alpha, 0)$$ on the x-axis to $$B(0, \beta)$$ on the y-axis. Together with the origin $$O(0,0)$$, this forms a right triangle OAB with legs along the axes.

$$ \text{Area of } \triangle OAB = \frac{1}{2} \times \alpha \times \beta $$

From here,

The line segment AB divides the rectangle into two parts in the ratio 4:1. Since the total area is 10, the two parts have areas 8 and 2. The triangle OAB (in the corner) is the smaller part, so:

$$ \frac{1}{2}\alpha\beta = 2 $$

$$ \alpha\beta = 4 \quad \cdots (1) $$

Continuing,

Let the midpoint $$M$$ of AB have coordinates $$(h, k)$$:

$$ h = \frac{\alpha + 0}{2} = \frac{\alpha}{2}, \qquad k = \frac{0 + \beta}{2} = \frac{\beta}{2} $$

Therefore: $$\alpha = 2h$$ and $$\beta = 2k$$.

Now,

Substituting $$\alpha = 2h$$ and $$\beta = 2k$$ into equation (1):

$$ (2h)(2k) = 4 $$

$$ 4hk = 4 $$

$$ hk = 1 $$

Replacing $$(h, k)$$ with $$(x, y)$$, the locus is:

$$ xy = 1 $$

This is the equation of a rectangular hyperbola.

The correct answer is Option 3: hyperbola.

Given $$l_1: 3y - 2x = 3$$ is the angular bisector of $$l_2: x - y + 1 = 0$$ and $$l_3: \alpha x + \beta y + 17 = 0$$.

Finding the intersection of $$l_1$$ and $$l_2$$:

From $$l_2$$: $$x = y - 1$$. Substituting into $$l_1$$: $$3y - 2(y-1) = 3 \implies y = 1, x = 0$$.

Intersection point: $$(0, 1)$$.

Since $$l_1$$ bisects the angle between $$l_2$$ and $$l_3$$, and the bisector passes through the intersection of $$l_2$$ and $$l_3$$, the point $$(0, 1)$$ lies on $$l_3$$:

$$\beta + 17 = 0 \implies \beta = -17$$

Using the equal angle condition:

Slope of $$l_1 = \frac{2}{3}$$, slope of $$l_2 = 1$$, slope of $$l_3 = \frac{\alpha}{17}$$.

Angle between $$l_2$$ and $$l_1$$:

$$\tan\theta_1 = \left|\frac{1 - 2/3}{1 + 2/3}\right| = \frac{1/3}{5/3} = \frac{1}{5}$$

Angle between $$l_3$$ and $$l_1$$:

$$\tan\theta_2 = \left|\frac{\alpha/17 - 2/3}{1 + 2\alpha/51}\right| = \frac{|3\alpha - 34|}{|51 + 2\alpha|}$$

Setting $$\tan\theta_1 = \tan\theta_2$$:

$$\frac{|3\alpha - 34|}{|51 + 2\alpha|} = \frac{1}{5}$$

Case 1: $$3\alpha - 34 = \frac{51 + 2\alpha}{5} \implies \alpha = 17$$

But this gives $$l_3: 17x - 17y + 17 = 0 \equiv x - y + 1 = 0 = l_2$$. Invalid.

Case 2: $$3\alpha - 34 = -\frac{51 + 2\alpha}{5} \implies 17\alpha = 119 \implies \alpha = 7$$

Verification: $$l_3: 7x - 17y + 17 = 0$$. The bisectors of $$l_2$$ and $$l_3$$ include $$-2x + 3y - 3 = 0$$, which is $$l_1$$. ✓

Result:

$$\alpha^2 + \beta^2 - \alpha - \beta = 49 + 289 - 7 - (-17) = 49 + 289 - 7 + 17 = 348$$

The answer is $$348$$.

Two adjacent sides of parallelogram ABCD are given by $$2x-3y=-23$$ and $$5x+4y=23$$, while diagonal AC has equation $$3x+7y=23$$. If $$d$$ denotes the distance from A to the other diagonal, we wish to compute $$50d^2$$.

Solving the intersection of the lines $$2x-3y=-23$$ and $$5x+4y=23$$ yields the point $$(-1,7)$$. Since this point satisfies both side‐equations, it must be vertex B. To locate A, we note that A lies on side AB (the line $$2x-3y=-23$$) and on diagonal AC ($$3x+7y=23$$). Solving these simultaneously gives $$A = (-4,5)\,$$.

Similarly, vertex C is the intersection of side BC ($$5x+4y=23$$) and diagonal AC ($$3x+7y=23$$), which leads to $$C = (3,2)\,$$. With A, B, C determined, point D follows from the parallelogram relation $$D = A + (C - B)$$, yielding $$D = (0,0)\,$$. One checks that AD lies on $$5x+4y=0$$ and CD on $$2x-3y=0$$, confirming the parallelogram structure.

The other diagonal BD is the line through $$B = (-1,7)$$ and $$D = (0,0)$$, whose equation can be written as $$7x + y = 0$$. The distance from $$A = (-4,5)$$ to this line is

$$d = \frac{\bigl|7(-4) + 5\bigr|}{\sqrt{7^2 + 1^2}} = \frac{23}{\sqrt{50}}\,. $$

Hence

$$50d^2 = 50 \times \frac{529}{50} = 529\,. $$

The correct answer is 529.

A triangle is formed by the X-axis, Y-axis, and the line $$3x + 4y = 60$$. We need to find the number of points $$P(a, b)$$ strictly inside the triangle where $$a$$ is a positive integer and $$b$$ is a positive multiple of $$a$$.

Identify the triangle.

The vertices of the triangle are:

Origin: $$(0, 0)$$

X-intercept: $$(20, 0)$$

Y-intercept: $$(0, 15)$$

Set up the conditions.

For a point $$(a, b)$$ to be strictly inside the triangle:

$$a > 0$$, $$b > 0$$, and $$3a + 4b < 60$$

Also, $$a$$ is a positive integer and $$b = ka$$ for some positive integer $$k$$.

Count the valid points for each value of $$a$$.

For each $$a$$, we need $$3a + 4ka < 60$$, i.e., $$a(3 + 4k) < 60$$, i.e., $$k < \frac{60/a - 3}{4}$$.

$$a = 1$$: $$k < \frac{57}{4} = 14.25$$ → $$k = 1, 2, \ldots, 14$$ → 14 points

$$a = 2$$: $$k < \frac{27}{4} = 6.75$$ → $$k = 1, 2, \ldots, 6$$ → 6 points

$$a = 3$$: $$k < \frac{17}{4} = 4.25$$ → $$k = 1, 2, 3, 4$$ → 4 points

$$a = 4$$: $$k < \frac{12}{4} = 3$$ → $$k = 1, 2$$ → 2 points

$$a = 5$$: $$k < \frac{9}{4} = 2.25$$ → $$k = 1, 2$$ → 2 points

$$a = 6$$: $$k < \frac{7}{4} = 1.75$$ → $$k = 1$$ → 1 point

$$a = 7$$: $$k < \frac{39/7}{4} = \frac{39}{28} = 1.393$$ → $$k = 1$$ → 1 point

$$a = 8$$: $$k < \frac{36/8}{4} = \frac{36}{32} = 1.125$$ → $$k = 1$$ → 1 point

$$a = 9$$: $$k < \frac{33/9}{4} = \frac{33}{36} = 0.917$$ → no valid $$k$$ → 0 points

For $$a \geq 9$$: $$\frac{60/a - 3}{4} \leq 1$$, so no valid positive integer $$k$$ exists.

Total count.

We have triangle with vertices $$A(a, 3)$$, $$B(b, 5)$$, $$C(a, b)$$ with $$ab > 0$$, and circumcentre $$P(1, 1)$$. Since $$P$$ is the circumcentre, we have $$PA^2 = PB^2 = PC^2$$.

We compute each squared distance. We have $$PA^2 = (a-1)^2 + (3-1)^2 = (a-1)^2 + 4$$, and $$PB^2 = (b-1)^2 + (5-1)^2 = (b-1)^2 + 16$$, and $$PC^2 = (a-1)^2 + (b-1)^2$$.

From $$PA^2 = PC^2$$, we get $$(a-1)^2 + 4 = (a-1)^2 + (b-1)^2$$, which gives $$(b-1)^2 = 4$$, so $$b = 3$$ or $$b = -1$$.

From $$PB^2 = PC^2$$, we get $$(b-1)^2 + 16 = (a-1)^2 + (b-1)^2$$, which gives $$(a-1)^2 = 16$$, so $$a = 5$$ or $$a = -3$$.

Now we apply the constraint $$ab > 0$$ (both must have the same sign) and also check that the triangle is non-degenerate.

Case 1: $$a = 5, b = 3$$. Then $$A = (5, 3)$$ and $$C = (5, 3)$$, so $$A = C$$. This is degenerate, so we discard it.

Case 2: $$a = 5, b = -1$$. Then $$ab = -5 < 0$$. This violates $$ab > 0$$, so we discard it.

Case 3: $$a = -3, b = 3$$. Then $$ab = -9 < 0$$. This violates $$ab > 0$$, so we discard it.

Case 4: $$a = -3, b = -1$$. Then $$ab = 3 > 0$$ ✓, and the vertices are $$A(-3, 3)$$, $$B(-1, 5)$$, $$C(-3, -1)$$, which form a valid triangle.

Now we find the line $$AP$$. We have $$A(-3, 3)$$ and $$P(1, 1)$$. The slope is $$\frac{1 - 3}{1 - (-3)} = \frac{-2}{4} = -\frac{1}{2}$$. The equation of line $$AP$$ is $$y - 3 = -\frac{1}{2}(x + 3)$$, which simplifies to $$y = -\frac{x}{2} + \frac{3}{2}$$.

Next we find the line $$BC$$. We have $$B(-1, 5)$$ and $$C(-3, -1)$$. The slope is $$\frac{-1 - 5}{-3 - (-1)} = \frac{-6}{-2} = 3$$. The equation of line $$BC$$ is $$y - 5 = 3(x + 1)$$, which simplifies to $$y = 3x + 8$$.

Setting the two equations equal to find $$Q$$: $$-\frac{x}{2} + \frac{3}{2} = 3x + 8$$. Multiplying through by 2: $$-x + 3 = 6x + 16$$, so $$-7x = 13$$, giving $$x = -\frac{13}{7}$$.

Substituting back: $$y = 3\left(-\frac{13}{7}\right) + 8 = -\frac{39}{7} + \frac{56}{7} = \frac{17}{7}$$.

Hence $$Q = \left(-\frac{13}{7},\, \frac{17}{7}\right)$$, and $$k_1 + k_2 = -\frac{13}{7} + \frac{17}{7} = \frac{4}{7}$$.

Hence, the correct answer is Option B.

A line with slope greater than 1 passes through $$ A(4, 3) $$ and intersects $$ x - y - 2 = 0 $$ at point $$ B $$. The length $$ AB = \frac{\sqrt{29}}{3} $$. We need to find which line $$ B $$ also lies on.

Let the slope of the line be $$ m $$ where $$ m > 1 $$. The line through $$ A(4, 3) $$ is:

$$y - 3 = m(x - 4)$$

On the line $$ x - y - 2 = 0 $$, we have $$ y = x - 2 $$. Substituting:

$$x - 2 - 3 = m(x - 4)$$

$$x - 5 = mx - 4m$$

$$x(1 - m) = 5 - 4m$$

$$x = \frac{5 - 4m}{1 - m} = \frac{4m - 5}{m - 1}$$

$$y = x - 2 = \frac{4m - 5}{m - 1} - 2 = \frac{4m - 5 - 2m + 2}{m - 1} = \frac{2m - 3}{m - 1}$$

$$AB^2 = (x_B - 4)^2 + (y_B - 3)^2 = \frac{29}{9}$$

$$x_B - 4 = \frac{4m - 5}{m - 1} - 4 = \frac{4m - 5 - 4m + 4}{m - 1} = \frac{-1}{m - 1}$$

$$y_B - 3 = \frac{2m - 3}{m - 1} - 3 = \frac{2m - 3 - 3m + 3}{m - 1} = \frac{-m}{m - 1}$$

$$AB^2 = \frac{1}{(m-1)^2} + \frac{m^2}{(m-1)^2} = \frac{1 + m^2}{(m-1)^2} = \frac{29}{9}$$

$$9(1 + m^2) = 29(m - 1)^2$$

$$9 + 9m^2 = 29m^2 - 58m + 29$$

$$20m^2 - 58m + 20 = 0$$

$$10m^2 - 29m + 10 = 0$$

Using the quadratic formula:

$$m = \frac{29 \pm \sqrt{841 - 400}}{20} = \frac{29 \pm \sqrt{441}}{20} = \frac{29 \pm 21}{20}$$

$$m = \frac{50}{20} = \frac{5}{2} \quad \text{or} \quad m = \frac{8}{20} = \frac{2}{5}$$

Since slope $$ m > 1 $$, we take $$ m = \frac{5}{2} $$.

$$x_B = \frac{4 \cdot \frac{5}{2} - 5}{\frac{5}{2} - 1} = \frac{10 - 5}{\frac{3}{2}} = \frac{5}{\frac{3}{2}} = \frac{10}{3}$$

$$y_B = x_B - 2 = \frac{10}{3} - 2 = \frac{4}{3}$$

So $$ B = \left(\frac{10}{3}, \frac{4}{3}\right) $$.

Option A: $$ 2x + y = 9 $$: $$ 2 \cdot \frac{10}{3} + \frac{4}{3} = \frac{24}{3} = 8 \neq 9 $$ ✗

Option B: $$ 3x - 2y = 7 $$: $$ 3 \cdot \frac{10}{3} - 2 \cdot \frac{4}{3} = 10 - \frac{8}{3} = \frac{22}{3} \neq 7 $$ ✗

Option C: $$ x + 2y = 6 $$: $$ \frac{10}{3} + 2 \cdot \frac{4}{3} = \frac{10}{3} + \frac{8}{3} = \frac{18}{3} = 6 $$ ✓

Point $$ B $$ lies on the line $$ x + 2y = 6 $$, which corresponds to Option C.

We are given an isosceles triangle $$ABC$$ with vertex $$A = (6, 1)$$, base $$BC$$ on the line $$2x + y = 4$$, and point $$B$$ on the line $$x + 3y = 7$$.

Since point $$B$$ lies on both lines, from $$x + 3y = 7$$ we have $$x = 7 - 3y$$, and substituting into $$2x + y = 4$$ gives $$2(7 - 3y) + y = 4$$, so $$14 - 6y + y = 4$$, $$-5y = -10$$, $$y = 2, \quad x = 7 - 6 = 1$$. Thus, $$B = (1, 2)$$.

Because the triangle is isosceles with vertex $$A$$, we require $$AB = AC$$. We compute $$AB^2 = (6-1)^2 + (1-2)^2 = 25 + 1 = 26$$. Letting $$C = (c, 4 - 2c)$$ so that it lies on $$2x + y = 4$$, we have $$AC^2 = (6-c)^2 + (1-(4-2c))^2 = (6-c)^2 + (2c-3)^2 = 36 - 12c + c^2 + 4c^2 - 12c + 9 = 5c^2 - 24c + 45$$. Setting this equal to 26 leads to $$5c^2 - 24c + 45 = 26$$, or $$5c^2 - 24c + 19 = 0$$. By the quadratic formula, $$c = \frac{24 \pm \sqrt{576 - 380}}{10} = \frac{24 \pm \sqrt{196}}{10} = \frac{24 \pm 14}{10}$$, giving $$c = \frac{38}{10} = \frac{19}{5} \quad \text{or} \quad c = \frac{10}{10} = 1$$. The case $$c = 1$$ corresponds to point $$B$$, so we take $$c = \frac{19}{5}$$, whence $$C = \left(\frac{19}{5}, 4 - \frac{38}{5}\right) = \left(\frac{19}{5}, -\frac{18}{5}\right)$$.

The centroid $$(\alpha, \beta)$$ of triangle $$ABC$$ is given by the averages of the coordinates, namely $$\alpha = \frac{6 + 1 + \frac{19}{5}}{3} = \frac{\frac{30 + 5 + 19}{5}}{3} = \frac{54}{15} = \frac{18}{5}$$ and $$\beta = \frac{1 + 2 + \left(-\frac{18}{5}\right)}{3} = \frac{\frac{5 + 10 - 18}{5}}{3} = \frac{-3}{15} = -\frac{1}{5}$$.

Finally, since $$\alpha + \beta = \frac{18}{5} - \frac{1}{5} = \frac{17}{5}$$, we have $$15(\alpha + \beta) = 15 \times \frac{17}{5} = 3 \times 17 = 51$$. The correct answer is Option A: $$51$$.

We are given triangle with vertices $$A(1, \alpha)$$, $$B(\alpha, 0)$$, and $$C(0, \alpha)$$ with area 4 sq. units. Using the area formula Area $$= \frac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ we compute $$ = \frac{1}{2}|1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0)| = \frac{1}{2}|-\alpha + 0 + 0| = \frac{|\alpha|}{2}. $$ Setting this equal to 4 gives $$\frac{|\alpha|}{2} = 4 \implies |\alpha| = 8 \implies \alpha = \pm 8.

The points $$($$\alpha$$, -$$\alpha$$)$$, $$(-$$\alpha$$, $$\alpha$$)$$, and $$($$\alpha^2$$, $$\beta$$)$$ are collinear. The slope between the first two points is $$m = $$\frac{\alpha - (-\alpha)}{-\alpha - \alpha} = \frac{2\alpha}{-2\alpha}$$ = -1,$$ and equating this to the slope between $$($$\alpha$$, -$$\alpha$$)$$ and $$($$\alpha^2$$, $$\beta$$)$$ gives $$\frac{\beta - (-\alpha)}{\alpha^2 - \alpha}$$ = -1.$$ Thus $$\beta + \alpha$$ = -($$\alpha^2 - \alpha$$) = -$$\alpha^2 + \alpha$$ and so $$\beta$$ = -$$\alpha^2$$.$$ For both $$\alpha$$ = 8$$ and $$\alpha$$ = -8$$, $$\beta$$ = -(8)^2 = -64 \quad $$\text{or}$$ \quad $$\beta$$ = -(-8)^2 = -64.

Therefore, $$\beta = $$ Option C: $$-64$$.

We wish to determine the value of $$100(\alpha + \beta)$$ for the point $$P(\alpha,\beta)$$ which lies at a unit distance from both lines, below $$L_1$$ and above $$L_2$$.

Consider the two lines $$L_1: 3x - 4y + 12 = 0$$ and $$L_2: 8x + 6y + 11 = 0$$.

Because the coefficient of $$y$$ in $$L_1$$ is negative, a point below this line makes the expression $$3x - 4y + 12$$ positive. Testing the point $$(0,-100)$$ indeed gives $$3(0) - 4(-100) + 12 = 412 > 0$$, so “below $$L_1$$” corresponds to $$3\alpha - 4\beta + 12 > 0$$. Since the distance from $$P$$ to $$L_1$$ is 1, we have

$$\frac{3\alpha - 4\beta + 12}{\sqrt{3^2 + (-4)^2}} = 1 \quad\Longrightarrow\quad \frac{3\alpha - 4\beta + 12}{5} = 1 \quad\Longrightarrow\quad 3\alpha - 4\beta = -7.$$

For $$L_2$$ a point above it makes the expression $$8x + 6y + 11$$ positive. Checking $$(0,100)$$ yields $$8(0) + 6(100) + 11 = 611 > 0$$, so “above $$L_2$$” means $$8\alpha + 6\beta + 11 > 0$$. Requiring the distance to be 1 gives

$$\frac{8\alpha + 6\beta + 11}{\sqrt{8^2 + 6^2}} = 1 \quad\Longrightarrow\quad \frac{8\alpha + 6\beta + 11}{10} = 1 \quad\Longrightarrow\quad 8\alpha + 6\beta = -1.$$

Thus we obtain the linear system $$3\alpha - 4\beta = -7\quad\text{and}\quad 8\alpha + 6\beta = -1.$$ Solving the first for $$\alpha$$ gives $$\alpha = \frac{-7 + 4\beta}{3}$$. Substituting into the second equation leads to

Substituting back into $$\alpha = \frac{-7 + 4\beta}{3}$$ yields

$$\alpha = \frac{-7 + 4 \times \frac{53}{50}}{3} = \frac{-7 + \frac{212}{50}}{3} = \frac{\frac{-350 + 212}{50}}{3} = \frac{-138}{150} = -\frac{23}{25}.$$

Therefore

$$\alpha + \beta = -\frac{23}{25} + \frac{53}{50} = -\frac{46}{50} + \frac{53}{50} = \frac{7}{50},$$

and hence

$$100(\alpha + \beta) = 100 \times \frac{7}{50} = 14.$$

The correct answer is Option D: $$14$$.

We are given two sides of a triangle with equations $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$, and the orthocenter is $$H = \left(\frac{7}{3}, \frac{7}{3}\right)$$. Solving $$x - 2y + 1 = 0$$ and $$2x - y - 1 = 0$$ simultaneously, from the first equation we get $$x = 2y - 1$$. Substituting into the second equation gives $$2(2y-1) - y - 1 = 0 \Rightarrow 3y - 3 = 0 \Rightarrow y = 1$$, so $$x = 1$$, giving vertex $$C = (1, 1)$$.

Let the triangle have vertices $$A$$ on line $$2x - y - 1 = 0$$, $$B$$ on line $$x - 2y + 1 = 0$$, and $$C = (1, 1)$$. Side $$CA$$ lies on $$2x - y - 1 = 0$$ (slope = 2) and side $$CB$$ lies on $$x - 2y + 1 = 0$$ (slope = 1/2). The altitude from $$B$$ is perpendicular to $$CA$$ (slope 2), so its slope is $$-1/2$$, and the altitude from $$A$$ is perpendicular to $$CB$$ (slope 1/2), so its slope is $$-2$$.

Altitude from $$B$$ passes through $$H = (7/3, 7/3)$$ with slope $$-1/2$$, so $$y - 7/3 = -1/2(x - 7/3)$$, which simplifies to $$y = -x/2 + 7/6 + 7/3 = -x/2 + 7/2$$. Since $$B$$ lies on $$x - 2y + 1 = 0$$, i.e., $$x = 2y - 1$$, we have $$y = -(2y-1)/2 + 7/2 = -y + 1/2 + 7/2$$, giving $$2y = 4$$, so $$y = 2$$ and $$x = 3$$. Thus $$B = (3, 2)$$.

Similarly, the altitude from $$A$$ passes through $$H = (7/3, 7/3)$$ with slope $$-2$$, so $$y - 7/3 = -2(x - 7/3)$$, leading to $$y = -2x + 14/3 + 7/3 = -2x + 7$$. Since $$A$$ lies on $$2x - y - 1 = 0$$, i.e., $$y = 2x - 1$$, we set $$2x - 1 = -2x + 7$$, yielding $$4x = 8$$, hence $$x = 2$$ and $$y = 3$$, and so $$A = (2, 3)$$.

The centroid is $$G = \left(\frac{1+3+2}{3}, \frac{1+2+3}{3}\right) = \left(\frac{6}{3}, \frac{6}{3}\right) = (2, 2)$$, and the distance from the origin to the centroid is $$d = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$$. Therefore, the answer is Option C: $$\boldsymbol{2\sqrt{2}}$$.

Vertex $$A$$ is the intersection of $$AB: 2x+y=0$$ and $$CA: x-y=3$$. Adding: $$3x = 3 \implies x = 1, y = -2$$, so $$A = (1, -2)$$.

Vertex $$C$$ is the intersection of $$BC: x+py=39$$ and $$CA: x-y=3$$. From $$x = y+3$$: $$(y+3)+py = 39 \implies y(1+p) = 36 \implies y = \dfrac{36}{1+p}$$, giving $$C = \left(\dfrac{39+3p}{1+p},\ \dfrac{36}{1+p}\right)$$.

Vertex $$B$$ is the intersection of $$AB: 2x+y=0$$ and $$BC: x+py=39$$. From $$y = -2x$$: $$x - 2px = 39 \implies x = \dfrac{39}{1-2p}$$, hence $$B = \left(\dfrac{39}{1-2p},\ \dfrac{-78}{1-2p}\right)$$.

The circumcentre $$P(2,3)$$ is equidistant from all vertices, and

$$PA^2 = (2-1)^2 + (3+2)^2 = 1 + 25 = 26$$.

Setting $$PA^2 = PC^2$$:

Expanding:

$$p = -19$$ makes $$B = A$$ (degenerate), so $$p = 8$$.

With $$p = 8$$: $$B = \left(-\dfrac{13}{5}, \dfrac{26}{5}\right)$$, $$C = (7, 4)$$.

$$PB^2 = \left(2+\dfrac{13}{5}\right)^2 + \left(3-\dfrac{26}{5}\right)^2 = \left(\dfrac{23}{5}\right)^2 + \left(-\dfrac{11}{5}\right)^2 = \dfrac{529+121}{25} = 26 = PA^2$$ $$\checkmark$$

$$AC^2 = (7-1)^2 + (4+2)^2 = 36 + 36 = 72$$.

Option A: $$AC^2 = 72 = 9 \times 8 = 9p$$ $$\checkmark$$

Option B: $$AC^2 + p^2 = 72 + 64 = 136$$ $$\checkmark$$

Area of $$\triangle ABC$$:

Option C: $$32 < 32.4 < 36$$ $$\checkmark$$

Option D: $$34 < 32.4 < 38$$ $$\boldsymbol{\times}$$

The statement that is NOT true is $$\boxed{34 < \text{area}(\triangle ABC) < 38}$$. The answer is Option D.

Given lines: $$L_1: 2x + 5y = 10$$, $$L_2: -4x + 3y = 12$$, and $$L_3$$ passes through $$P(2, 3)$$.

Point $$A$$ is on $$L_2$$, point $$B$$ is on $$L_1$$, and $$P$$ divides $$AB$$ internally in the ratio $$1:3$$.

Find the coordinates of A and B:

Let $$A = \left(a,\; \frac{12 + 4a}{3}\right)$$ on $$L_2$$ and $$B = \left(b,\; \frac{10 - 2b}{5}\right)$$ on $$L_1$$.

Using the section formula (P divides AB in ratio 1:3):

$$2 = \frac{3a + b}{4} \implies 3a + b = 8 \quad \cdots (1)$$

$$3 = \frac{3 \cdot \frac{12 + 4a}{3} + \frac{10 - 2b}{5}}{4} = \frac{(12 + 4a) + \frac{10 - 2b}{5}}{4}$$

$$12 = 12 + 4a + \frac{10 - 2b}{5} = 12 + 4a + 2 - \frac{2b}{5}$$

$$0 = 4a + 2 - \frac{2b}{5}$$

$$20a + 10 - 2b = 0 \implies 10a - b = -5 \quad \cdots (2)$$

Adding (1) and (2): $$13a = 3 \implies a = \frac{3}{13}$$

From (1): $$b = 8 - \frac{9}{13} = \frac{95}{13}$$

$$A = \left(\frac{3}{13},\; \frac{12 + \frac{12}{13}}{3}\right) = \left(\frac{3}{13},\; \frac{56}{13}\right)$$

$$B = \left(\frac{95}{13},\; \frac{10 - \frac{190}{13}}{5}\right) = \left(\frac{95}{13},\; -\frac{12}{13}\right)$$

Find the intersection of $$L_1$$ and $$L_2$$:

Solving $$2x + 5y = 10$$ and $$-4x + 3y = 12$$ simultaneously:

Multiply the first equation by 2: $$4x + 10y = 20$$

Adding: $$13y = 32 \implies y = \frac{32}{13}$$

$$x = \frac{10 - \frac{160}{13}}{2} = \frac{-\frac{30}{13}}{2} = -\frac{15}{13}$$

So $$C = \left(-\frac{15}{13},\; \frac{32}{13}\right)$$

Calculate the area of triangle ABC:

Area $$= \frac{1}{2}\left|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\right|$$

$$= \frac{1}{2}\left|\frac{3}{13}\left(-\frac{12}{13} - \frac{32}{13}\right) + \frac{95}{13}\left(\frac{32}{13} - \frac{56}{13}\right) + \left(-\frac{15}{13}\right)\left(\frac{56}{13} + \frac{12}{13}\right)\right|$$

$$= \frac{1}{2}\left|\frac{3}{13} \cdot \frac{-44}{13} + \frac{95}{13} \cdot \frac{-24}{13} + \frac{-15}{13} \cdot \frac{68}{13}\right|$$

$$= \frac{1}{2}\left|\frac{-132 - 2280 - 1020}{169}\right| = \frac{1}{2} \cdot \frac{3432}{169} = \frac{1716}{169} = \frac{132}{13}$$

The correct answer is Option B: $$\dfrac{132}{13}$$.

We have a square of side $$a$$ with adjacent side slopes $$m_1, m_2$$ satisfying $$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$$. One vertex is $$P = 10(\cos\alpha - \sin\alpha,\; \sin\alpha + \cos\alpha)$$ and one diagonal has equation $$(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$$.

We note that vertex $$P$$ lies on the diagonal line. Substituting $$P$$ into the diagonal equation: $$(\cos\alpha - \sin\alpha) \cdot 10(\cos\alpha - \sin\alpha) + (\sin\alpha + \cos\alpha) \cdot 10(\sin\alpha + \cos\alpha) = 10(\cos\alpha - \sin\alpha)^2 + 10(\sin\alpha + \cos\alpha)^2 = 10[(\cos^2\alpha - 2\sin\alpha\cos\alpha + \sin^2\alpha) + (\sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha)] = 10[1 + 1] = 20 \neq 10$$.

So $$P$$ does not lie on this diagonal. Since a square has two diagonals, $$P$$ is a vertex, and one diagonal is given, $$P$$ must be a vertex not on this diagonal. The distance from $$P$$ to the diagonal equals half the other diagonal, which is $$\dfrac{a}{\sqrt{2}}$$ (since both diagonals of a square have length $$a\sqrt{2}$$ and the distance from a vertex to the opposite diagonal is $$\dfrac{a\sqrt{2}}{2}$$).

The distance from $$P$$ to the line $$(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$$ is: $$d = \dfrac{|20 - 10|}{\sqrt{(\cos\alpha - \sin\alpha)^2 + (\sin\alpha + \cos\alpha)^2}} = \dfrac{10}{\sqrt{2}}$$

For a square, the distance from any vertex to the opposite diagonal is $$\dfrac{a\sqrt{2}}{2}$$. So $$\dfrac{10}{\sqrt{2}} = \dfrac{a\sqrt{2}}{2}$$, giving $$\dfrac{10}{\sqrt{2}} = \dfrac{a}{\sqrt{2}}$$, so $$a = 10$$.

Now, since adjacent sides of a square are perpendicular, $$m_1 \cdot m_2 = -1$$. The diagonal direction is along $$(-(\sin\alpha + \cos\alpha),\; \cos\alpha - \sin\alpha)$$ (normal to the line). The diagonal slope is $$\dfrac{\cos\alpha - \sin\alpha}{-(\sin\alpha + \cos\alpha)}$$. The sides of the square make 45° with the diagonal, so the slopes of the sides are related to the diagonal slope.

The diagonal direction vector is proportional to $$(\sin\alpha + \cos\alpha,\; -(\cos\alpha - \sin\alpha)) = (\sin\alpha + \cos\alpha,\; \sin\alpha - \cos\alpha)$$. The slope of this diagonal is $$\dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}$$.

For the sides making $$45°$$ with this diagonal, using the formula $$\tan 45° = \left|\dfrac{m - m_d}{1 + m \cdot m_d}\right|$$, and since both sides are perpendicular to each other, one slope comes from $$+45°$$ and the other from $$-45°$$ rotation.

Let $$m_d = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}$$. Then $$m_1 = \dfrac{m_d + 1}{1 - m_d}$$ and $$m_2 = \dfrac{m_d - 1}{1 + m_d}$$.

Now $$m_1 = \dfrac{\sin\alpha - \cos\alpha + \sin\alpha + \cos\alpha}{\sin\alpha + \cos\alpha - \sin\alpha + \cos\alpha} = \dfrac{2\sin\alpha}{2\cos\alpha} = \tan\alpha$$.

Similarly, $$m_2 = \dfrac{\sin\alpha - \cos\alpha - \sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha + \sin\alpha - \cos\alpha} = \dfrac{-2\cos\alpha}{2\sin\alpha} = -\cot\alpha$$.

So $$m_1^2 + m_2^2 = \tan^2\alpha + \cot^2\alpha$$.

Substituting into $$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$$ with $$a = 10$$: $$100 + 110 + 3(\tan^2\alpha + \cot^2\alpha) = 220$$, so $$3(\tan^2\alpha + \cot^2\alpha) = 10$$.

Now we need $$72(\sin^4\alpha + \cos^4\alpha) + a^2 - 3a + 13$$. With $$a = 10$$: $$a^2 - 3a + 13 = 100 - 30 + 13 = 83$$.

We have $$\tan^2\alpha + \cot^2\alpha = \dfrac{\sin^4\alpha + \cos^4\alpha}{\sin^2\alpha\cos^2\alpha} = \dfrac{10}{3}$$.

Let $$p = \sin^2\alpha + \cos^2\alpha = 1$$ and $$q = \sin^2\alpha\cos^2\alpha$$. Then $$\sin^4\alpha + \cos^4\alpha = 1 - 2q$$. The condition becomes $$\dfrac{1-2q}{q} = \dfrac{10}{3}$$, so $$3 - 6q = 10q$$, giving $$16q = 3$$, so $$q = \dfrac{3}{16}$$. Therefore $$\sin^4\alpha + \cos^4\alpha = 1 - \dfrac{6}{16} = 1 - \dfrac{3}{8} = \dfrac{5}{8}$$.

Finally, $$72 \cdot \dfrac{5}{8} + 83 = 45 + 83 = 128$$.

Hence, the correct answer is Option B: $$128$$.

We have $$A(\alpha, -2)$$, $$B(\alpha, 6)$$, and $$C\left(\dfrac{\alpha}{4}, -2\right)$$ as vertices of $$\triangle ABC$$, with circumcentre at $$\left(5, \dfrac{\alpha}{4}\right)$$.

We observe that $$A$$ and $$B$$ have the same $$x$$-coordinate $$\alpha$$, so $$AB$$ is vertical with length $$|6 - (-2)| = 8$$. Also, $$A$$ and $$C$$ have the same $$y$$-coordinate $$-2$$, so $$AC$$ is horizontal with length $$\left|\alpha - \dfrac{\alpha}{4}\right| = \dfrac{3|\alpha|}{4}$$. Since $$AB \perp AC$$, this is a right triangle with the right angle at $$A$$.

For a right triangle, the circumcentre is the midpoint of the hypotenuse $$BC$$. The midpoint of $$BC$$ is $$\left(\dfrac{\alpha + \frac{\alpha}{4}}{2},\; \dfrac{6 + (-2)}{2}\right) = \left(\dfrac{5\alpha}{8},\; 2\right)$$.

Setting this equal to the given circumcentre $$\left(5, \dfrac{\alpha}{4}\right)$$: from the $$y$$-coordinate, $$\dfrac{\alpha}{4} = 2$$, so $$\alpha = 8$$. Let us verify with the $$x$$-coordinate: $$\dfrac{5 \cdot 8}{8} = 5$$. This checks out.

Now with $$\alpha = 8$$: $$A = (8, -2)$$, $$B = (8, 6)$$, $$C = (2, -2)$$, and the circumcentre is $$(5, 2)$$.

We compute the sides: $$AB = 8$$ (vertical), $$AC = 8 - 2 = 6$$ (horizontal), and $$BC = \sqrt{(8-2)^2 + (6-(-2))^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$.

Now we check each option:

Area: $$\dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times 8 \times 6 = 24$$. This is correct.

Perimeter: $$8 + 6 + 10 = 24$$, not 25. So "perimeter is 25" is NOT correct.

Circumradius: For a right triangle, $$R = \dfrac{\text{hypotenuse}}{2} = \dfrac{10}{2} = 5$$. This is correct.

Inradius: $$r = \dfrac{\text{Area}}{s}$$ where $$s = \dfrac{24}{2} = 12$$. So $$r = \dfrac{24}{12} = 2$$. This is correct.

Hence, the correct answer is Option B: perimeter is 25 (this statement is NOT correct).

A ray from $$P(2, 3)$$ reflects off the X-axis at point $$A$$ and passes through $$Q(5, 4)$$.

By the reflection principle, the reflected ray from $$A$$ to $$Q$$ corresponds to a direct ray from the reflection of $$P$$ in the X-axis, which is $$P'(2, -3)$$, to $$Q(5, 4)$$.

Line $$P'Q$$: slope $$= \frac{4-(-3)}{5-2} = \frac{7}{3}$$.

Equation: $$y + 3 = \frac{7}{3}(x - 2) \implies 3y + 9 = 7x - 14 \implies 7x - 3y = 23$$

Point $$A$$ is where this line meets the X-axis ($$y = 0$$):

$$7x = 23 \implies x = \frac{23}{7}$$

So $$A = \left(\frac{23}{7}, 0\right)$$.

Point $$R$$ divides $$AQ$$ internally in ratio $$2:1$$:

$$R = \left(\frac{2 \times 5 + 1 \times \frac{23}{7}}{3}, \frac{2 \times 4 + 1 \times 0}{3}\right) = \left(\frac{10 + \frac{23}{7}}{3}, \frac{8}{3}\right)$$

$$= \left(\frac{\frac{70+23}{7}}{3}, \frac{8}{3}\right) = \left(\frac{93}{21}, \frac{8}{3}\right) = \left(\frac{31}{7}, \frac{8}{3}\right)$$

The bisector of angle $$PAQ$$: Since $$A$$ is on the X-axis and the ray reflects off the X-axis, the angle bisector of $$\angle PAQ$$ is the Y-axis direction at $$A$$, i.e., the vertical line $$x = \frac{23}{7}$$.

The foot of perpendicular from $$R = \left(\frac{31}{7}, \frac{8}{3}\right)$$ to the line $$x = \frac{23}{7}$$ is:

$$M = \left(\frac{23}{7}, \frac{8}{3}\right)$$

So $$\alpha = \frac{23}{7}$$ and $$\beta = \frac{8}{3}$$.

$$7\alpha + 3\beta = 7 \times \frac{23}{7} + 3 \times \frac{8}{3} = 23 + 8 = 31$$

Hence the answer is $$\boxed{31}$$.

Equation of side $$AB$$ is $$2x + y = 0 \; \Rightarrow \; y = -2x$$, so its slope is $$m_{AB} = -2$$.

Equation of side $$CA$$ is $$x - y = 3 \; \Rightarrow \; y = x - 3$$, so its slope is $$m_{CA} = 1$$.

Equation of side $$BC$$ is $$x + py = 15a$$ (slope will be obtained later).

Vertices are obtained by intersecting pairs of sides:
A is the intersection of $$AB$$ and $$CA$$.
  From $$y = -2x$$ and $$y = x - 3$$, we get $$-2x = x - 3 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x_A = 1, \; y_A = -2.$$

C is the intersection of $$CA$$ and $$BC$$.
  From $$x - y = 3 \;\Rightarrow\; x = 3 + y$$ and $$x + py = 15a$$:
  $$3 + y + py = 15a \;\Rightarrow\; y_C = \dfrac{15a - 3}{1 + p},$$
  $$x_C = 3 + y_C = 3 + \dfrac{15a - 3}{1 + p}.$$

B is the intersection of $$AB$$ and $$BC$$.
  Using $$y = -2x$$ in $$x + py = 15a$$:
  $$x - 2px = 15a \;\Rightarrow\; x_B = \dfrac{15a}{\,1 - 2p\,}, \quad y_B = -\,\dfrac{30a}{\,1 - 2p\,}.$$

The orthocentre is given as $$H(2,\,a)$$. Altitudes pass through $$H$$ and the corresponding vertex, and each altitude is perpendicular to the opposite side.

Slope of $$AB$$ is $$-2$$, hence the altitude through C has slope $$\dfrac12$$.
Therefore, $$\dfrac{a - y_C}{\,2 - x_C\,} = \dfrac12.$$

Substituting $$y_C, x_C$$:
$$\dfrac{a - \dfrac{15a - 3}{1 + p}}{\,2 - \Bigl(3 + \dfrac{15a - 3}{1 + p}\Bigr)} = \dfrac12.$$

Simplifying numerator and denominator gives
$$\dfrac{a(p - 14) + 3}{\,2 - p - 15a} = \dfrac12,$$
which reduces to
$$a(2p - 13) + p + 4 = 0 \quad -(1).$$

Slope of $$CA$$ is $$1$$, so the perpendicular slope is $$-1$$.
Thus $$\dfrac{a - y_B}{\,2 - x_B\,} = -1.$$

Substituting $$y_B, x_B$$:
$$\dfrac{a + \dfrac{30a}{1 - 2p}}{\,2 - \dfrac{15a}{1 - 2p}} = -1.$$

Simplifying gives
$$a(8 - p) - 2p + 1 = 0 \quad -(2).$$

From $$(2)$$, solve for $$a$$:
$$a = \dfrac{2p - 1}{8 - p}, \qquad p \neq 8.$$

Substitute this value of $$a$$ into $$(1)$$:

$$\bigl(2p - 1\bigr)(2p - 13) + (p + 4)(8 - p) = 0.$$

Expand and combine like terms:
$$(4p^2 - 28p + 13) + (-p^2 + 4p + 32) = 0,$$
$$3p^2 - 24p + 45 = 0,$$
$$p^2 - 8p + 15 = 0,$$
$$(p - 3)(p - 5) = 0.$$

Possible values are $$p = 3$$ or $$p = 5$$.

Corresponding $$a$$ values:
For $$p = 3$$: $$a = \dfrac{2(3) - 1}{8 - 3} = \dfrac{5}{5} = 1.$$

For $$p = 5$$: $$a = \dfrac{2(5) - 1}{8 - 5} = \dfrac{9}{3} = 3.$$

The given condition is $$-\dfrac12 \lt a \lt 2$$, which only $$a = 1$$ satisfies.
Hence $$p = 3$$ is the required value.

Final Answer: $$p = 3$$

We need to find the distance of the point $$(7, 1)$$ from the line $$6x + y - 46 = 0$$.

The formula for the distance from a point $$(x_0, y_0)$$ to a line $$ax + by + c = 0$$ is $$d = \dfrac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$.

Here $$a = 6$$, $$b = 1$$, $$c = -46$$, and $$(x_0, y_0) = (7, 1)$$.

$$d = \dfrac{|6(7) + 1(1) - 46|}{\sqrt{36 + 1}} = \dfrac{|42 + 1 - 46|}{\sqrt{37}} = \dfrac{|-3|}{\sqrt{37}} = \dfrac{3}{\sqrt{37}}$$

Since the distance is given as $$\dfrac{a}{\sqrt{37}}$$, comparing we get $$a = 3$$.

The answer is $$3$$.

We are given $$A(-1, 1)$$, $$B(3, 4)$$, $$C(2, 0)$$ and a line $$y = mx$$ with $$m > 0$$ that intersects line $$AC$$ at $$P$$ and line $$BC$$ at $$Q$$. We need $$A_1 = 3A_2$$ where $$A_1$$ and $$A_2$$ are areas of $$\triangle ABC$$ and $$\triangle PQC$$.

The equation of line $$AC$$ (passing through $$A(-1,1)$$ and $$C(2,0)$$) is $$y = \frac{2-x}{3}$$. Setting $$y = mx$$: $$mx = \frac{2-x}{3}$$, giving $$x = \frac{2}{3m+1}$$. So $$P = \left(\frac{2}{3m+1}, \frac{2m}{3m+1}\right)$$.

The equation of line $$BC$$ (passing through $$B(3,4)$$ and $$C(2,0)$$) is $$y = 4(x-2)$$. Setting $$y = mx$$: $$mx = 4x - 8$$, giving $$x = \frac{8}{4-m}$$. So $$Q = \left(\frac{8}{4-m}, \frac{8m}{4-m}\right)$$.

Since $$P$$ lies on $$AC$$ and $$Q$$ lies on $$BC$$, triangles $$PQC$$ and $$ABC$$ share the angle at $$C$$, so $$\frac{A_2}{A_1} = \frac{CP}{CA} \cdot \frac{CQ}{CB}$$.

Computing $$\frac{CP}{CA}$$: $$CP^2 = \frac{36m^2 + 4m^2}{(3m+1)^2} = \frac{40m^2}{(3m+1)^2}$$. Since $$CA^2 = 10$$, we get $$\frac{CP}{CA} = \frac{2m}{3m+1}$$.

Computing $$\frac{CQ}{CB}$$: $$CQ^2 = \frac{4m^2 + 64m^2}{(4-m)^2} = \frac{68m^2}{(4-m)^2}$$. Since $$CB^2 = 17$$, we get $$\frac{CQ}{CB} = \frac{2m}{4-m}$$.

Setting $$\frac{A_2}{A_1} = \frac{1}{3}$$: $$\frac{4m^2}{(3m+1)(4-m)} = \frac{1}{3}$$. This gives $$12m^2 = -3m^2 + 11m + 4$$, so $$15m^2 - 11m - 4 = 0$$.

Using the quadratic formula: $$m = \frac{11 \pm \sqrt{121 + 240}}{30} = \frac{11 \pm 19}{30}$$. This gives $$m = 1$$ or $$m = -\frac{8}{30}$$. Since $$m > 0$$, we have $$m = 1$$.

We have the three vertices $$A(a,0)$$, $$B(b,2b+1)$$ and $$C(0,b)$$.

The area of a triangle with vertices $$\bigl(x_1$$, $$y_1\bigr)$$, $$\;(x_2$$, $$y_2)$$, $$\;(x_3$$, $$y_3)$$ is given by the determinant formula

Area $$\;=\;\frac12\,\Bigl|\,x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\Bigr|.$$

Substituting $$x_1=a$$, $$\;y_1=0$$, $$\;x_2=b$$, $$\;y_2=2b+1$$, $$\;x_3=0$$, $$\;y_3=b$$ we obtain

$$\begin{aligned} \text{Area}&=\frac12\Bigl|\,a\bigl((2b+1)-b\bigr)+b\bigl(b-0\bigr)+0\bigl(0-(2b+1)\bigr)\Bigr| \\[4pt] &=\frac12\Bigl|\,a(b+1)+b^2\Bigr|. \end{aligned}$$

We are told that the area equals $$1$$ square unit, so

$$\frac12\Bigl|\,a(b+1)+b^2\Bigr|=1.$$

Multiplying by $$2$$, this becomes

$$\Bigl|\,a(b+1)+b^2\Bigr|=2.$$

Removing the absolute value gives the two possible linear equations

$$\begin{cases} a(b+1)+b^2=2,\\ a(b+1)+b^2=-2. \end{cases}$$

Because $$b\neq -1$$ (since $$|b|\neq 1$$), we can safely divide by $$(b+1)$$ in each equation.

From the first equation:

$$a(b+1)=2-b^2\quad\Longrightarrow\quad a=\dfrac{2-b^2}{b+1}.$$

From the second equation:

$$a(b+1)=-(b^2+2)\quad\Longrightarrow\quad a=-\dfrac{b^2+2}{b+1}.$$

The problem asks for the sum of all possible values of $$a$$, so we add the two expressions:

$$\begin{aligned} a_{\text{sum}}&=\frac{2-b^2}{b+1}+\left(-\frac{b^2+2}{b+1}\right)\\[4pt] &=\frac{\,2-b^2-(b^2+2)\,}{b+1}\\[4pt] &=\frac{-2b^2}{b+1}. \end{aligned}$$

This value matches Option C.

Hence, the correct answer is Option C.

The equation of a line in intercept form is $$\frac{x}{a} + \frac{y}{b} = 1$$, where $$a$$ and $$b$$ are the intercepts on the coordinate axes.

The arithmetic mean of the reciprocals of the intercepts is given as $$\frac{1}{4}$$. So $$\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{1}{4}$$, which gives $$\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$$.

Now we check which of the given points lies on the line $$\frac{x}{a} + \frac{y}{b} = 1$$.

For point A$$(1, 1)$$: $$\frac{1}{a} + \frac{1}{b} = \frac{1}{2} \neq 1$$, so A does not lie on the line.

For point B$$(2, 2)$$: $$\frac{2}{a} + \frac{2}{b} = 2\left(\frac{1}{a} + \frac{1}{b}\right) = 2 \times \frac{1}{2} = 1$$. So B lies on the line.

For point C$$(4, 4)$$: $$\frac{4}{a} + \frac{4}{b} = 4\left(\frac{1}{a} + \frac{1}{b}\right) = 4 \times \frac{1}{2} = 2 \neq 1$$, so C does not lie on the line.

Hence, the correct answer is Option C.

The first line is  $$x\csc\alpha - y\sec\alpha = k\cot 2\alpha.$$  We rewrite it in the standard straight-line form $$Ax+By+C=0$$ by bringing the right hand side to the left:

$$x\csc\alpha - y\sec\alpha - k\cot 2\alpha = 0.$$

Here $$A=\csc\alpha,\; B=-\sec\alpha,\; C=-k\cot 2\alpha.$$

For a line $$Ax+By+C=0,$$ the perpendicular distance of the origin $$(0,0)$$ from the line is given by the well-known formula  $$\displaystyle d=\frac{|C|}{\sqrt{A^{2}+B^{2}}}\;.$$  Stating this first keeps every step clear.

Applying the formula to the first line, the length of the perpendicular from the origin is

$$p=\frac{|\, -k\cot 2\alpha\, |}{\sqrt{\csc^{2}\alpha+(-\sec\alpha)^{2}}} =\frac{|k|\;|\cot 2\alpha|}{\sqrt{\csc^{2}\alpha+\sec^{2}\alpha}}.$$

Now we express the denominator completely in $$\sin\alpha$$ and $$\cos\alpha$$:

$$\csc^{2}\alpha=\frac{1}{\sin^{2}\alpha},\qquad \sec^{2}\alpha=\frac{1}{\cos^{2}\alpha}.$$

Therefore

$$\csc^{2}\alpha+\sec^{2}\alpha =\frac{1}{\sin^{2}\alpha}+\frac{1}{\cos^{2}\alpha} =\frac{\cos^{2}\alpha+\sin^{2}\alpha}{\sin^{2}\alpha\cos^{2}\alpha} =\frac{1}{\sin^{2}\alpha\cos^{2}\alpha}.$$

So

$$\sqrt{\csc^{2}\alpha+\sec^{2}\alpha} =\frac{1}{\sin\alpha\cos\alpha}.$$

Substituting this back, we obtain

$$p=|k|\;|\cot 2\alpha|\;(\sin\alpha\cos\alpha).$$

But $$\sin\alpha\cos\alpha=\dfrac{\sin 2\alpha}{2}$$ and also $$\cot 2\alpha=\dfrac{\cos 2\alpha}{\sin 2\alpha}.$$ Hence

$$p=|k|\;\Bigl|\frac{\cos 2\alpha}{\sin 2\alpha}\Bigr|\;\frac{\sin 2\alpha}{2} =\frac{|k|\;|\cos 2\alpha|}{2}.$$

Squaring to remove the absolute value,

$$p^{2}=\frac{k^{2}\cos^{2}2\alpha}{4}.\qquad(1)$$

Now we treat the second line $$x\sin\alpha+y\cos\alpha=k\sin 2\alpha.$$  Writing it as $$x\sin\alpha+y\cos\alpha-k\sin 2\alpha=0,$$ we have

$$A=\sin\alpha,\;B=\cos\alpha,\;C=-k\sin 2\alpha.$$

The distance formula gives

$$q=\frac{|\, -k\sin 2\alpha\, |}{\sqrt{\sin^{2}\alpha+\cos^{2}\alpha}} =\frac{|k|\;|\sin 2\alpha|}{\sqrt{1}} =|k|\;|\sin 2\alpha|.$$

Squaring,

$$q^{2}=k^{2}\sin^{2}2\alpha.\qquad(2)$$

We now possess two equations connecting $$p^{2},q^{2}$$ and $$k^{2}$$:

$$p^{2}=\dfrac{k^{2}\cos^{2}2\alpha}{4},\qquad q^{2}=k^{2}\sin^{2}2\alpha.$$

From the first, $$k^{2}=4p^{2}/\cos^{2}2\alpha,$$ and from the second, $$k^{2}=q^{2}/\sin^{2}2\alpha.$$ Equating these two expressions for $$k^{2}$$ (because they are the same quantity) we have

$$\frac{4p^{2}}{\cos^{2}2\alpha}=\frac{q^{2}}{\sin^{2}2\alpha}.$$

Cross-multiplying,

$$4p^{2}\sin^{2}2\alpha=q^{2}\cos^{2}2\alpha.$$

Bring the terms involving $$\sin^{2}2\alpha$$ to one side:

$$4p^{2}\sin^{2}2\alpha+q^{2}\sin^{2}2\alpha=q^{2}.$$

Factorising,

$$\sin^{2}2\alpha\,(4p^{2}+q^{2})=q^{2}.$$

Solve for $$\sin^{2}2\alpha$$:

$$\sin^{2}2\alpha=\frac{q^{2}}{4p^{2}+q^{2}}.$$

Finally, substitute this value into $$k^{2}=q^{2}/\sin^{2}2\alpha$$ obtained from equation (2):

$$k^{2}=\frac{q^{2}}{\dfrac{q^{2}}{4p^{2}+q^{2}}} =q^{2}\;\frac{4p^{2}+q^{2}}{q^{2}} =4p^{2}+q^{2}.$$

Thus we have reached the required result:

$$k^{2}=4p^{2}+q^{2}.$$

Hence, the correct answer is Option D.

The centroid of equilateral triangle $$ABC$$ is at the origin and one side lies along $$x + y = 3$$. The distance from the origin to the line $$x + y = 3$$ is $$\frac{|0 + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$$.

For an equilateral triangle, the centroid coincides with both the circumcenter and incenter. The distance from the centroid to a side equals the inradius $$r$$, so $$r = \frac{3}{\sqrt{2}}$$. For an equilateral triangle, the circumradius is $$R = 2r$$, so $$R = \frac{6}{\sqrt{2}}$$.

Therefore $$R + r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$$.

We have triangle $$PQR$$ with $$P(-2, 4)$$ and $$Q(4, -2)$$. The perpendicular bisector of $$PR$$ is $$2x - y + 2 = 0$$. We need the circumcentre.

The circumcentre lies on the perpendicular bisector of every side of the triangle.

The midpoint of $$PQ$$ is $$\left(\frac{-2+4}{2}, \frac{4+(-2)}{2}\right) = (1, 1)$$.

The slope of $$PQ$$ is $$\frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1$$. So the perpendicular bisector of $$PQ$$ has slope $$1$$ and passes through $$(1, 1)$$: $$y - 1 = 1(x - 1)$$, i.e., $$y = x$$ $$-(1)$$.

The perpendicular bisector of $$PR$$ is given as $$2x - y + 2 = 0$$ $$-(2)$$.

The circumcentre lies on both $$(1)$$ and $$(2)$$. From $$(1)$$: $$y = x$$. Substituting into $$(2)$$: $$2x - x + 2 = 0$$, so $$x = -2$$ and $$y = -2$$.

The circumcentre is $$(-2, -2)$$, which matches Option B.

We have two fixed points $$P(5,6)$$ and $$Q(3,2)$$. If we take a variable point $$R(\alpha,\beta)$$, the area of $$\triangle PQR$$ is given by the coordinate-geometry formula

Area $$=\frac12\left|\,x_P\,(y_Q-y_R)+x_Q\,(y_R-y_P)+x_R\,(y_P-y_Q)\,\right|.$$

Substituting $$P(5,6)$$ and $$Q(3,2)$$ we get

$$ \frac12\left|\,5(2-\beta)+3(\beta-6)+\alpha(6-2)\,\right|=12. $$

Simplifying step by step, first expand the terms inside the absolute value:

$$5(2-\beta)=10-5\beta$$, $$\qquad 3(\beta-6)=3\beta-18$$, $$\qquad \alpha(6-2)=4\alpha.$$

Adding them, we have

$$ (10-5\beta)+(3\beta-18)+4\alpha \;=\;4\alpha-2\beta-8. $$

So the area condition becomes

$$ \frac12\left|\,4\alpha-2\beta-8\,\right|=12 \;\;\Longrightarrow\;\; \left|\,4\alpha-2\beta-8\,\right|=24. $$

Dividing by $$2$$ gives

$$ \left|\,2\alpha-\beta-4\,\right|=12. $$

Removing the absolute value yields two linear equations:

$$ 2\alpha-\beta-4=12 \quad\text{or}\quad 2\alpha-\beta-4=-12. $$

Simplifying each, we obtain the pair of straight lines

$$ 2\alpha-\beta+8=0 \quad\text{and}\quad 2\alpha-\beta-16=0. $$

Thus the locus $$A$$ consists of all points on these two lines, each of which is parallel to the original line $$PQ$$. To proceed, let us confirm the equation of $$PQ$$ itself. The slope of $$PQ$$ is

$$ m=\frac{2-6}{3-5}=\frac{-4}{-2}=2, $$

hence $$PQ$$ is $$y=2x-4$$ or equivalently $$2x-y-4=0$$. The two locus lines are obtained by shifting this line by a perpendicular distance such that the triangle’s area is $$12$$. We shall compute that distance explicitly.

The length of the segment $$PQ$$ is

$$ |PQ|=\sqrt{(5-3)^2+(6-2)^2}=\sqrt{2^2+4^2}=\sqrt{4+16}=2\sqrt5. $$

Now, for any triangle,

Area $$=\frac12\times\text{(base)}\times\text{(height)},$$

where “height’’ is the perpendicular distance from the third vertex to the base. Using the base $$PQ$$ and the given area $$12$$, the required height $$h$$ is

$$ h=\frac{2\times\text{Area}}{|PQ|} =\frac{2\times12}{2\sqrt5} =\frac{24}{2\sqrt5} =\frac{12}{\sqrt5}. $$

Indeed, the two new lines must therefore lie a perpendicular distance $$h=12/\sqrt5$$ above and below the line $$PQ:2x-y-4=0$$. The standard distance of a point $$(x,y)$$ from a line $$ax+by+c=0$$ is

$$ \frac{|ax+by+c|}{\sqrt{a^2+b^2}}. $$

For $$2x-y-4=0$$ we have $$a=2,\,b=-1$$ and $$\sqrt{a^2+b^2}=\sqrt{4+1}=\sqrt5$$. Requiring the distance to be $$12/\sqrt5$$ means

$$ \frac{|2x-y-4+c|}{\sqrt5}=\frac{12}{\sqrt5} \;\;\Longrightarrow\;\; |c|=12. $$

Hence $$c=+12$$ or $$c=-12$$, exactly giving the earlier two parallel lines

$$ 2x-y+8=0 \quad\text{and}\quad 2x-y-16=0. $$

We now find the least distance from the origin $$O(0,0)$$ to these two lines, because any point $$(\alpha,\beta)$$ on either line belongs to the set $$A$$.

First line: $$2x-y+8=0$$ Distance from origin:

$$ D_1=\frac{|2\cdot0-1\cdot0+8|}{\sqrt5} =\frac{8}{\sqrt5}. $$

Second line: $$2x-y-16=0$$ Distance from origin:

$$ D_2=\frac{|2\cdot0-1\cdot0-16|}{\sqrt5} =\frac{16}{\sqrt5}. $$

Clearly $$D_1<D_2$$, so the minimum possible length of the segment joining the origin to a point in $$A$$ is

$$ \boxed{\dfrac{8}{\sqrt5}}. $$

Hence, the correct answer is Option A.

We have the fixed point $$A(0,\,6)$$ and the moving point $$B(2t,\,0)$$, where $$t$$ is a real parameter.

First we find the midpoint $$M$$ of $$AB$$. The midpoint formula states: if the endpoints are $$(x_1,\,y_1)$$ and $$(x_2,\,y_2)$$, then the midpoint is $$\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right).$$ Hence

$$M\;=\;\left(\dfrac{0+2t}{2},\;\dfrac{6+0}{2}\right)\;=\;\left(t,\,3\right).$$

Now we write the equation of the perpendicular bisector of $$AB$$. The slope of $$AB$$ is found from the two-point slope formula $$m=\dfrac{y_2-y_1}{x_2-x_1}$$:

$$m_{AB}\;=\;\dfrac{0-6}{2t-0}\;=\;-\dfrac{6}{2t}\;=\;-\dfrac{3}{t}.$$

The slope of a line perpendicular to another is the negative reciprocal, so

$$m_{\perp}\;=\;-\dfrac{1}{m_{AB}}\;=\;-\dfrac{1}{-\dfrac{3}{t}}\;=\;\dfrac{t}{3}.$$

The perpendicular bisector passes through $$M(t,\,3)$$, so using the point-slope form $$y-y_0=m(x-x_0),$$ we get

$$y-3=\dfrac{t}{3}\,(x-t).$$

Next we locate its intersection $$C$$ with the $$y$$-axis. On the $$y$$-axis we have $$x=0$$, so we substitute $$x=0$$:

$$y-3=\dfrac{t}{3}\,(0-t)=-\dfrac{t^2}{3}\;\;\Longrightarrow\;\;y=3-\dfrac{t^2}{3}.$$

Thus $$C\;(0,\,3-\dfrac{t^2}{3}).$$

We are asked for the locus of the midpoint $$P$$ of $$MC$$. Again using the midpoint formula between $$M(t,\,3)$$ and $$C\!\left(0,\,3-\dfrac{t^2}{3}\right)$$, we obtain

$$x_P=\dfrac{t+0}{2}=\dfrac{t}{2},$$ $$y_P=\dfrac{\,3+\left(3-\dfrac{t^2}{3}\right)}{2}=\dfrac{\,6-\dfrac{t^2}{3}}{2}=3-\dfrac{t^2}{6}.$$

Let $$P(x,\,y)$$ denote a general position of this midpoint. Then

$$x=\dfrac{t}{2}\;\;\Longrightarrow\;\;t=2x.$$

Substituting this value of $$t$$ into the expression for $$y$$ gives

$$y=3-\dfrac{(2x)^2}{6}=3-\dfrac{4x^2}{6}=3-\dfrac{2x^2}{3}.$$

To eliminate fractions, multiply by $$3$$:

$$3y=9-2x^2.$$

Rearranging all terms to one side, we finally obtain the Cartesian equation of the locus:

$$2x^2+3y-9=0.$$

This equation coincides with Option D in the given list. Hence, the correct answer is Option D.

Let the vertices be $$A(-3,1),\;B(x_B,y_B),\;C(x_C,y_C)$$.

The given data translate to the following two straight-line conditions:

• Median through $$B$$: $$2x+y-3=0$$ $$\;-(1)$$ (passes through $$B$$ and the midpoint of $$AC$$).
• Angle bisector at $$C$$: $$7x-4y-1=0$$ $$\;-(2)$$ (passes through $$C$$).

Step 1: Find the coordinates of $$C$$

The midpoint of $$AC$$ is $$M\Bigl(\dfrac{-3+x_C}{2},\,\dfrac{1+y_C}{2}\Bigr)$$. Because the median through $$B$$ also contains $$M$$, substitute $$M$$ in $$(1)$$:

$$2\left(\dfrac{-3+x_C}{2}\right)+\dfrac{1+y_C}{2}-3=0$$ $$\Longrightarrow\;(-3+x_C)+\dfrac{1+y_C}{2}-3=0$$ $$\Longrightarrow\;x_C-6+\dfrac{1+y_C}{2}=0$$ $$\Longrightarrow\;2x_C+y_C-11=0$$ $$\;-(3)$$

The point $$C$$ also lies on the bisector $$(2)$$: $$7x_C-4y_C-1=0$$ $$\;-(4)$$.

Solve $$(3)$$ and $$(4)$$ simultaneously:
From $$(3)$$, $$y_C=11-2x_C$$.
Insert in $$(4)$$: $$7x_C-4(11-2x_C)-1=0$$ $$\Rightarrow 7x_C-44+8x_C-1=0$$ $$\Rightarrow 15x_C=45\;\Longrightarrow\;x_C=3$$ $$\Rightarrow y_C=11-2(3)=5$$.

Hence $$C(3,5)$$.

Step 2: Slope of side $$CA$$

$$m_{CA}=\dfrac{5-1}{\,3-(-3)\,}= \dfrac{4}{6}= \dfrac{2}{3}.$$

Step 3: Slope of the angle-bisector at $$C$$

Rewrite $$(2)$$: $$4y=7x-1 \;\Longrightarrow\; y=\dfrac{7}{4}x-\dfrac{1}{4}$$, so $$m_{\text{bis}}=\dfrac{7}{4}.$$

Step 4: Angle between $$CA$$ and the bisector

For two lines with slopes $$m_1,\,m_2$$ the acute angle $$\phi$$ between them is $$\tan\phi=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|.$$ Here $$\tan\phi=\left|\dfrac{\dfrac{7}{4}-\dfrac{2}{3}}{1+\dfrac{2}{3}\cdot\dfrac{7}{4}}\right| =\dfrac{\dfrac{13}{12}}{\dfrac{13}{6}}=\dfrac{1}{2}.$$

Thus $$\phi=\arctan\!\left(\dfrac{1}{2}\right).$$

Step 5: Obtain $$\theta$$ from the angle bisector property

The line $$(2)$$ is the internal angle bisector at $$C$$; hence it divides the angle $$\angle ACB$$ into two equal parts. Therefore $$\theta=2\phi$$.

Using the double-angle identity $$\tan(2\phi)=\dfrac{2\tan\phi}{1-\tan^2\phi}$$, take $$\tan\phi=\dfrac{1}{2}$$:

$$\tan\theta=\dfrac{2\cdot\dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2} =\dfrac{1}{\,1-\dfrac{1}{4}\,}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}.$$

Answer: $$\displaystyle \tan\theta=\dfrac{4}{3}$$ (Option B).

To find the image of the point $$(3, 5)$$ in the line $$x - y + 1 = 0$$, we first find the foot of the perpendicular from $$(3, 5)$$ to this line.

The line $$x - y + 1 = 0$$ has normal direction $$(1, -1)$$. The perpendicular from $$(3, 5)$$ can be parametrized as $$(3 + t, 5 - t)$$. For this point to lie on the line: $$(3 + t) - (5 - t) + 1 = 0$$, which gives $$2t - 1 = 0$$, so $$t = \frac{1}{2}$$.

The foot of the perpendicular is $$\left(\frac{7}{2}, \frac{9}{2}\right)$$.

The image is obtained by reflecting through the foot, so the image point is $$(3 + 2 \times \frac{1}{2},\; 5 - 2 \times \frac{1}{2}) = (4, 4)$$.

Now we check which circle contains $$(4, 4)$$. For the circle $$(x - 2)^2 + (y - 4)^2 = 4$$: $$(4 - 2)^2 + (4 - 4)^2 = 4 + 0 = 4$$. This is satisfied.

Therefore, the image of $$(3, 5)$$ lies on $$(x - 2)^2 + (y - 4)^2 = 4$$.

We are told that two adjacent sides of a parallelogram lie along the straight lines $$4x + 5y = 0$$ and $$7x + 2y = 0$$. Because the two lines intersect, their point of intersection is one vertex of the parallelogram.

Solving the simultaneous equations $$4x + 5y = 0 \quad\text{and}\quad 7x + 2y = 0$$ we first express one variable in terms of the other from the first equation:

$$4x + 5y = 0 \;\Longrightarrow\; 5y = -4x \;\Longrightarrow\; y = -\dfrac{4x}{5}.$$

Substituting this value of $$y$$ in the second equation gives

$$7x + 2\!\left(-\dfrac{4x}{5}\right)=0 \;\Longrightarrow\; 7x - \dfrac{8x}{5}=0 \;\Longrightarrow\; \dfrac{35x - 8x}{5}=0 \;\Longrightarrow\; \dfrac{27x}{5}=0 \;\Longrightarrow\; x = 0.$$

Putting $$x = 0$$ into $$y = -\dfrac{4x}{5}$$ yields $$y = 0$$. So the two lines meet at the origin $$O(0,0)$$, and we take this as the first vertex of the parallelogram.

Next, we need direction vectors for the two given sides. For $$4x + 5y = 0$$, set $$x = t$$; then $$y = -\dfrac{4}{5}t$$, giving the direction vector $$(5,-4).$$ For $$7x + 2y = 0$$, set $$x = s$$; then $$y = -\dfrac{7}{2}s$$, giving the direction vector $$(2,-7).$$

Let the adjacent sides from the origin be

$$\overrightarrow{OA} = a\,(5,-4)\quad\text{and}\quad \overrightarrow{OB} = b\,(2,-7),$$

where $$a\neq 0,\, b\neq 0.$$ Thus the coordinates of the next two vertices are

$$A(5a,\,-4a), \qquad B(2b,\,-7b).$$

The fourth vertex $$C$$ is obtained by the parallelogram law:

$$\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB},$$ so $$C\bigl(5a+2b,\,-4a-7b\bigr).$$

The two diagonals are $$\overline{OC}$$ and $$\overline{AB}$$. We are told that one diagonal satisfies $$11x + 7y = 9$$. Notice that $$O(0,0)$$ does not satisfy this equation because $$11\cdot0 + 7\cdot0 = 0 \neq 9$$. Therefore the diagonal that lies on $$11x + 7y = 9$$ cannot be $$\overline{OC}$$ (which passes through the origin); it must be the other diagonal $$\overline{AB}$$.

Because every point on $$\overline{AB}$$ satisfies $$11x + 7y = 9$$, both end-points $$A$$ and $$B$$ must satisfy it. So we substitute the coordinates of $$A$$ and $$B$$ into the equation.

For $$A(5a,\,-4a):$$ $$11(5a) + 7(-4a) = 55a - 28a = 27a.$$ Hence $$27a = 9 \;\Longrightarrow\; a = \dfrac{1}{3}.$$

For $$B(2b,\,-7b):$$ $$11(2b) + 7(-7b) = 22b - 49b = -27b.$$ Hence $$-27b = 9 \;\Longrightarrow\; b = -\dfrac{1}{3}.$$

With these values we now have explicit coordinates for all the vertices:

$$A\!\left(5\!\left(\dfrac13\right),\,-4\!\left(\dfrac13\right)\right)=\left(\dfrac{5}{3},\,-\dfrac{4}{3}\right),$$ $$B\!\left(2\!\left(-\dfrac13\right),\,-7\!\left(-\dfrac13\right)\right)=\left(-\dfrac{2}{3},\;\dfrac{7}{3}\right),$$ $$C = A + B = \left(\dfrac{5}{3}-\dfrac{2}{3},\,-\dfrac{4}{3}+\dfrac{7}{3}\right)=(1,1).$$

Now consider the other diagonal $$\overline{OC}$$ connecting $$O(0,0)$$ and $$C(1,1).$$ The two-point form of the equation of a straight line is $$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1}\,(x - x_1).$$ Using $$(x_1,y_1)=(0,0)$$ and $$(x_2,y_2)=(1,1)$$, we get

$$y - 0 = \dfrac{1 - 0}{1 - 0}\,(x - 0) \;\Longrightarrow\; y = x.$$

Therefore every point on the second diagonal satisfies $$y = x$$, or equivalently $$x - y = 0.$

We now check which of the four given points satisfies $$x = y$$:

• $$(1,2):\; 1 \neq 2$$ (reject) • $$(2,2):\; 2 = 2$$ (accept) • $$(2,1):\; 2 \neq 1$$ (reject) • $$(1,3):\; 1 \neq 3$$ (reject)

Only $$(2,2)$$ lies on the required diagonal.

Hence, the correct answer is Option 2.

We start with the given pair of lines

$$x^{2}-4xy-5y^{2}=0.$$

Because the constant term is zero, the curve passes through the origin, so it must represent two straight lines through the origin. We factor the quadratic:

$$x^{2}-4xy-5y^{2}= (x-5y)(x+y)=0.$$

Hence the two individual lines are

$$L_{1}:x-5y=0 \quad\Longrightarrow\quad y=\frac{1}{5}x,$$

$$L_{2}:x+y=0 \quad\Longrightarrow\quad y=-x.$$

Now we wish to obtain the equations of the angle-bisector lines of these two lines. For two lines through the origin,

$$a_{1}x+b_{1}y=0 \quad\text{and}\quad a_{2}x+b_{2}y=0,$$

the angle-bisectors are obtained from the formula

$$\frac{a_{1}x+b_{1}y}{\sqrt{a_{1}^{2}+b_{1}^{2}}}= \pm \frac{a_{2}x+b_{2}y}{\sqrt{a_{2}^{2}+b_{2}^{2}}}.$$

Here we have

$$L_{1}:a_{1}=1,\;b_{1}=-5, \qquad L_{2}:a_{2}=1,\;b_{2}=1.$$

Therefore

$$\frac{x-5y}{\sqrt{1^{2}+(-5)^{2}}}= \pm \frac{x+y}{\sqrt{1^{2}+1^{2}}}.$$

Simplifying the square-roots,

$$\frac{x-5y}{\sqrt{26}} = \pm \frac{x+y}{\sqrt{2}}.$$

Cross-multiplying gives

$$\sqrt{2}\,(x-5y)= \pm \sqrt{26}\,(x+y).$$

To obtain a single equation containing both bisectors, we square both sides and then bring all terms to one side:

$$\bigl[\sqrt{2}\,(x-5y)\bigr]^{2} - \bigl[\sqrt{26}\,(x+y)\bigr]^{2}=0.$$

That is,

$$2\,(x-5y)^{2} - 26\,(x+y)^{2}=0.$$

Dividing every term by $$2$$ for convenience,

$$(x-5y)^{2} - 13\,(x+y)^{2}=0.$$

Now we expand each square separately:

First square: $$ (x-5y)^{2}=x^{2}-10xy+25y^{2}. $$

Second square: $$ (x+y)^{2}=x^{2}+2xy+y^{2}. $$

Substituting these into the equation, we get

$$x^{2}-10xy+25y^{2}-13\bigl(x^{2}+2xy+y^{2}\bigr)=0.$$

Distribute the $$-13$$:

$$x^{2}-10xy+25y^{2}-13x^{2}-26xy-13y^{2}=0.$$

Combine like terms for each power:

For $$x^{2}: \;\; x^{2}-13x^{2}= -12x^{2},$$

For $$xy:\;\; -10xy-26xy= -36xy,$$

For $$y^{2}:\;\; 25y^{2}-13y^{2}= 12y^{2}.$$

Thus the equation becomes

$$-12x^{2}-36xy+12y^{2}=0.$$

Every coefficient is divisible by $$-12$$. Dividing through by $$-12$$ gives the simplest form:

$$x^{2}+3xy-y^{2}=0.$$

This is exactly the required combined equation of the two angle-bisector lines. Matching this with the options, we see it corresponds to Option C.

Hence, the correct answer is Option C.

The vertices of the triangle are $$A(-2, 3)$$, $$B(1, 9)$$, and $$C(3, 8)$$. We first find the circumcentre of the triangle.

The circumcentre $$O = (h, k)$$ is equidistant from all three vertices. From $$|OA|^2 = |OB|^2$$: $$(h+2)^2 + (k-3)^2 = (h-1)^2 + (k-9)^2$$ $$h^2 + 4h + 4 + k^2 - 6k + 9 = h^2 - 2h + 1 + k^2 - 18k + 81$$ $$6h + 12k = 69 \implies 2h + 4k = 23 \quad \cdots (1)$$

From $$|OB|^2 = |OC|^2$$: $$(h-1)^2 + (k-9)^2 = (h-3)^2 + (k-8)^2$$ $$h^2 - 2h + 1 + k^2 - 18k + 81 = h^2 - 6h + 9 + k^2 - 16k + 64$$ $$4h - 2k = -9 \quad \cdots (2)$$

From (1): $$h = \frac{23 - 4k}{2}$$. Substituting into (2): $$4 \cdot \frac{23 - 4k}{2} - 2k = -9$$ $$2(23 - 4k) - 2k = -9$$ $$46 - 8k - 2k = -9$$ $$10k = 55 \implies k = \frac{11}{2}$$

Then $$h = \frac{23 - 22}{2} = \frac{1}{2}$$. So the circumcentre is $$O = \left(\frac{1}{2}, \frac{11}{2}\right)$$.

The midpoint of $$BC$$ is $$M = \left(\frac{1+3}{2}, \frac{9+8}{2}\right) = \left(2, \frac{17}{2}\right)$$.

The line $$L$$ passes through $$O = \left(\frac{1}{2}, \frac{11}{2}\right)$$ and $$M = \left(2, \frac{17}{2}\right)$$. The slope is: $$m = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}} = \frac{3}{\frac{3}{2}} = 2$$

The equation of line $$L$$: $$y - \frac{11}{2} = 2\left(x - \frac{1}{2}\right) \Rightarrow y = 2x + \frac{11}{2} - 1 = 2x + \frac{9}{2}$$.

At $$x = 0$$: $$y = \frac{9}{2}$$. The y-intercept is $$\left(0, \frac{9}{2}\right) = \left(0, \frac{\alpha}{2}\right)$$, so $$\alpha = 9$$.

We have the three straight lines $$$x - y + 1 = 0\;,\; x - 2y + 3 = 0\;,\; 2x - 5y + 11 = 0$$$. Their pair-wise points of intersection will give us three points, and the question tells us that these very points are the mid-points of the sides of a triangle $$ABC$$.

First we find the intersection of $$x - y + 1 = 0$$ and $$x - 2y + 3 = 0$$. From $$x - y + 1 = 0$$ we get $$y = x + 1$$. Substituting this in $$x - 2y + 3 = 0$$, we obtain $$$x - 2(x + 1) + 3 = 0 \;\Longrightarrow\; x - 2x - 2 + 3 = 0 \;\Longrightarrow\; -x + 1 = 0 \;\Longrightarrow\; x = 1.$$$ Putting $$x = 1$$ back in $$y = x + 1$$ gives $$y = 2$$. So the first point is $$M_1(1,\,2).$$

Next we intersect $$x - 2y + 3 = 0$$ with $$2x - 5y + 11 = 0$$. We keep the first equation as it is and multiply it by $$2$$ to get a convenient pair: $$$2(x - 2y + 3) = 0 \;\Longrightarrow\; 2x - 4y + 6 = 0.$$$ Now we subtract this from $$2x - 5y + 11 = 0$$: $$$(2x - 5y + 11) - (2x - 4y + 6) = 0 \;\Longrightarrow\; -y + 5 = 0 \;\Longrightarrow\; y = 5.$$$ Putting $$y = 5$$ in $$x - 2y + 3 = 0$$ gives $$$x - 2(5) + 3 = 0 \;\Longrightarrow\; x - 10 + 3 = 0 \;\Longrightarrow\; x = 7.$$$ Hence the second point is $$M_2(7,\,5).$$

Finally, we intersect $$2x - 5y + 11 = 0$$ with $$x - y + 1 = 0$$. Again, from $$x - y + 1 = 0$$ we have $$y = x + 1$$. Substituting in $$2x - 5y + 11 = 0$$, we get $$$2x - 5(x + 1) + 11 = 0 \;\Longrightarrow\; 2x - 5x - 5 + 11 = 0 \;\Longrightarrow\; -3x + 6 = 0 \;\Longrightarrow\; x = 2.$$$ Putting $$x = 2$$ in $$y = x + 1$$ yields $$y = 3.$$ So the third point is $$M_3(2,\,3).$$

Thus the mid-points of $$\triangle ABC$$ are $$$M_1(1,\,2),\; M_2(7,\,5),\; M_3(2,\,3).$$$

Now, for any triangle, the triangle formed by the three mid-points (called the medial triangle) has exactly one-fourth the area of the original triangle. Hence $$$\text{Area of } \triangle ABC = 4 \times \text{Area of } \triangle M_1M_2M_3.$$$ So we must first compute the area of $$$\triangle M_1M_2M_3.$

The coordinate-geometry formula for the area of a triangle with vertices $$$(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)$$ is $$\text{Area} = \frac12\;\bigl|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\bigr|.$$ Taking $$(x_1,y_1) = (1,2),\; (x_2,y_2) = (7,5),\; (x_3,y_3) = (2,3)$$$, we substitute step by step:

$$$\begin{aligned} \text{Area}(M_1M_2M_3) &= \frac12\;\Bigl|\, 1\,(5 - 3) + 7\,(3 - 2) + 2\,(2 - 5) \Bigr| \\ &= \frac12\;\Bigl|\, 1 \times 2 \;+\; 7 \times 1 \;+\; 2 \times (-3) \Bigr| \\ &= \frac12\;\bigl|\,2 + 7 - 6\bigr| \\ &= \frac12\;\bigl|\,3\bigr| \\ &= \frac{3}{2}. \end{aligned}$$

Therefore $$\text{Area of } \triangle ABC = 4 \times \frac{3}{2} = 6.$$

So, the answer is $$6$$.

We have a starting point $$P(-3,4)$$ on the coordinate plane. The man must first reach some point $$R$$ on the x-axis, so we can denote $$R(x,0)$$, and then go to the fixed point $$Q(0,2)$$. The speed of the man is constant, so the time taken is directly proportional to the total distance travelled. Therefore minimising the time is exactly the same as minimising the total distance $$PR+RQ$$.

To minimise a broken-line path that meets a straight line (here, the x-axis) at an intermediate point, we use the well-known reflection idea: reflect one terminal point in the line and join the reflection to the other terminal point by a straight segment. So, we reflect the point $$Q(0,2)$$ in the x-axis. The x-axis is the line $$y=0$$, and reflection in it changes the sign of the y-coordinate while keeping the x-coordinate unchanged. Thus the reflection of $$Q(0,2)$$ is

$$Q'(0,-2).$$

Now the broken path $$P \to R \to Q$$ has exactly the same length as the broken path $$P \to R \to Q'$$ because $$RQ = RQ'$$ (both are vertical mirror images). A straight line is the shortest path between two points, so the shortest possible broken path occurs when $$P,\,R,\,Q'$$ are collinear. Hence $$R$$ must be the point where the straight line through $$P$$ and $$Q'$$ cuts the x-axis.

First we determine the equation of the line through $$P(-3,4)$$ and $$Q'(0,-2)$$. The slope formula is

$$m = \dfrac{y_2 - y_1}{x_2 - x_1}.$$

Substituting $$P(-3,4)$$ as $$(x_1,y_1)$$ and $$Q'(0,-2)$$ as $$(x_2,y_2)$$ we get

$$m = \dfrac{-2 - 4}{\,0 - (-3)\,} = \dfrac{-6}{3} = -2.$$

Using the point-slope form $$y - y_1 = m(x - x_1)$$ with point $$P(-3,4)$$, we have

$$y - 4 = -2\bigl(x - (-3)\bigr) = -2(x + 3).$$

Simplifying,

$$y - 4 = -2x - 6 \quad\Longrightarrow\quad y = -2x - 2.$$

The x-axis is $$y=0$$, so we set $$y=0$$ to find the coordinates of $$R$$:

$$0 = -2x - 2 \quad\Longrightarrow\quad -2x = 2 \quad\Longrightarrow\quad x = -1.$$

Thus

$$R(-1,0).$$

We now calculate the distances $$PR$$ and $$RQ$$ using the distance formula

$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

Distance $$PR$$:

$$PR = \sqrt{(-1 - (-3))^2 + (0 - 4)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}.$$

Distance $$RQ$$:

$$RQ = \sqrt{(0 - (-1))^2 + (2 - 0)^2} = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$

Next we compute the quantity requested in the question, namely $$50\bigl[(PR)^2 + (RQ)^2\bigr]$$. First we find the squared lengths:

$$(PR)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20,$$ $$(RQ)^2 = (\sqrt{5})^2 = 5.$$

Adding,

$$(PR)^2 + (RQ)^2 = 20 + 5 = 25.$$

Finally, multiplying by 50:

$$50\bigl[(PR)^2 + (RQ)^2\bigr] = 50 \times 25 = 1250.$$

So, the answer is $$1250$$.

We have the fixed points $$A(1,-1)$$ and $$B(0,2)$$ and a variable point $$P(x',y')$$ that lies on the straight line

$$3x+y-4\lambda = 0.$$

Therefore every such point must satisfy

$$3x' + y' = 4\lambda. \quad -(1)$$

The area of the triangle formed by the three points is given to be $$5$$ square units. For three points $$P(x_1,y_1),\;A(x_2,y_2),\;B(x_3,y_3)$$ the area formula is

$$\text{Area}=\frac12\left|\,x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\,\right|.$$

Putting $$P(x',y'),\;A(1,-1),\;B(0,2)$$ into the formula, we get

$$\text{Area}=\frac12\left|\,x'( -1-2) + 1(2-y') + 0(y' +1)\,\right|.$$

Simplifying the expression inside the modulus step by step,

$$x'( -1-2)=x'(-3)=-3x',$$ $$1(2-y')=2-y',$$ and the last term is zero. Hence

$$\text{Area}=\frac12\left|\, -3x' + 2 - y'\,\right|.$$

Given that this area equals $$5$$, we write

$$\frac12\left| -3x' + 2 - y' \right| = 5.$$

Multiplying by $$2$$ gives

$$\left| -3x' + 2 - y' \right| = 10.$$

Because a modulus is unchanged by an overall sign, we may reverse the order to obtain

$$\left|\,3x' + y' - 2\,\right| = 10. \quad -(2)$$

Now we substitute the line condition (1), namely $$3x'+y'=4\lambda,$$ into (2):

$$\left|\,4\lambda - 2\,\right| = 10.$$

Removing the absolute value leads to the two equations

$$4\lambda - 2 = 10 \quad\text{or}\quad 4\lambda - 2 = -10.$$

Solving each, we get

First case: $$4\lambda = 12 \;\Rightarrow\; \lambda = 3.$$ Second case: $$4\lambda = -8 \;\Rightarrow\; \lambda = -2.$$

Among the answer choices only $$\lambda = 3$$ is listed.

Hence, the correct answer is Option B.

First we list the coordinates of the three vertices of the triangle:

$$A(-1,\;7), \qquad B(-7,\;1), \qquad C(5,\;-5).$$

The orthocentre is the common point of the three altitudes. An altitude is a line drawn from one vertex and perpendicular to the opposite side. We shall find two such altitudes and then compute their intersection; that point will automatically be the orthocentre.

Altitude from vertex A

We first need the slope of side $$BC$$. Using the slope formula $$m=\dfrac{y_2-y_1}{x_2-x_1}$$, we have

$$m_{BC}=\dfrac{-5-1}{\,5-(-7)\,}=\dfrac{-6}{12}=-\dfrac12.$$

An altitude from $$A$$ is perpendicular to $$BC$$. For two perpendicular lines the product of their slopes is $$-1$$, that is

$$m_{BC}\;m_{\perp}=-1.$$

Substituting $$m_{BC}=-\dfrac12$$, we get

$$\Bigl(-\dfrac12\Bigr)\,m_{\perp}=-1 \;\Rightarrow\; m_{\perp}=2.$$

The altitude through $$A(-1,7)$$ therefore has slope $$2$$ and its equation (point-slope form $$y-y_1=m(x-x_1)$$) is

$$y-7=2\,(x+1).$$

Simplifying,

$$y-7=2x+2 \;\Longrightarrow\; y=2x+9.$$

Altitude from vertex B

Next we need the slope of side $$AC$$:

$$m_{AC}=\dfrac{-5-7}{\,5-(-1)\,}=\dfrac{-12}{6}=-2.$$

The altitude from $$B$$ is perpendicular to $$AC$$, so using $$m_{AC}\,m_{\perp}=-1$$ we get

$$(-2)\,m_{\perp}=-1 \;\Rightarrow\; m_{\perp}=\dfrac12.$$

This altitude passes through $$B(-7,1)$$ and has slope $$\dfrac12$$, giving

$$y-1=\dfrac12\,(x+7).$$

Simplifying,

$$y-1=\dfrac12x+\dfrac72 \;\Longrightarrow\; y=\dfrac12x+\dfrac72+1=\dfrac12x+\dfrac92.$$

Writing the constant in decimal for convenience, $$\dfrac92=4.5$$, so

$$y=\dfrac12x+4.5.$$

Intersection of the two altitudes

The orthocentre is where the lines

$$y=2x+9 \qquad\text{and}\qquad y=\dfrac12x+4.5$$

meet. Equating their right-hand sides,

$$2x+9=\dfrac12x+4.5.$$

To eliminate the fraction, multiply every term by $$2$$:

$$4x+18 = x+9.$$

Now bring all terms in $$x$$ to one side and constants to the other:

$$4x - x = 9 - 18 \;\Longrightarrow\; 3x = -9.$$

Dividing by $$3$$ gives

$$x = -3.$$

Substituting $$x=-3$$ into either altitude equation, say $$y=2x+9$$, we obtain

$$y = 2(-3)+9 = -6+9 = 3.$$

Thus the coordinates of the orthocentre are

$$(-3,\;3).$$

Comparing with the given options, this matches option A.

Hence, the correct answer is Option A.

We have two points $$P(1,\,4)$$ and $$Q(k,\,3)$$. Our task is to use the given condition on the perpendicular bisector of $$\overline{PQ}$$ to find the unknown $$k$$.

First, recall the midpoint formula. If $$P(x_1,\,y_1)$$ and $$Q(x_2,\,y_2)$$ are the endpoints of a segment, then the midpoint $$M$$ has coordinates

$$M\Bigl(\,\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\Bigr).$$

Substituting $$x_1=1,\;y_1=4,\;x_2=k,\;y_2=3$$ we obtain

$$M\Bigl(\dfrac{1+k}{2},\;\dfrac{4+3}{2}\Bigr)=\Bigl(\dfrac{1+k}{2},\;\dfrac{7}{2}\Bigr).$$

Next, recall the slope formula. For the line through $$P(x_1,\,y_1)$$ and $$Q(x_2,\,y_2)$$ the slope is

$$m_{PQ}=\dfrac{y_2-y_1}{x_2-x_1}.$$

Thus

$$m_{PQ}=\dfrac{3-4}{k-1}=\dfrac{-1}{k-1}=-\,\dfrac{1}{k-1}.$$

Two lines are perpendicular if the product of their slopes is $$-1$$. Equivalently, the slope of the perpendicular bisector is the negative reciprocal of $$m_{PQ}$$. Hence

$$m_{\text{perp}}=-(1/m_{PQ})=-(k-1)=-1\cdot\Bigl(-\,\dfrac{k-1}{1}\Bigr)\quad\Longrightarrow\quad m_{\text{perp}}=k-1.$$

Now we write the point-slope form for the perpendicular bisector, using point $$M\Bigl(\dfrac{1+k}{2},\,\dfrac{7}{2}\Bigr)$$:

$$y-\dfrac{7}{2}=(k-1)\Bigl(x-\dfrac{1+k}{2}\Bigr).$$

The problem states that the perpendicular bisector has a $$y$$-intercept of $$-4$$. This means the point $$(0,\,-4)$$ lies on the line. Substituting $$x=0,\;y=-4$$ into the equation of the perpendicular bisector gives

$$-4-\dfrac{7}{2}=(k-1)\Bigl(0-\dfrac{1+k}{2}\Bigr).$$

We simplify the left side:

$$-4=\frac{-8}{2}\quad\Longrightarrow\quad -4-\frac{7}{2}=\frac{-8}{2}-\frac{7}{2}=\frac{-15}{2}.$$

Hence

$$\frac{-15}{2}=(k-1)\Bigl(-\,\dfrac{1+k}{2}\Bigr).$$

Notice the right side contains a product with a negative sign. We can write

$$(k-1)\Bigl(-\,\dfrac{1+k}{2}\Bigr)=-\,\dfrac{(k-1)(1+k)}{2}=-\,\dfrac{k^2-1}{2}.$$

The equation therefore becomes

$$\frac{-15}{2}=-\,\dfrac{k^{2}-1}{2}.$$

Both sides have a common denominator $$2$$, so we multiply the entire equation by $$2$$ to clear the fraction:

$$-15=-(k^{2}-1).$$

Multiplying both sides by $$-1$$ gives

$$15=k^{2}-1.$$

Adding $$1$$ to both sides yields

$$16=k^{2}.$$

Taking the square root of both sides, we obtain two possible values:

$$k=4 \quad\text{or}\quad k=-4.$$

We now compare these with the offered options $$\bigl(-2,\,-4,\,\sqrt{14},\,\sqrt{15}\bigr).$$ Only $$k=-4$$ appears among the choices.

Hence, the correct answer is Option B.

We start by writing the equation of the given line. A line whose intercepts on the coordinate axes are $$a$$ on the $$x$$-axis and $$b$$ on the $$y$$-axis can be expressed in the intercept form $$\dfrac{x}{a} + \dfrac{y}{b} = 1$$. Here the intercepts are $$3$$ and $$1$$, so we have $$\dfrac{x}{3} + \dfrac{y}{1} = 1$$.

Multiplying every term by $$3$$ gives $$x + 3y = 3$$, which we rearrange to the standard form $$x + 3y - 3 = 0$$. Thus, for the line $$L$$ we identify $$a = 1$$, $$b = 3$$ and $$c = -3$$ in the general equation $$ax + by + c = 0$$.

To find the reflection (image) of a point $$(x_0,\,y_0)$$ across a line $$ax + by + c = 0$$, we use the well-known formula

$$ x' \;=\; x_0 \;-\; \dfrac{2a\,(ax_0 + by_0 + c)}{a^2 + b^2}, \qquad y' \;=\; y_0 \;-\; \dfrac{2b\,(ax_0 + by_0 + c)}{a^2 + b^2}. $$

For the point $$(-1,\,-4)$$ we substitute $$x_0 = -1$$ and $$y_0 = -4$$. First, we evaluate the expression $$ax_0 + by_0 + c$$:

$$ ax_0 + by_0 + c \;=\; (1)(-1) + (3)(-4) + (-3) \;=\; -1 - 12 - 3 \;=\; -16. $$

Next, we compute $$a^2 + b^2$$:

$$ a^2 + b^2 \;=\; 1^2 + 3^2 \;=\; 1 + 9 \;=\; 10. $$

Now we substitute everything into the reflection formulas.

For the reflected $$x$$-coordinate:

$$ x' \;=\; -1 \;-\; \dfrac{2 \cdot 1 \cdot (-16)}{10} \;=\; -1 \;+\; \dfrac{32}{10} \;=\; -1 + \dfrac{16}{5} \;=\; -\dfrac{5}{5} + \dfrac{16}{5} \;=\; \dfrac{11}{5}. $$

For the reflected $$y$$-coordinate:

$$ y' \;=\; -4 \;-\; \dfrac{2 \cdot 3 \cdot (-16)}{10} \;=\; -4 \;+\; \dfrac{96}{10} \;=\; -4 + \dfrac{48}{5} \;=\; -\dfrac{20}{5} + \dfrac{48}{5} \;=\; \dfrac{28}{5}. $$

Hence the image of $$(-1,\,-4)$$ in the line $$x + 3y - 3 = 0$$ is the point $$\left(\dfrac{11}{5},\,\dfrac{28}{5}\right)$$.

Among the given options, this corresponds to Option A.

Hence, the correct answer is Option A.

We have a source point $$S(2,\,2\sqrt3)$$. The vertical mirror is the straight line $$x = 1$$. Let the point of incidence on the mirror be $$A(1,\,y)$$, where $$y$$ is to be determined.

The incident ray is the segment joining $$S$$ and $$A$$, so its slope is obtained in the usual way:

$$m_{\text{inc}}=\frac{y-2\sqrt3}{1-2}=2\sqrt3-y.$$

The question states that the ray strikes the line $$x=1$$ at an angle of $$30^\circ$$. Because the line $$x=1$$ is vertical, “angle with the mirror” means the angle between the ray and the vertical line. Hence the same ray makes an angle of $$90^\circ-30^\circ=60^\circ$$ with the positive $$x$$-axis (i.e. with the horizontal). For a line which makes an angle $$\theta$$ with the positive $$x$$-axis, the magnitude of its slope is

$$|m|=\tan\theta.$$

Putting $$\theta=60^\circ$$ we obtain

$$|m_{\text{inc}}|=\tan60^\circ=\sqrt3.$$ Therefore

$$|\,2\sqrt3-y\,|=\sqrt3.$$

This single modulus equation splits into two ordinary equations:

1. $$2\sqrt3-y=\sqrt3\qquad\Longrightarrow\qquad y=\sqrt3,$$

2. $$2\sqrt3-y=-\sqrt3\qquad\Longrightarrow\qquad y=3\sqrt3.$$

Thus the point of incidence can be either $$A_1(1,\sqrt3)\quad\text{or}\quad A_2(1,3\sqrt3).$$

Next we impose the law of reflection: the angle of incidence equals the angle of reflection. For a vertical mirror this simply reverses the sign of the horizontal component of any direction vector, so the slope of the reflected ray is the negative of the incident slope, i.e.

$$m_{\text{ref}}=-m_{\text{inc}}.$$

We now treat the two candidate points separately to see which one produces a reflected ray that passes through one of the points given in the options.

Case 1: $$A_1(1,\sqrt3).$$

For this point $$m_{\text{inc}}=2\sqrt3-\sqrt3=\sqrt3,$$ so $$m_{\text{ref}}=-\sqrt3.$$ The equation of the reflected ray through $$A_1$$ is written in point-slope form:

$$y-\sqrt3=-\sqrt3\,(x-1).$$

We test the four candidate points one by one.

For $$\bigl(3,-\sqrt3\bigr):$$ $$\text{LHS}=y-\sqrt3=-\sqrt3-\sqrt3=-2\sqrt3,$$ $$\text{RHS}=-\sqrt3\,(3-1)=-2\sqrt3,$$ so the point satisfies the equation. Hence line $$A_1B$$ (with $$m=-\sqrt3$$) indeed passes through $$\bigl(3,-\sqrt3\bigr).$$

Because a single straight line cannot pass through two non-collinear points, the other three options need not be checked further. Therefore the reflected ray that starts at $$A_1$$ includes the point $$\bigl(3,-\sqrt3\bigr).$$

Case 2: $$A_2(1,3\sqrt3).$$

For this point $$m_{\text{inc}}=2\sqrt3-3\sqrt3=-\sqrt3,$$ so $$m_{\text{ref}}=\sqrt3.$$ The equation of the reflected ray is

$$y-3\sqrt3=\sqrt3\,(x-1).$$

Substituting each of the four option points shows that none of them satisfies this second equation, so Case 2 yields no admissible answer.

Consequently, only the ray that reflects from $$A_1(1,\sqrt3)$$ is consistent with the data and the given options, and that ray passes through the point $$\bigl(3,-\sqrt3\bigr).$$

Hence, the correct answer is Option C.

We have a right-angled triangle $$\triangle ABC$$ such that the right angle is at vertex $$A$$, that is, $$\angle BAC = 90^{\circ}$$. The coordinates of the fixed vertices are $$A(1,2)$$ and $$B(3,1)$$, while the coordinates of vertex $$C$$ are unknown and may be written as $$C(x,\;y)$$. Because the triangle lies in the first quadrant, both $$x$$ and $$y$$ must be positive.

First we use the fact that the vectors $$\overrightarrow{AB}$$ and $$\overrightarrow{AC}$$ are perpendicular (since the angle between them is $$90^{\circ}$$). The dot-product criterion for perpendicularity is: two vectors $$\mathbf{u}(u_{1},u_{2})$$ and $$\mathbf{v}(v_{1},v_{2})$$ are perpendicular iff $$u_{1}v_{1}+u_{2}v_{2}=0$$.

We calculate the components of the required vectors:

$$\overrightarrow{AB} = (3-1,\;1-2) = (2,\,-1)$$

$$\overrightarrow{AC} = (x-1,\;y-2)$$

Now we impose the dot-product condition:

$$\overrightarrow{AB}\cdot\overrightarrow{AC}=0$$

$$\Rightarrow 2\,(x-1)+(-1)\,(y-2)=0$$

$$\Rightarrow 2x-2-y+2=0$$

$$\Rightarrow 2x-y=0$$

$$\Rightarrow y = 2x$$

Thus, the coordinates of vertex $$C$$ must lie on the straight line $$y=2x$$.

Next we use the given area. For a right-angled triangle with the right angle at $$A$$, the two perpendicular sides are $$AB$$ and $$AC$$. The area formula for a right-angled triangle is

$$\text{Area} = \dfrac{1}{2}\,(\text{length of }AB)\,(\text{length of }AC).$$

The length of $$AB$$ is found by the distance formula:

$$|AB| = \sqrt{(3-1)^{2} + (1-2)^{2}} = \sqrt{2^{2}+(-1)^{2}} = \sqrt{4+1} = \sqrt{5}.$$

The area is given to be $$5\sqrt{5}$$ square units, so we write

$$5\sqrt{5} = \dfrac{1}{2}\,|AB|\,|AC|.$$

Substituting $$|AB|=\sqrt{5}$$, we get

$$5\sqrt{5} = \dfrac{1}{2}\,(\sqrt{5})\,|AC|.$$

Multiplying both sides by $$2$$ and dividing by $$\sqrt{5}$$ gives

$$|AC| = \dfrac{2\,(5\sqrt{5})}{\sqrt{5}} = 10.$$

Thus the distance from $$A(1,2)$$ to $$C(x,2x)$$ must be exactly $$10$$. Applying the distance formula again:

$$|AC|^{2} = (x-1)^{2} + (2x-2)^{2}.$$

Because $$|AC|=10$$, we have

$$10^{2} = (x-1)^{2} + (2x-2)^{2}.$$

Expanding the right side:

$$100 = (x-1)^{2} + 4(x-1)^{2}$$

$$\;= 5(x-1)^{2}.$$

Therefore,

$$(x-1)^{2} = \dfrac{100}{5} = 20.$$

Taking square roots,

$$x - 1 = \pm\sqrt{20} = \pm\,2\sqrt{5}.$$

This gives two possible values:

$$x = 1 + 2\sqrt{5}\quad \text{or}\quad x = 1 - 2\sqrt{5}.$$

Since the triangle lies in the first quadrant, $$x$$ must be positive. The second choice yields a negative number (because $$2\sqrt{5}\approx4.472$$), so it is inadmissible. Hence, the only viable abscissa is

$$x = 1 + 2\sqrt{5}.$$

Looking at the given options, this value corresponds to Option B.

Hence, the correct answer is Option B.

Let us begin by choosing an arbitrary point on the given line $$x = 2y$$. If we denote the ordinate of this point by the parameter $$k$$, then the abscissa must be twice the ordinate. So the point can be written as $$P\,(2k,\;k)$$.

From this point $$P$$ we drop a perpendicular to the line $$x = y$$. To find the foot of this perpendicular we recall the standard formula:

For a line written as $$ax + by + c = 0$$ and an external point $$(x_1,y_1)$$, the coordinates of the foot of the perpendicular $$(x',y')$$ are given by

$$x' \;=\; x_1 - \dfrac{a\,(ax_1 + by_1 + c)}{a^2 + b^2}, \qquad y' \;=\; y_1 - \dfrac{b\,(ax_1 + by_1 + c)}{a^2 + b^2}.$$

Writing $$x = y$$ as $$x - y = 0$$ we identify $$a = 1,\;b = -1,\;c = 0$$. For the point $$P(2k,k)$$ we now compute

$$ax_1 + by_1 + c \;=\; 1(2k) + (-1)(k) + 0 \;=\; k,$$ $$a^2 + b^2 \;=\; 1^2 + (-1)^2 \;=\; 2.$$

Substituting these values into the formula, the coordinates of the foot $$Q$$ are

$$x_Q \;=\; 2k \;-\; \dfrac{1\,(k)}{2} \;=\; 2k - \dfrac{k}{2} \;=\; \dfrac{3k}{2},$$ $$y_Q \;=\; k \;-\; \dfrac{(-1)\,(k)}{2} \;=\; k + \dfrac{k}{2} \;=\; \dfrac{3k}{2}.$$

Hence $$Q\Bigl(\dfrac{3k}{2},\;\dfrac{3k}{2}\Bigr)$$ lies on the line $$x = y$$, as expected.

We now seek the midpoint $$M$$ of the segment $$PQ$$. The midpoint formula is

$$M\bigl(x_M,\,y_M\bigr)\;=\;\left(\dfrac{x_P + x_Q}{2},\;\dfrac{y_P + y_Q}{2}\right).$$

Using $$P(2k,k)$$ and $$Q\bigl(\dfrac{3k}{2},\dfrac{3k}{2}\bigr)$$ we obtain

$$x_M \;=\; \dfrac{\,2k + \dfrac{3k}{2}\,}{2} \;=\; \dfrac{\dfrac{4k + 3k}{2}}{2} \;=\; \dfrac{\dfrac{7k}{2}}{2} \;=\; \dfrac{7k}{4},$$

$$y_M \;=\; \dfrac{\,k + \dfrac{3k}{2}\,}{2} \;=\; \dfrac{\dfrac{2k + 3k}{2}}{2} \;=\; \dfrac{\dfrac{5k}{2}}{2} \;=\; \dfrac{5k}{4}.$$

Thus the running midpoint has coordinates $$M\Bigl(\dfrac{7k}{4},\;\dfrac{5k}{4}\Bigr).$$

To find the locus we eliminate the parameter $$k$$. From the coordinates we have

$$k \;=\; \dfrac{4x_M}{7}, \qquad k \;=\; \dfrac{4y_M}{5}.$$

Equating these two expressions for $$k$$ gives

$$\dfrac{4x_M}{7} \;=\; \dfrac{4y_M}{5}\;\;\Longrightarrow\;\; \dfrac{x_M}{7} \;=\; \dfrac{y_M}{5}\;\;\Longrightarrow\;\; 5x_M - 7y_M = 0.$$

Since $$(x_M,y_M)$$ is a general point on the required locus, the locus is represented by the straight‐line equation

$$5x - 7y = 0.$$

Hence, the correct answer is Option B.

For any point $$P(x,y)$$ the algebraic expression $$x + y - 1$$ tells us on which side of the line $$x + y = 1$$ the point lies, because the line itself is described by $$x + y - 1 = 0$$.

We first evaluate this expression for the fixed point $$(1,2)$$.

Substituting $$x = 1,\; y = 2$$ we get

$$1 + 2 - 1 = 2.$$

The result $$2$$ is positive, so the point $$(1,2)$$ lies on the side of the line where the value of $$x + y - 1$$ is positive.

Now take the variable point $$(\sin\theta,\; \cos\theta)$$ with $$\theta \in (0,\pi)$$. Substituting $$x = \sin\theta,\; y = \cos\theta$$ in the same expression we obtain

$$\sin\theta + \cos\theta - 1.$$

For the two points to lie on the same side of the line, the signs of their respective expressions must be identical. Because the first expression is positive, the second must also be positive:

$$\sin\theta + \cos\theta - 1 > 0.$$

Re-arranging gives the key inequality

$$\sin\theta + \cos\theta > 1.$$

To solve this, we use the standard trigonometric identity

$$\sin\theta + \cos\theta = \sqrt{2}\,\sin\!\Bigl(\theta + \frac{\pi}{4}\Bigr).$$

Substituting the identity, the inequality becomes

$$\sqrt{2}\,\sin\!\Bigl(\theta + \frac{\pi}{4}\Bigr) > 1.$$

Dividing both sides by $$\sqrt{2}$$ (which is positive and hence preserves the inequality sign) we get

$$\sin\!\Bigl(\theta + \frac{\pi}{4}\Bigr) > \frac{1}{\sqrt{2}}.$$

Notice that $$\frac{1}{\sqrt{2}} = \sin\frac{\pi}{4},$$ so we can write

$$\sin\!\Bigl(\theta + \frac{\pi}{4}\Bigr) > \sin\frac{\pi}{4}.$$

Let $$\phi = \theta + \frac{\pi}{4}.$$ Because $$\theta \in (0,\pi),$$ adding $$\frac{\pi}{4}$$ shifts the interval, giving

$$\phi \in \Bigl(\frac{\pi}{4},\, \frac{5\pi}{4}\Bigr).$$

The inequality $$\sin\phi > \sin\frac{\pi}{4}$$ holds when $$\phi$$ is in the interval where the sine curve stays above its value at $$\frac{\pi}{4}.$$ On the principal interval $$[0,\pi]$$ the sine function rises from $$0$$ at $$0$$ to $$1$$ at $$\frac{\pi}{2}$$ and then falls back to $$0$$ at $$\pi$$. Thus

$$\sin\phi > \sin\frac{\pi}{4} \quad\Longleftrightarrow\quad \phi \in \Bigl(\frac{\pi}{4},\,\frac{3\pi}{4}\Bigr).$$

We now intersect this with the allowable range of $$\phi$$ obtained earlier:

$$\Bigl(\frac{\pi}{4},\,\frac{5\pi}{4}\Bigr)\;\cap\;\Bigl(\frac{\pi}{4},\,\frac{3\pi}{4}\Bigr) \;=\; \Bigl(\frac{\pi}{4},\,\frac{3\pi}{4}\Bigr).$$

Returning to $$\theta = \phi - \frac{\pi}{4},$$ we translate the interval back:

$$\theta \in \Bigl(\frac{\pi}{4} - \frac{\pi}{4},\, \frac{3\pi}{4} - \frac{\pi}{4}\Bigr) = (0,\frac{\pi}{2}).$$

Thus the required set of $$\theta$$ values is the open interval $$(0,\frac{\pi}{2}).$$

Hence, the correct answer is Option A.

Let us begin by choosing a convenient Cartesian coordinate system so that every calculation can be carried out algebraically.

We place the origin on the ground at the foot of the taller pole A. Thus we take

$$A=(0,0).$$

The line AC is the horizontal ground, so the x-axis is along AC. Let the horizontal distance between the two poles be $$d\;{\rm m}$$ (its actual value will not matter; we shall soon see it cancels out). Hence the foot of the shorter pole is

$$C=(d,0).$$

The poles are vertical, therefore their tops have the same x-coordinates as their feet. Using the given heights, we write

$$B=(0,15) \quad\text{and}\quad D=(d,10).$$

Now we require the coordinates of the intersection $$P$$ of the two straight lines $$BC$$ and $$AD$$ and, finally, its y-coordinate (the height above the ground).

First we find a parametric equation for the line $$BC$$. Starting at $$B$$ and moving a fraction $$t$$ of the way toward $$C$$, we have

$$\bigl(x,y\bigr)=B+t(C-B) =(0,15)+t\bigl(d,0-15\bigr) =(td,\;15-15t).$$

So every point of $$BC$$ can be written as

$$\bigl(x,y\bigr)=\bigl(td,\;15(1-t)\bigr),\qquad 0\le t\le1.$$

Next we write a parametric equation for the line $$AD$$. Starting at $$A$$ and moving a fraction $$s$$ of the way toward $$D$$, we get

$$\bigl(x,y\bigr)=A+s(D-A) =(0,0)+s\bigl(d,10\bigr) =(sd,\;10s).$$

Thus every point of $$AD$$ can be expressed as

$$\bigl(x,y\bigr)=\bigl(sd,\;10s\bigr),\qquad 0\le s\le1.$$

At the point $$P$$ the coordinates coming from the two descriptions are equal; hence we equate them:

$$td = sd \quad\text{and}\quad 15(1-t) = 10s.$$

From the first equality (the x-coordinates) we have

$$td = sd \implies t = s,$$

because $$d\neq0$$. Substituting $$s=t$$ into the second equality (the y-coordinates) gives

$$15(1-t) = 10t.$$

Expanding and gathering like terms:

$$15 - 15t = 10t \\ \Rightarrow 15 = 25t \\ \Rightarrow t = \frac{15}{25} = \frac{3}{5}.$$

Because $$s=t$$, we also have

$$s=\frac{3}{5}.$$

Now we obtain the height of $$P$$ by inserting this value into the y-coordinate formula of either line (both give the same result). Using the expression from $$AD$$,

$$y = 10s = 10\left(\frac{3}{5}\right) = 6.$$

Thus the intersection point $$P$$ is $$6\;{\rm m}$$ above the ground line $$AC$$. Notice that the horizontal distance $$d$$ indeed cancelled out, confirming that the answer is independent of how far apart the poles stand.

Hence, the correct answer is Option D.

We have the curve $$y = x^{2}+7x+2$$ and the straight line $$y = 3x-3$$. To find the point $$P(h,k)$$ on the curve that is nearest to the line, we minimise the perpendicular distance from a general point $$(x,\,x^{2}+7x+2)$$ of the curve to the given line.

First we write the line in the standard form $$Ax+By+C=0$$. Starting from $$y = 3x-3$$ we bring all terms to the left:

$$3x-y-3 = 0$$

Thus $$A = 3,\;B = -1,\;C = -3$$.

The perpendicular distance $$D$$ of a point $$(x_1,y_1)$$ from $$Ax+By+C=0$$ is given by the formula $$D = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}}\,.$$

Substituting $$(x_1,y_1)=\left(x,\;x^{2}+7x+2\right)$$ and $$(A,B,C)=(3,-1,-3)$$ we obtain

$$D =\dfrac{\bigl|\,3x-\,\bigl(x^{2}+7x+2\bigr)\,-3\bigr|}{\sqrt{3^{2}+(-1)^{2}}} =\dfrac{|\, -x^{2}-4x-5\,|}{\sqrt{10}}.$$

Because the denominator $$\sqrt{10}$$ is constant, minimising $$D$$ is the same as minimising the numerator’s absolute value. Let us denote

$$g(x) = -x^{2}-4x-5 = -(x^{2}+4x+5).$$

The square of the distance is proportional to $$g(x)^{2}$$, and minimising $$g(x)^{2}$$ is equivalent to minimising the non-negative quadratic

$$h(x)=x^{2}+4x+5.$$ Now we differentiate $$h(x)$$ with respect to $$x$$ and set the derivative to zero:

$$\dfrac{dh}{dx}=2x+4,$$ $$2x+4=0\;\Longrightarrow\;x=-2.$$

This value indeed gives the minimum because $$\dfrac{d^{2}h}{dx^{2}}=2>0.$$ Hence the abscissa of the required point is $$h=-2$$.

Substituting $$x=-2$$ in the equation of the curve to find the ordinate:

$$k = (-2)^{2}+7(-2)+2 = 4-14+2 = -8.$$

Therefore the nearest point is $$P(h,k)=(-2,-8).$$

Next we need the equation of the normal to the curve at this point. The slope of the tangent to the curve $$y=x^{2}+7x+2$$ is obtained by differentiation:

$$\frac{dy}{dx}=2x+7.$$

At $$x=-2$$, the tangent slope is $$m_{\text{tangent}} = 2(-2)+7 = -4+7 = 3.$$

The normal is perpendicular to the tangent, so its slope is the negative reciprocal:

$$m_{\text{normal}} = -\dfrac{1}{3}.$$

Using the point-slope form of a line through $$(-2,-8)$$ with slope $$-\dfrac{1}{3}$$: $$y-(-8)= -\dfrac{1}{3}\,\bigl(x-(-2)\bigr),$$ $$y+8 = -\dfrac{1}{3}(x+2).$$

Multiplying by $$3$$ to clear the fraction:

$$3(y+8)=-(x+2),$$ $$3y + 24 = -x - 2.$$

Bringing all terms to the left:

$$x + 3y + 26 = 0.$$

This is precisely option A.

Hence, the correct answer is Option A.

We have the three vertices of the triangle as $$A(3,-1),\;B(1,3),\;C(2,4)$$. The centroid of a triangle with vertices $$\bigl(x_1,y_1\bigr),\;\bigl(x_2,y_2\bigr),\;\bigl(x_3,y_3\bigr)$$ is given by the well-known formula

$$\Bigl(\dfrac{x_1+x_2+x_3}{3},\;\dfrac{y_1+y_2+y_3}{3}\Bigr).$$

Substituting the coordinates of $$A,B,C$$ we obtain

$$x\text{-coordinate of }D=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2,$$ $$y\text{-coordinate of }D=\dfrac{-1+3+4}{3}=\dfrac{6}{3}=2.$$

So the centroid is $$D(2,2).$$

Next we determine the point $$P$$ where the two straight lines intersect. The equations of the lines are

$$x+3y-1=0\quad\text{and}\quad3x-y+1=0.$$

From the first equation we isolate $$x$$:

$$x = 1 - 3y.$$

Now we substitute this expression for $$x$$ in the second equation:

$$3(1-3y) - y + 1 = 0.$$

Simplifying step by step,

$$3 - 9y - y + 1 = 0,$$ $$4 - 10y = 0,$$ $$10y = 4,$$ $$y = \dfrac{4}{10} = \dfrac{2}{5}.$$

Substituting $$y=\dfrac{2}{5}$$ back into $$x = 1 - 3y$$ gives

$$x = 1 - 3\left(\dfrac{2}{5}\right) = 1 - \dfrac{6}{5} = -\,\dfrac{1}{5}.$$

Thus the intersection point is $$P\!\left(-\dfrac{1}{5},\,\dfrac{2}{5}\right).$$

We now require the equation of the straight line passing through $$D(2,2)$$ and $$P\!\left(-\dfrac{1}{5},\,\dfrac{2}{5}\right).$$ The slope (gradient) of a line through two points $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr)$$ is given by

$$m=\dfrac{y_2-y_1}{x_2-x_1}.$$

So,

$$m=\dfrac{2-\dfrac{2}{5}}{2-\left(-\dfrac{1}{5}\right)}=\dfrac{\dfrac{10}{5}-\dfrac{2}{5}}{\dfrac{10}{5}+\dfrac{1}{5}}=\dfrac{\dfrac{8}{5}}{\dfrac{11}{5}}=\dfrac{8}{11}.$$

Using the point-slope form $$y-y_1=m(x-x_1)$$ with the point $$D(2,2)$$, we write

$$y-2=\dfrac{8}{11}\,\bigl(x-2\bigr).$$

To clear the denominator, multiply every term by $$11$$:

$$11(y-2)=8(x-2).$$

Expanding the brackets,

$$11y-22=8x-16.$$

Bringing all terms to the left side,

$$8x-11y+6=0.$$

This is the equation of the line through $$D$$ and $$P$$. Now we test which of the given option points satisfies $$8x-11y+6=0$$.

Option A: $$(-9,-6)$$ Substituting, $$8(-9)-11(-6)+6=-72+66+6=0.$$ So $$(-9,-6)$$ lies on the line.

Option B: $$(9,7)$$ $$8(9)-11(7)+6=72-77+6=1\neq0.$$

Option C: $$(7,6)$$ $$8(7)-11(6)+6=56-66+6=-4\neq0.$$

Option D: $$(-9,-7)$$ $$8(-9)-11(-7)+6=-72+77+6=11\neq0.$$

Only Option A satisfies the line equation. Hence, the correct answer is Option A.

We are given the vertices of the triangle as $$A(1,0),\;B(6,2),\;C\left(\dfrac32,6\right).$$ Inside this triangle there is a point $$P$$ such that the areas of the three triangles $$\triangle APC,\;\triangle APB,\;\triangle BPC$$ are equal.

First recall a well-known fact: for any point $$P$$ inside a triangle, the ratios of the areas of the sub-triangles to the whole triangle give the barycentric coordinates of $$P$$. If all three areas are equal, each is one third of the total area, so the barycentric coordinates of $$P$$ are $$\left(\dfrac13,\dfrac13,\dfrac13\right).$$ A point with equal barycentric coordinates is nothing but the centroid of the triangle. Hence $$P$$ must be the centroid of $$\triangle ABC.$$

The centroid formula states that for vertices $$\bigl(x_1,y_1\bigr),\bigl(x_2,y_2\bigr),\bigl(x_3,y_3\bigr)$$ the centroid $$G(x_G,y_G)$$ is

$$x_G=\dfrac{x_1+x_2+x_3}{3},\qquad y_G=\dfrac{y_1+y_2+y_3}{3}.$$

Applying this to our triangle:

$$x_P=\dfrac{1+6+\dfrac32}{3}=\dfrac{1+6+1.5}{3}=\dfrac{8.5}{3}=\dfrac{17}{6},$$

$$y_P=\dfrac{0+2+6}{3}=\dfrac{8}{3}.$$

Thus $$P\left(\dfrac{17}{6},\dfrac83\right).$$

We are asked for the length of segment $$PQ$$ where $$Q\!\left(-\dfrac76,-\dfrac13\right).$$ We now use the distance formula between two points $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr):$$

$$\text{Distance}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$

Compute the differences:

$$\Delta x = x_P - x_Q = \dfrac{17}{6}-\left(-\dfrac76\right)=\dfrac{17+7}{6}=\dfrac{24}{6}=4,$$

$$\Delta y = y_P - y_Q = \dfrac83-\left(-\dfrac13\right)=\dfrac{8+1}{3}=\dfrac{9}{3}=3.$$

Substituting these into the distance formula gives

$$PQ=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5.$$

So, the answer is $$5$$.

We have the given line $$2x - y + 3 = 0$$ and two other lines $$4x - 2y + \alpha = 0$$ and $$6x - 3y + \beta = 0$$ which are both obviously parallel to the first line because their left-hand sides are constant multiples of $$2x - y$$. For parallel lines we use the formula for the perpendicular distance between the pair of lines

$$\text{Distance} \;=\; \frac{\lvert C_1 - C_2\rvert}{\sqrt{A^{2} + B^{2}}},$$

where the lines are written in the common normal form $$Ax + By + C_1 = 0$$ and $$Ax + By + C_2 = 0$$.

First we compare the first line with $$4x - 2y + \alpha = 0$$. To make the two equations share the same normal vector, we divide the entire second equation by $$2$$:

$$\frac{4x - 2y + \alpha}{2} = 0 \;\;\Longrightarrow\;\; 2x - y + \frac{\alpha}{2} = 0.$$

Now both lines are in the form $$2x - y + C = 0$$, so we can directly apply the distance formula. The required distance is given to be $$\dfrac{1}{\sqrt{5}}$$, and the denominator $$\sqrt{A^{2} + B^{2}}$$ equals $$\sqrt{2^{2} + (-1)^{2}} = \sqrt{5}$$. Substituting these into the formula we get

$$\frac{\lvert 3 - \dfrac{\alpha}{2}\rvert}{\sqrt{5}} = \frac{1}{\sqrt{5}} \;\;\Longrightarrow\;\; \lvert 3 - \dfrac{\alpha}{2}\rvert = 1.$$

This absolute-value equation splits into two linear equations:

1. $$3 - \dfrac{\alpha}{2} = 1 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2} = 2 \;\;\Longrightarrow\;\; \alpha = 4,$$

2. $$3 - \dfrac{\alpha}{2} = -1 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2} = 4 \;\;\Longrightarrow\;\; \alpha = 8.$$

Hence the possible values of $$\alpha$$ are $$4$$ and $$8$$.

Next we compare $$2x - y + 3 = 0$$ with the line $$6x - 3y + \beta = 0$$. We again normalise by dividing by $$3$$:

$$\frac{6x - 3y + \beta}{3} = 0 \;\;\Longrightarrow\;\; 2x - y + \frac{\beta}{3} = 0.$$

The distance between these two parallel lines is stated to be $$\dfrac{2}{\sqrt{5}}$$. Using the same denominator $$\sqrt{5}$$, we write

$$\frac{\lvert 3 - \dfrac{\beta}{3}\rvert}{\sqrt{5}} = \frac{2}{\sqrt{5}} \;\;\Longrightarrow\;\; \lvert 3 - \dfrac{\beta}{3}\rvert = 2.$$

Again we split the absolute value:

1. $$3 - \dfrac{\beta}{3} = 2 \;\;\Longrightarrow\;\; \dfrac{\beta}{3} = 1 \;\;\Longrightarrow\;\; \beta = 3,$$

2. $$3 - \dfrac{\beta}{3} = -2 \;\;\Longrightarrow\;\; \dfrac{\beta}{3} = 5 \;\;\Longrightarrow\;\; \beta = 15.$$

Thus the possible values of $$\beta$$ are $$3$$ and $$15$$.

The problem asks for the sum of all possible values of $$\alpha$$ and $$\beta$$ together. Adding these distinct values we get

$$\underbrace{4 + 8}_{\alpha\text{-values}} + \underbrace{3 + 15}_{\beta\text{-values}} = 12 + 18 = 30.$$

Hence, the correct answer is 30.

We are told that ABDC is a parallelogram and that the coordinates of the three known vertices are $$A(1,2),\; B(3,4),\; C(2,5).$$ The fourth vertex $$D(x,y)$$ is not given. Our goal is to find the equation of the diagonal $$AD\;.$$

In any parallelogram, a very important fact is that the diagonals bisect each other. Stating this property mathematically: if the diagonals join the opposite vertices $$A\text{ to }D$$ and $$B\text{ to }C,$$ then the mid-point of $$AD$$ is exactly the same as the mid-point of $$BC$$.

First we write the co-ordinates of the mid-point of $$AD$$ in terms of the unknown co-ordinates of $$D(x,y).$$ By the mid-point formula, $$\text{Mid-point of }AD = \left(\frac{1+x}{2},\; \frac{2+y}{2}\right).$$

Next we find the mid-point of the completely known diagonal $$BC.$$ Again, using the mid-point formula, $$\text{Mid-point of }BC = \left(\frac{3+2}{2},\; \frac{4+5}{2}\right) = \left(\frac{5}{2},\; \frac{9}{2}\right).$$

Because the two mid-points are equal, we can equate the corresponding co-ordinates:

$$\frac{1+x}{2} = \frac{5}{2} \quad\text{and}\quad \frac{2+y}{2} = \frac{9}{2}.$$

Multiplying both equations by $$2$$ to clear the denominators, we get

$$1 + x = 5 \quad\text{and}\quad 2 + y = 9.$$

Now we solve for $$x$$ and $$y$$ separately:

$$x = 5 - 1 = 4,$$ $$y = 9 - 2 = 7.$$

So we have located the fourth vertex: $$D(4,7).$$

With both end-points of the required diagonal $$AD$$ known as $$A(1,2)\; \text{and}\; D(4,7),$$ we now find the equation of the straight line passing through these two points.

First calculate the slope. The slope (or gradient) formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}.$$ Substituting $$A(1,2)$$ as $$(x_1,y_1)$$ and $$D(4,7)$$ as $$(x_2,y_2)$$ gives

$$m = \frac{7 - 2}{4 - 1} = \frac{5}{3}.$$

Next we use the point-slope form of the equation of a line, $$y - y_1 = m(x - x_1).$$ Choosing the point $$A(1,2),$$ we substitute:

$$y - 2 = \frac{5}{3}\,(x - 1).$$

To eliminate the fraction, multiply every term by $$3$$:

$$3(y - 2) = 5(x - 1).$$

Expand both sides:

$$3y - 6 = 5x - 5.$$

Now bring all terms to one side to obtain the standard linear form:

$$5x - 5 - 3y + 6 = 0,$$

$$5x - 3y + 1 = 0.$$

This is already simplified, and it exactly matches Option A.

Hence, the correct answer is Option A.

Let us denote the vertices of the triangle by $$A(0,\,2),\;B(4,\,3)\; \text{and}\; C(x,\,y),$$ and let the orthocenter be $$H(0,\,0).$$

The orthocenter is the common point of the three altitudes, so every altitude must pass through the origin.

Altitude through $$A$$
We have the two points $$A(0,\,2)\quad\text{and}\quad H(0,\,0).$$ Because their abscissae are equal, the line $$AH$$ is the vertical line $$x=0.$$ An altitude is perpendicular to the opposite side, so the side $$BC$$ must be horizontal, that is, it must have slope $$0$$. Consequently,

$$y_B = y_C \;\Longrightarrow\; y = 3.$$

Altitude through $$B$$
The points of this altitude are $$B(4,\,3)\quad\text{and}\quad H(0,\,0).$$ Its slope is therefore

$$m_{BH} = \frac{0-3}{0-4}= \frac{-3}{-4}= \frac{3}{4}.$$

This altitude is perpendicular to side $$AC$$. If the slope of $$AC$$ is $$m_{AC}$$, the product of the slopes of two perpendicular lines in the plane is $$-1$$. Hence

$$m_{AC}\,(\text{slope of } AC)\times m_{BH}= -1.$$ That is, $$m_{AC}\,\cdot\frac{3}{4}= -1 \;\Longrightarrow\; m_{AC}= -\frac{4}{3}.$$

Now, since $$A(0,\,2)$$ and $$C(x,\,3)$$ lie on side $$AC$$, we can write the slope formula:

$$m_{AC}= \frac{3-2}{x-0}= \frac{1}{x}.$$

Substituting the required slope, we have

$$\frac{1}{x}= -\frac{4}{3} \;\Longrightarrow\; x= -\frac{3}{4}.$$

Coordinates of the third vertex
Combining the results for $$x$$ and $$y$$, we obtain

$$C\!\left(-\frac{3}{4},\,3\right).$$

Quadrant of the point $$C$$
The x-coordinate is negative and the y-coordinate is positive, so the point lies in the second quadrant.

Hence, the correct answer is Option B.

We are told that the three points $$(h,k),\,(1,2)$$ and $$(-3,4)$$ all lie on the same straight line, which we shall call $$L_1$$. We first recall the formula for the slope of a line joining two points. If the points are $$\bigl(x_1,y_1\bigr)$$ and $$\bigl(x_2,y_2\bigr)$$, then the slope is

$$m \;=\;\dfrac{y_2-y_1}{x_2-x_1}.$$

Using the two fixed points $$(1,2)$$ and $$(-3,4)$$ on $$L_1$$, we obtain

$$m_1 \;=\;\dfrac{\,4-2\,}{\,(-3)-1\,} \;=\;\dfrac{2}{-4} \;=\;-\dfrac12.$$ Hence the slope of $$L_1$$ is $$m_1=-\dfrac12.$$

Because the unknown point $$(h,k)$$ also lies on $$L_1$$, the slope of the segment joining $$(1,2)$$ and $$(h,k)$$ must equal $$-\dfrac12$$ as well. Applying the slope formula once more, we have

$$\dfrac{k-2}{h-1}= -\dfrac12.$$

Cross-multiplying gives

$$2(k-2)=-(h-1).$$

Expanding both sides,

$$2k-4 = -h+1.$$

Bringing every term to one side, we rewrite this as

$$h + 2k = 5.\qquad(1)$$

Now consider the second line $$L_2$$, which passes through $$(h,k)$$ and $$(4,3)$$ and is stated to be perpendicular to $$L_1$$. The slope of $$L_2$$, using the two points that define it, is

$$m_2 \;=\;\dfrac{3-k}{4-h}.$$

For two lines to be perpendicular, the product of their slopes equals $$-1$$. That is, if $$m_1$$ is the slope of the first line and $$m_2$$ is the slope of the second, then

$$m_1\,m_2=-1.$$

Substituting $$m_1=-\dfrac12$$ into this condition, we obtain

$$\Bigl(-\dfrac12\Bigr)\,m_2 = -1.$$

Dividing by $$-\dfrac12$$ (that is, multiplying by $$-2$$) yields

$$m_2 = 2.$$

But we already have $$m_2=\dfrac{3-k}{4-h}$$, so we must have

$$\dfrac{3-k}{4-h}=2.$$

Cross-multiplying once more,

$$3-k = 2(4-h).$$

Expanding the right-hand side,

$$3-k = 8-2h.$$

Collecting terms by moving every variable to the left and constants to the right, we get

$$2h - k = 5.\qquad(2)$$

We now have two linear equations in $$h$$ and $$k$$: $$\begin{aligned} (1)&\;\; h + 2k &= 5,\\ (2)&\;\; 2h - k &= 5. \end{aligned}$$

We solve this system step by step. From equation (1) we isolate $$h$$:

$$h = 5 - 2k.$$

Substituting this expression for $$h$$ into equation (2), we obtain

$$2(5 - 2k) - k = 5.$$

Multiplying out the bracket,

$$10 - 4k - k = 5.$$

Combining the $$k$$ terms,

$$10 - 5k = 5.$$

Subtracting $$10$$ from both sides,

$$-5k = -5.$$

Dividing by $$-5$$ gives

$$k = 1.$$

Returning to $$h = 5 - 2k$$ and substituting $$k=1$$, we find

$$h = 5 - 2(1) = 3.$$

Therefore,

$$\dfrac{k}{h} = \dfrac{1}{3}.$$

Hence, the correct answer is Option D.

We start from the given relation

$$y=\sin x\,\sin (x+2)\;-\;\sin^{2}(x+1).$$

First we simplify the product $$\sin x\,\sin (x+2)$$ by recalling the product-to-sum identity

$$\sin A\,\sin B=\frac{\cos(A-B)-\cos(A+B)}{2}.$$

Taking $$A=x,\;B=x+2,$$ we get

$$\sin x\,\sin (x+2)=\frac{\cos\!\bigl(x-(x+2)\bigr)-\cos\!\bigl(x+(x+2)\bigr)}{2} =\frac{\cos(-2)-\cos(2x+2)}{2} =\frac{\cos 2-\cos(2x+2)}{2}.$$

Substituting this result back, the expression for y becomes

$$y=\frac{\cos 2-\cos(2x+2)}{2}-\sin^{2}(x+1).$$

Next we simplify $$\sin^{2}(x+1)$$ with the double-angle formula

$$\sin^{2}\theta=\frac{1-\cos 2\theta}{2}.$$

Putting $$\theta=x+1,$$ we have

$$\sin^{2}(x+1)=\frac{1-\cos\!\bigl(2x+2\bigr)}{2}.$$

Inserting this into the previous expression gives

$$ \begin{aligned} y &= \frac{\cos 2-\cos(2x+2)}{2}\;-\;\frac{1-\cos(2x+2)}{2} \\[4pt] &=\frac{\cos 2-\cos(2x+2)-1+\cos(2x+2)}{2} \\[4pt] &=\frac{\cos 2-1}{2}. \end{aligned} $$

Observe that every term containing x has cancelled out; hence y is a constant. To see its sign, rewrite

$$\cos 2-1=-(1-\cos 2).$$

Using $$1-\cos\theta=2\sin^{2}\dfrac{\theta}{2},$$ with $$\theta=2,$$ we get

$$1-\cos 2=2\sin^{2}1,$$

so

$$\cos 2-1=-2\sin^{2}1.$$

Therefore

$$y=\frac{\cos 2-1}{2}= \frac{-2\sin^{2}1}{2}=-\sin^{2}1.$$

The quantity $$\sin^{2}1$$ is positive (because $$1\text{ rad}\approx57^\circ$$ lies in the first quadrant). Hence

$$y=-\sin^{2}1\lt 0\quad\text{for every }x\in\mathbb R.$$

Thus every point of the straight line has a negative ordinate. Since x can take both positive and negative values, points on the line occur with

• $$x \lt 0,\;y \lt 0$$  ⇒ third quadrant, and
• $$x \gt 0,\;y \lt 0$$  ⇒ fourth quadrant.

No point can appear in the first or second quadrants because there $$y\gt 0.$$

Hence, the correct answer is Option D.

Let us denote the two given non-parallel lines as

$$L_1:\;x+y=3\quad\text{and}\quad L_2:\;x-y+3=0\;(\text{or }x-y=-3).$$

These are the directions of two adjacent sides of the required parallelogram. For any straight line written in the form $$ax+by=c$$ a direction vector is $$\bigl(b,-a\bigr).$$ So, for the two lines we obtain

Direction of $$L_1: (1,-1),\qquad$$ Direction of $$L_2: (1,1).$$

Choose two vectors along the sides

$$\vec{v_1}=k(1,-1)=\,(k,\,-k),\qquad \vec{v_2}=m(1,1)=\,(m,\,m),$$

where the non-zero scalars $$k$$ and $$m$$ are still unknown.

Let $$A(x_1,y_1)$$ be one vertex of the parallelogram. With the standard Labelling $$AB\parallel L_1,\;AD\parallel L_2$$ we get

$$B=A+\vec{v_1},\quad D=A+\vec{v_2},\quad C=A+\vec{v_1}+\vec{v_2}.$$

In every parallelogram the diagonals bisect each other. Hence the common midpoint $$M$$ of the diagonals satisfies the midpoint formula

$$M=\frac{A+C}{2}=\frac{A+(A+\vec{v_1}+\vec{v_2})}{2} =\frac{2A+\vec{v_1}+\vec{v_2}}{2}.$$

The problem states that the diagonals meet at $$M(2,4).$$ Putting $$M(2,4)$$ into the last relation gives

$$2M=(4,8)=2A+\vec{v_1}+\vec{v_2}.$$

Re-arranging for the unknown vertex $$A$$ we obtain

$$2A=(4,8)-(\vec{v_1}+\vec{v_2})\;\Longrightarrow\; A=(2,4)-\frac{\vec{v_1}+\vec{v_2}}{2}.$$

Compute $$\vec{v_1}+\vec{v_2}:$$

$$\vec{v_1}+\vec{v_2}=(k+m,\,-k+m).$$

Therefore

$$A=\Bigl(2-\dfrac{k+m}{2},\;4-\dfrac{m-k}{2}\Bigr).$$

We now look among the four given options to see whether any one of them can be expressed by suitable integers $$k$$ and $$m.$$ Start with Option A, the point $$(3,6).$$ Equating coordinate-wise,

$$2-\dfrac{k+m}{2}=3\quad\text{and}\quad 4-\dfrac{m-k}{2}=6.$$

First equation:

$$2-\dfrac{k+m}{2}=3 \;\Longrightarrow\; -\dfrac{k+m}{2}=1 \;\Longrightarrow\; k+m=-2.$$

Second equation:

$$4-\dfrac{m-k}{2}=6 \;\Longrightarrow\; -\dfrac{m-k}{2}=2 \;\Longrightarrow\; m-k=-4.$$

Now solve the simultaneous linear equations

$$\begin{cases} k+m=-2\\ m-k=-4 \end{cases}\qquad\Longrightarrow\qquad \begin{aligned} \;2m&=-6&\Longrightarrow&m=-3,\\ k&=1. \end{aligned}$$

Both $$k=1$$ and $$m=-3$$ are admissible (they merely determine the lengths and directions of the sides). Hence with these values the point $$(3,6)$$ is indeed a vertex of a parallelogram whose sides are parallel to the two given lines and whose diagonals meet at $$(2,4).$$

Having found one working choice, the answer is fixed because a parallelogram is fully determined by the two direction lines and the point of intersection of its diagonals; only four possible vertices exist, and among the options only $$(3,6)$$ satisfies the required relations.

Hence, the correct answer is Option A.

We have been asked to locate those points which simultaneously satisfy the straight-line equation $$3x + 5y = 15$$ and are equidistant from the two coordinate axes.

First, we recall the mathematical fact that the perpendicular distance of a point $$P(x , y)$$ from the $$y$$-axis is $$|x|$$, while the perpendicular distance from the $$x$$-axis is $$|y|$$. A point is equidistant from the two axes exactly when these two absolute values are equal. Hence we must have

$$|x| = |y|.$$

This equality of absolute values can occur in two algebraic ways:

(i) $$y = x$$    or    (ii) $$y = -x.$$

Thus every point that is equidistant from the axes lies on one of the two lines $$y = x$$ or $$y = -x$$. We now examine each possibility in turn, always keeping the additional requirement that the point must also lie on the given straight line $$3x + 5y = 15$$.

Case 1 : $$y = x$$

We substitute $$y = x$$ in the equation $$3x + 5y = 15$$.

Using substitution we get

$$3x + 5x = 15.$$

Combining like terms on the left side,

$$8x = 15.$$

Solving for $$x$$, we divide both sides by $$8$$:

$$x = \dfrac{15}{8}.$$

Because $$y = x$$ in this case, we have

$$y = \dfrac{15}{8}.$$

Both coordinates are positive, i.e. $$x > 0$$ and $$y > 0$$, so the point

$$\left(\dfrac{15}{8},\,\dfrac{15}{8}\right)$$

lies in the first quadrant.

Case 2 : $$y = -x$$

Again we substitute, now using $$y = -x$$ in the same line equation:

$$3x + 5(-x) = 15.$$

That gives

$$3x - 5x = 15.$$

Combining like terms,

$$-2x = 15.$$

Dividing by $$-2$$, we obtain

$$x = -\dfrac{15}{2}.$$

Since $$y = -x$$ here,

$$y = -\left(-\dfrac{15}{2}\right) = \dfrac{15}{2}.$$

Now we observe that $$x < 0$$ while $$y > 0$$; hence the point

$$\left(-\dfrac{15}{2},\,\dfrac{15}{2}\right)$$

falls in the second quadrant.

There are no valid solutions in the third or fourth quadrants, because our two exhaustive cases have already accounted for all possibilities of $$|x| = |y|$$.

Therefore the points satisfying both conditions—lying on $$3x + 5y = 15$$ and being equidistant from the coordinate axes—occur only in the first and second quadrants.

Hence, the correct answer is Option A.

Let us denote the required triangle by $$\triangle ABC$$ and assume, without loss of generality, that $$A(1,\,2)$$ is the given vertex. We are further told that the mid-points of the two sides through this vertex are $$M_1(-1,\,1)$$ and $$M_2(2,\,3).$$ Therefore $$M_1$$ is the mid-point of side $$AB$$ and $$M_2$$ is the mid-point of side $$AC.$$

We now recall the Mid-point Formula: for the segment joining $$(x_1,\,y_1)$$ and $$(x_2,\,y_2),$$ the co-ordinates of the mid-point are $$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right).$$

Applying this formula to $$AB,$$ we write $$M_1(-1,\,1)=\left(\dfrac{1+x_B}{2},\ \dfrac{2+y_B}{2}\right).$$ Equating the individual co-ordinates, we obtain each step explicitly:

For the $$x$$-coordinate $$\dfrac{1+x_B}{2}=-1$$ $$\Longrightarrow\ 1+x_B = -2$$ $$\Longrightarrow\ x_B = -3.$$

For the $$y$$-coordinate $$\dfrac{2+y_B}{2}=1$$ $$\Longrightarrow\ 2+y_B = 2$$ $$\Longrightarrow\ y_B = 0.$$

Hence we have found $$B(-3,\,0).$$

In the same manner, using $$M_2$$ as the mid-point of $$AC,$$ we write $$M_2(2,\,3)=\left(\dfrac{1+x_C}{2},\ \dfrac{2+y_C}{2}\right).$$

Again, matching co-ordinates one by one:

For the $$x$$-coordinate $$\dfrac{1+x_C}{2}=2$$ $$\Longrightarrow\ 1+x_C = 4$$ $$\Longrightarrow\ x_C = 3.$$

For the $$y$$-coordinate $$\dfrac{2+y_C}{2}=3$$ $$\Longrightarrow\ 2+y_C = 6$$ $$\Longrightarrow\ y_C = 4.$$

Thus we have $$C(3,\,4).$$

With all three vertices now known as $$A(1,2),\ B(-3,0),\ C(3,4),$$ we proceed to the centroid. The Centroid Formula states that the centroid $$G(x_G,\,y_G)$$ of the triangle whose vertices are $$(x_A,\,y_A),\ (x_B,\,y_B),\ (x_C,\,y_C)$$ is

$$\bigl(x_G,\,y_G\bigr)=\left(\dfrac{x_A+x_B+x_C}{3},\ \dfrac{y_A+y_B+y_C}{3}\right).$$

Substituting the specific values we have gathered:

$$x_G=\dfrac{1+(-3)+3}{3}=\dfrac{1-3+3}{3}=\dfrac{1}{3},$$

$$y_G=\dfrac{2+0+4}{3}=\dfrac{6}{3}=2.$$

So the centroid is $$\left(\dfrac{1}{3},\,2\right).$$

Hence, the correct answer is Option C.

We have the family of lines

$$p\,x+q\,y+r=0$$

subject to the linear condition

$$3p+2q+4r=0.$$

From the second relation we can express r in terms of p and q. Solving for r,

$$4r=-(3p+2q)\;\Longrightarrow\; r=-\dfrac{3p+2q}{4}.$$

Substituting this value of r back into the equation of the line,

$$p\,x+q\,y-\dfrac{3p+2q}{4}=0.$$

To clear the fraction, multiply every term by 4:

$$4p\,x+4q\,y-3p-2q=0.$$

Group the terms containing p and those containing q:

$$p\,(4x-3)+q\,(4y-2)=0.$$

In this expression p and q are arbitrary real numbers, not both zero. For the left-hand side to vanish for every possible choice of (p,q) that satisfies the original constraint, each of the factors that multiplies p and q must itself be zero. Thus we require

$$4x-3=0 \quad\text{and}\quad 4y-2=0.$$

Solving these two simple linear equations gives

$$x=\dfrac34, \qquad y=\dfrac12.$$

So every line in the given set passes through the single fixed point $$\left(\dfrac34,\dfrac12\right).$$

Because all the lines meet at one common point, they are concurrent, and the point of concurrency is exactly $$\left(\dfrac34,\dfrac12\right).$$

Hence, the correct answer is Option B.

We have a straight line that cuts the coordinate axes at two points: let the x-intercept be $$A(a,0)$$ and the y-intercept be $$B(0,b)$$. In coordinate geometry, the equation of such a line in “intercept form” is first stated as

$$\frac{x}{a}+\frac{y}{b}=1.$$

The problem states that the fixed point $$P(-3,4)$$ lies on the line and, moreover, that the segment $$AB$$ is bisected at $$P$$. To translate “bisected at $$P$$”, we recall the mid-point formula: if $$M(x_m, y_m)$$ is the midpoint of $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$, then

$$x_m=\frac{x_1+x_2}{2},\qquad y_m=\frac{y_1+y_2}{2}.$$

Applying this to $$A(a,0)$$ and $$B(0,b)$$ with midpoint $$P(-3,4)$$, we write

$$\frac{a+0}{2}=-3,\qquad\frac{0+b}{2}=4.$$

Solving these two simple equations one by one:

First, $$\dfrac{a}{2}=-3 \;\Longrightarrow\; a=-6.$$

Second, $$\dfrac{b}{2}=4 \;\Longrightarrow\; b=8.$$

So the line meets the x-axis at $$(-6,0)$$ and the y-axis at $$(0,8)$$. We now substitute these values of $$a$$ and $$b$$ back into the intercept form of the line.

Starting with $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ and inserting $$a=-6$$, $$b=8$$, we get

$$\frac{x}{-6}+\frac{y}{8}=1.$$

To clear the denominators, multiply both sides by the LCM $$24$$:

$$24\left(\frac{x}{-6}\right)+24\left(\frac{y}{8}\right)=24(1).$$

Simplifying term by term,

$$24\left(\frac{x}{-6}\right)=-4x,\qquad 24\left(\frac{y}{8}\right)=3y,\qquad 24(1)=24.$$

So the equation becomes

$$-4x+3y=24.$$

For a tidier appearance—and to match one of the given options—we multiply through by $$-1$$:

$$4x-3y=-24.$$

Finally, bringing all terms to the left-hand side gives the standard form

$$4x-3y+24=0.$$

Hence, the correct answer is Option B.

We have the straight line $$3x + 4y - 24 = 0$$. To find where it cuts the $$x$$-axis, we put $$y = 0$$.

Substituting $$y = 0$$ gives $$3x + 4(0) - 24 = 0 \;\Rightarrow\; 3x - 24 = 0 \;\Rightarrow\; x = \dfrac{24}{3} = 8$$.

So the intersection with the $$x$$-axis is the point $$A(8,\,0)$$.

Next, to find where the same line meets the $$y$$-axis, we set $$x = 0$$.

Putting $$x = 0$$ yields $$3(0) + 4y - 24 = 0 \;\Rightarrow\; 4y - 24 = 0 \;\Rightarrow\; y = \dfrac{24}{4} = 6$$.

Hence the intersection with the $$y$$-axis is the point $$B(0,\,6)$$.

The origin is $$O(0,\,0)$$, so the three vertices of the triangle $$OAB$$ are now fixed: $$O(0,0),\; A(8,0),\; B(0,6)$$.

Let us calculate the three side lengths. Using the distance formula we obtain

$$OA = \sqrt{(8-0)^2 + (0-0)^2} = \sqrt{64} = 8,$$

$$OB = \sqrt{(0-0)^2 + (6-0)^2} = \sqrt{36} = 6,$$

$$AB = \sqrt{(8-0)^2 + (0-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.$$

Because $$OA$$ lies along the $$x$$-axis and $$OB$$ lies along the $$y$$-axis, they are perpendicular; so the triangle is right-angled at $$O$$.

In a right triangle with the right angle at the origin and legs on the coordinate axes, the angle bisector of the right angle is the line $$y = x$$. Therefore the incentre must have equal $$x$$ and $$y$$ coordinates. Let us denote the incentre by $$I(r,\,r)$$; the value $$r$$ will in fact be the inradius.

To obtain $$r$$, we first recall the formulas:

1. Semiperimeter: $$s = \dfrac{a + b + c}{2},$$ where $$a, b, c$$ are the side lengths.

2. Relation between area, semiperimeter and inradius: $$\Delta = r\,s.$$

The side lengths we just found are $$a = 8,\; b = 6,\; c = 10$$. Hence

$$s = \dfrac{8 + 6 + 10}{2} = \dfrac{24}{2} = 12.$$

The area of a right triangle is $$\dfrac12 \times (\text{product of the perpendicular sides})$$, so

$$\Delta = \dfrac12 \times 8 \times 6 = 24.$$

Using $$\Delta = r\,s$$, we get

$$24 = r \times 12 \;\Longrightarrow\; r = \dfrac{24}{12} = 2.$$

Thus the incentre is $$I(2,\,2)$$.

Hence, the correct answer is Option D.

We are given the two straight-line equations $$x + (a-1)y = 1$$ and $$2x + a^{2}y = 1$$, where the real number $$a$$ is restricted by $$a \neq 0,\,1$$. Because the lines are stated to be perpendicular, we first express each line in slope-intercept form so that we can compare their slopes.

Starting with the first line, we isolate $$y$$:

$$x + (a-1)y = 1$$

$$\Rightarrow (a-1)y = 1 - x$$

$$\Rightarrow y = \dfrac{1 - x}{\,a-1\,}$$

Hence the slope of the first line is

$$m_{1} = \dfrac{-1}{\,a-1\,}.$$

Now we treat the second line in the same way:

$$2x + a^{2}y = 1$$

$$\Rightarrow a^{2}y = 1 - 2x$$

$$\Rightarrow y = \dfrac{1 - 2x}{\,a^{2}}$$

So the slope of the second line is

$$m_{2} = \dfrac{-2}{\,a^{2}}.$$

For two lines to be perpendicular, the product of their slopes must be $$-1$$. Therefore we impose

$$m_{1}\,m_{2} = -1.$$

Substituting the expressions found above, we get

$$\left(\dfrac{-1}{\,a-1\,}\right)\!\left(\dfrac{-2}{\,a^{2}}\right) = -1.$$

Because the two minus signs give a plus sign, the left-hand side becomes

$$\dfrac{2}{(a-1)a^{2}} = -1.$$

Now we cross-multiply to clear the denominator:

$$2 = -(a-1)a^{2}.$$

Removing the minus sign by multiplying both sides by $$-1$$ yields

$$(a-1)a^{2} = -2.$$

Expanding the left-hand side gives

$$a^{3} - a^{2} + 2 = 0.$$

To solve this cubic, we try simple integral values. Substituting $$a = -1$$ gives

$$(\,-1)^{3} - (\,-1)^{2} + 2 = -1 - 1 + 2 = 0,$$

so $$a = -1$$ is indeed a root. Hence we can factor the cubic as

$$(a + 1)\bigl(a^{2} - 2a + 2\bigr) = 0.$$

The quadratic factor $$a^{2} - 2a + 2 = 0$$ has discriminant $$\Delta = (-2)^{2} - 4\cdot1\cdot2 = 4 - 8 = -4 < 0$$, giving no real roots. Therefore the only real value that satisfies the perpendicularity condition is

$$a = -1.$$

This value also respects the given restriction $$a \neq 0,\,1$$, so we accept it.

With $$a = -1$$, the two original lines become

$$x + (-1 - 1)y = 1 \;\Longrightarrow\; x - 2y = 1,$$

$$2x + (-1)^{2}y = 1 \;\Longrightarrow\; 2x + y = 1.$$

We now find their point of intersection by solving these two equations simultaneously. From the second equation we express $$y$$ in terms of $$x$$:

$$2x + y = 1 \;\Longrightarrow\; y = 1 - 2x.$$

Substituting this value of $$y$$ into the first equation gives

$$x - 2(1 - 2x) = 1.$$

On expanding we get

$$x - 2 + 4x = 1,$$

$$5x - 2 = 1,$$

$$5x = 3,$$

$$x = \dfrac{3}{5}.$$

Now we find $$y$$:

$$y = 1 - 2x = 1 - 2\left(\dfrac{3}{5}\right) = 1 - \dfrac{6}{5} = -\,\dfrac{1}{5}.$$

Thus the point of intersection is $$\left(\dfrac{3}{5},\, -\dfrac{1}{5}\right).$$

Finally, we calculate its distance from the origin. The distance formula from the point $$(x,y)$$ to the origin $$(0,0)$$ is

$$d = \sqrt{x^{2} + y^{2}}.$$

Substituting $$x = \dfrac{3}{5}$$ and $$y = -\dfrac{1}{5}$$ gives

$$d = \sqrt{\left(\dfrac{3}{5}\right)^{2} + \left(-\dfrac{1}{5}\right)^{2}} = \sqrt{\dfrac{9}{25} + \dfrac{1}{25}} = \sqrt{\dfrac{10}{25}} = \sqrt{\dfrac{2}{5}}.$$

The numerical value $$\sqrt{\dfrac{2}{5}}$$ matches Option D in the list provided.

Hence, the correct answer is Option D.

We have the given line $$4x - 3y + 2 = 0$$. Any line that is parallel to this line must have the same coefficients of $$x$$ and $$y$$; therefore its general equation can be written as

$$4x - 3y + c = 0,$$

where $$c$$ is a constant that we are free to choose.

The problem states that such a parallel line must be at a distance $$\dfrac{3}{5}$$ units from the origin $$O(0,0)$$. We recall the distance formula from a point $$(x_0,y_0)$$ to a line $$ax + by + c = 0$$:

$$\text{Distance} = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$$

Here the point is the origin, so $$x_0 = 0$$ and $$y_0 = 0$$. Substituting these, the distance from the origin to the line $$4x - 3y + c = 0$$ is

$$\frac{|4(0) - 3(0) + c|}{\sqrt{4^2 + (-3)^2}} \;=\; \frac{|c|}{\sqrt{16 + 9}} \;=\; \frac{|c|}{\sqrt{25}} \;=\; \frac{|c|}{5}.$$

The required distance is $$\dfrac{3}{5}$$, so

$$\frac{|c|}{5} = \frac{3}{5} \quad\Longrightarrow\quad |c| = 3.$$

Hence $$c = 3$$ or $$c = -3$$. Therefore the only two lines that satisfy the condition are

$$4x - 3y + 3 = 0 \quad\text{and}\quad 4x - 3y - 3 = 0.$$

Now we check each option to see whether the point lies on either of these two lines.

Option A: Point $$\left(\dfrac{1}{4}, -\dfrac{1}{3}\right).$$
We compute $$4x - 3y$$:

$$4\!\left(\frac{1}{4}\right) - 3\!\left(-\frac{1}{3}\right) = 1 + 1 = 2.$$

Add $$c = 3$$: $$2 + 3 = 5 \neq 0.$$
Add $$c = -3$$: $$2 - 3 = -1 \neq 0.$$
So the point is on neither line.

Option B: Point $$\left(-\dfrac{1}{4}, \dfrac{2}{3}\right).$$
We compute $$4x - 3y$$:

$$4\!\left(-\frac{1}{4}\right) - 3\!\left(\frac{2}{3}\right) = -1 - 2 = -3.$$

Add $$c = 3$$: $$-3 + 3 = 0.$$
Thus this point satisfies $$4x - 3y + 3 = 0$$ and therefore lies on the required line.

Option C: Point $$\left(-\dfrac{1}{4}, -\dfrac{2}{3}\right).$$
Compute $$4x - 3y$$:

$$4\!\left(-\frac{1}{4}\right) - 3\!\left(-\frac{2}{3}\right) = -1 + 2 = 1.$$

Add $$c = 3$$: $$1 + 3 = 4 \neq 0.$$
Add $$c = -3$$: $$1 - 3 = -2 \neq 0.$$
So this point is not on either line.

Option D: Point $$\left(\dfrac{1}{4}, \dfrac{1}{3}\right).$$
Compute $$4x - 3y$$:

$$4\!\left(\frac{1}{4}\right) - 3\!\left(\frac{1}{3}\right) = 1 - 1 = 0.$$

Add $$c = 3$$: $$0 + 3 = 3 \neq 0.$$
Add $$c = -3$$: $$0 - 3 = -3 \neq 0.$$
So this point also fails to lie on the required lines.

Only Option B satisfies the equation of one of the lines at the specified distance.

Hence, the correct answer is Option B.

Let the required line pass through the fixed point $$P(2,3)$$ and have slope $$m$$. The two-point form of the equation of a line tells us that a line through $$P(x_1,y_1)$$ with slope $$m$$ is written as $$y-y_1=m(x-x_1)$$. Using $$x_1=2,\;y_1=3$$ we therefore write

$$y-3=m(x-2).$$

This same line meets the given line $$x+y=7$$ at some point $$Q(x,y)$$. At the point of intersection both equations are satisfied, so we equate the expressions for $$y$$.

From the first line we have $$y=3+m(x-2)$$, while from the second we have $$y=7-x$$. Setting them equal,

$$3+m(x-2)=7-x.$$

Expanding and rearranging,

$$m(x-2)+3=7-x$$

$$m(x-2)=4-x$$

$$mx-2m+x=4$$

$$(m+1)\,x=4+2m.$$

So, provided $$m\neq-1$$, the abscissa of $$Q$$ is

$$x=\frac{4+2m}{m+1}.$$

The ordinate is obtained from $$y=7-x$$:

$$y=7-\frac{4+2m}{m+1}.$$

We now compute the vector $$\overrightarrow{PQ}=(x-2,\;y-3)$$, because the distance $$PQ$$ must be $$4$$ units. Substituting the expressions just found:

$$x-2=\frac{4+2m}{m+1}-2 =\frac{4+2m-2(m+1)}{m+1} =\frac{4+2m-2m-2}{m+1} =\frac{2}{m+1},$$

and

$$y-3=7-\frac{4+2m}{m+1}-3 =\frac{7(m+1)-(4+2m)-3(m+1)}{m+1} =\frac{7m+7-4-2m-3m-3}{m+1} =\frac{2m}{m+1}.$$

Thus

$$\overrightarrow{PQ}=\left(\frac{2}{m+1},\;\frac{2m}{m+1}\right).$$

The distance formula states $$PQ=\sqrt{(x-2)^2+(y-3)^2}$$, so

$$PQ^2=\left(\frac{2}{m+1}\right)^2+\left(\frac{2m}{m+1}\right)^2 =\frac{4(1+m^2)}{(m+1)^2}.$$

This distance must equal $$4$$, so its square must equal $$16$$:

$$\frac{4(1+m^2)}{(m+1)^2}=16.$$

Dividing both sides by $$4$$,

$$\frac{1+m^2}{(m+1)^2}=4.$$

Cross-multiplying,

$$1+m^2=4(m+1)^2 =4(m^2+2m+1) =4m^2+8m+4.$$

Bringing all terms to one side,

$$0=4m^2+8m+4-(1+m^2) =3m^2+8m+3.$$

Hence the required slope(s) satisfy the quadratic equation

$$3m^2+8m+3=0.$$

Using the quadratic formula $$m=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with $$a=3,\;b=8,\;c=3$$, we obtain

$$m=\frac{-8\pm\sqrt{8^2-4\cdot3\cdot3}}{2\cdot3} =\frac{-8\pm\sqrt{64-36}}{6} =\frac{-8\pm\sqrt{28}}{6} =\frac{-8\pm2\sqrt7}{6} =\frac{-4\pm\sqrt7}{3}.$$

This gives two numerical values. To match them with the options we simplify one of them:

$$m_1=\frac{-4+\sqrt7}{3}=-\frac{4-\sqrt7}{3},$$

$$m_2=\frac{-4-\sqrt7}{3}.$$

Now compare $$m_1$$ with the alternatives. Multiplying the numerator and denominator of Option B by $$(1-\sqrt7)$$ we see

$$\frac{1-\sqrt7}{1+\sqrt7} =-\frac{\sqrt7-1}{1+\sqrt7} =-\frac{(\sqrt7-1)(1-\sqrt7)}{(1+\sqrt7)(1-\sqrt7)} =-\frac{(\sqrt7-1)(1-\sqrt7)}{1-7} =-\frac{(\sqrt7-1)(1-\sqrt7)}{-6} =\frac{(\sqrt7-1)(\sqrt7-1)}{6} =\frac{7-2\sqrt7+1}{6} =\frac{8-2\sqrt7}{6} =\frac{-4+\sqrt7}{3}=m_1.$$

Thus $$m_1$$ is exactly the value given in Option 2, while $$m_2$$ does not correspond to any listed option. Therefore the slope consistent with the problem statement is

$$m=\frac{1-\sqrt7}{1+\sqrt7}.$$

Hence, the correct answer is Option 2.

Let the moving point be $$P(x,\,y)$$. According to the question, $$P$$ always satisfies the straight-line equation $$2x - 3y + 4 = 0$$.

We also have two fixed points: $$Q(1,\,4)$$ and $$R(3,\,-2)$$.

For any triangle with vertices $$P(x,\,y)$$, $$Q(1,\,4)$$ and $$R(3,\,-2)$$, the centroid formula states:

$$\bigl(X,\,Y\bigr)\;=\;\Bigl(\dfrac{x + 1 + 3}{3},\;\dfrac{y + 4 - 2}{3}\Bigr).$$

So the centroid $$G$$ of $$\triangle PQR$$ has coordinates

$$X \;=\;\dfrac{x + 4}{3}, \qquad Y \;=\;\dfrac{y + 2}{3}.$$

Next, we use the condition that $$P(x,\,y)$$ lies on the line $$2x - 3y + 4 = 0$$. Rearranging that equation, we obtain

$$2x - 3y + 4 = 0 \;\;\Longrightarrow\;\; 2x = 3y - 4 \;\;\Longrightarrow\;\; x = \dfrac{3y - 4}{2}.$$

We substitute this expression for $$x$$ into the formula for $$X$$:

$$X \;=\;\dfrac{\dfrac{3y - 4}{2} + 4}{3}.$$

Combining the terms in the numerator,

$$\dfrac{3y - 4}{2} + 4 \;=\;\dfrac{3y - 4 + 8}{2} \;=\;\dfrac{3y + 4}{2}.$$

Thus

$$X \;=\;\dfrac{\dfrac{3y + 4}{2}}{3} \;=\;\dfrac{3y + 4}{6}.$$

From the expression for $$Y$$ we have

$$Y \;=\;\dfrac{y + 2}{3} \;\;\Longrightarrow\;\; y \;=\;3Y - 2.$$

We now eliminate $$y$$ by replacing it in the equation for $$X$$:

$$X \;=\;\dfrac{3(3Y - 2) + 4}{6}.$$

Expanding the numerator gives

$$3(3Y - 2) + 4 \;=\;9Y - 6 + 4 \;=\;9Y - 2,$$

so

$$X \;=\;\dfrac{9Y - 2}{6}.$$

To obtain the relation between $$X$$ and $$Y$$, we cross-multiply:

$$6X \;=\;9Y - 2.$$

Re-arranging,

$$9Y \;=\;6X + 2 \;\;\Longrightarrow\;\; Y = \dfrac{6}{9}X + \dfrac{2}{9}.$$

Simplifying the coefficient,

$$Y \;=\;\dfrac{2}{3}X + \dfrac{2}{9}.$$

This is the equation of a straight line whose slope is $$\dfrac{2}{3}$$. Therefore, the locus of the centroid $$G$$ is a line with slope $$\frac{2}{3}$$.

Hence, the correct answer is Option A.

Let the required line L cut the positive x-axis at  $$A(a,0)$$ and the positive y-axis at $$B(0,b)$$, where of course $$a \gt 0$$ and $$b \gt 0$$. In intercept form we therefore write

$$\frac{x}{a}\;+\;\frac{y}{b}=1\qquad \qquad(1)$$

Re-writing (1) in the general form $$Px+Qy+R=0$$ we obtain

$$bx+ay-ab=0\;.\qquad \qquad(2)$$

Formula for distance of a point $$(x_0,y_0)$$ from the line $$Px+Qy+R=0$$ is

$$\text{Distance}=\frac{\left|P\,x_0+Q\,y_0+R\right|}{\sqrt{P^{2}+Q^{2}}}\;.$$

The origin $$(0,0)$$ has to be 4 units away from L, hence using (2)

$$\frac{\left|-ab\right|}{\sqrt{b^{2}+a^{2}}}=4 \;\Longrightarrow\; \frac{ab}{\sqrt{a^{2}+b^{2}}}=4\;.$$

Squaring both sides gives

$$\frac{a^{2}b^{2}}{a^{2}+b^{2}}=16 \;\Longrightarrow\; \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{16}\;.\qquad \qquad(3)$$

Next we use the information about the angle. The perpendicular drawn from the origin to L is along the normal vector of L. From (2) that normal vector is $$\mathbf{n}=(b,a)\,.$$

The given line $$x+y=0$$ has direction vector $$\mathbf{v}=(1,-1)$$ (because its normal is $$(1,1)$$). The angle between the vectors $$\mathbf{n}$$ and $$\mathbf{v}$$ is 60°, so by the dot-product formula

$$\cos 60^{\circ}=\frac{|\mathbf{n}\cdot\mathbf{v}|}{|\mathbf{n}|\,|\mathbf{v}|} =\frac{|\,b-a\,|}{\sqrt{a^{2}+b^{2}}\;\sqrt{1^{2}+(-1)^{2}}} =\frac{|\,b-a\,|}{\sqrt{a^{2}+b^{2}}\;\sqrt2}\;.$$

Because $$\cos 60^{\circ}= \dfrac12$$, we have

$$|\,b-a\,|=\frac{\sqrt2}{2}\,\sqrt{a^{2}+b^{2}}\;.$$

Squaring and simplifying every term carefully,

$$\bigl(b-a\bigr)^{2}=\frac12\bigl(a^{2}+b^{2}\bigr) \;\Longrightarrow\; b^{2}-2ab+a^{2}=\frac12a^{2}+\frac12b^{2}$$

$$\Longrightarrow\;2b^{2}-4ab+2a^{2}=a^{2}+b^{2} \;\Longrightarrow\; a^{2}-4ab+b^{2}=0\;.$$

Dividing through by $$a^{2}$$ (which is positive) gives

$$\left(\frac{b}{a}\right)^{2}-4\left(\frac{b}{a}\right)+1=0\;.$$

Let $$k=\dfrac{b}{a}$$. Solving the quadratic $$k^{2}-4k+1=0$$,

$$k=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm2\sqrt3}{2}=2\pm\sqrt3\;.$$

Both values are positive, but we shall see that $$k=2+\sqrt3$$ will match the options. So we fix

$$b=k\,a,\quad\text{where }k=2+\sqrt3\;.\qquad \qquad(4)

Now substitute (4) into the distance relation (3):

$$\frac{1}{a^{2}}+\frac{1}{k^{2}a^{2}}=\frac{1}{16} \;\Longrightarrow\; \frac{1+\frac{1}{k^{2}}}{a^{2}}=\frac{1}{16} \;\Longrightarrow\; a^{2}=16\!\left(1+\frac{1}{k^{2}}\right).$$

First compute $$k^{2}=(2+\sqrt{3})^{2}=4+4\sqrt{3}+3=7+4\sqrt{3}\;.$$ Because $$7^{2}-(4\sqrt{3})^{2}=49-48=1,$$ we may rationalise neatly:

$$\frac{1}{k^{2}}=\frac{1}{7+4\sqrt{3}}=\frac{7-4\sqrt{3}}{(7)^{2}-(4\sqrt{3})^{2}}=7-4\sqrt{3}\;.$$

Hence

$$1+\frac{1}{k^{2}}=1+7-4\sqrt{3}=8-4\sqrt{3}\;,$$

so that

$$a^{2}=16\bigl(8-4\sqrt{3}\bigr),\qquad a=4\sqrt{8-4\sqrt{3}}\;.$$

From (4) we also have

$$b=k\,a=(2+\sqrt{3})\cdot 4\sqrt{8-4\sqrt{3}}\;.$$

Go back to the intercept form (1):

$$\frac{x}{a}+\frac{y}{b}=1 \;\Longrightarrow\; bx+ay=ab\;.$$

To obtain simpler numerical coefficients we multiply the whole equation by $$\lambda=\sqrt{3}-1$$. Observe that

$$\lambda\,b=(\sqrt{3}-1)b =(\sqrt{3}-1)(2+\sqrt{3})a =(\sqrt{3}+1)a$$

because $$(\sqrt{3}-1)(2+\sqrt{3})=\sqrt{3}+1.$$

Therefore

$$(\sqrt{3}+1)x+(\sqrt{3}-1)y=(\sqrt{3}+1)a\;.$$

It only remains to show that $$(\sqrt{3}+1)a=8\sqrt{2}.$$ Indeed, using $$a=4\sqrt{8-4\sqrt{3}},$$

$$(\sqrt{3}+1)a =(\sqrt{3}+1)\cdot 4\sqrt{8-4\sqrt{3}}.$$

Square both sides to verify it equals $$(8\sqrt{2})^{2}=128.$$ Left-hand side squared is

$$16\,(\sqrt{3}+1)^{2}\,(8-4\sqrt{3}) =16\,(4+2\sqrt{3})\,(8-4\sqrt{3})$$

$$=16\cdot 2(2+\sqrt{3})\cdot 4(2-\sqrt{3}) =16\cdot 8\,(2+\sqrt{3})(2-\sqrt{3}) =128\,(4-3)=128.$$

Since the squares match, the quantities themselves match in magnitude, and both sides are positive, so indeed

$$(\sqrt{3}+1)a = 8\sqrt{2}.$$

Thus the equation of L is

$$(\sqrt{3}+1)\,x + (\sqrt{3}-1)\,y = 8\sqrt{2}.$$

This coincides exactly with Option A.

Hence, the correct answer is Option A.

We have the first straight line $$2x - 3y + 17 = 0$$. In order to read its slope easily, we bring it to the slope-intercept form $$y = mx + c$$.

Starting with $$2x - 3y + 17 = 0,$$ we isolate the $$y$$-term:

$$-3y = -2x - 17.$$

Dividing every term by $$-3$$ gives

$$y = \frac{2}{3}x + \frac{17}{3}.$$

So the slope of this line is $$m_1 = \frac{2}{3}.$$

Now, two lines are perpendicular when the product of their slopes equals $$-1$$. Hence, if the slope of the required second line is called $$m_2$$, then

$$m_1 \, m_2 = -1 \quad\Longrightarrow\quad \frac{2}{3}\, m_2 = -1.$$

Solving for $$m_2$$:

$$m_2 = -\frac{3}{2}.$$

Next, the second line passes through the two given points $$(7,\,17)$$ and $$(15,\,\beta)$$. By the two-point slope formula, the slope of this line is

$$m_2 = \frac{\beta - 17}{15 - 7}.$$

Simplifying the denominator:

$$m_2 = \frac{\beta - 17}{8}.$$

We have already found that $$m_2 = -\dfrac{3}{2}$$, so we equate the two expressions for the same slope:

$$\frac{\beta - 17}{8} = -\frac{3}{2}.$$

To clear the fractions, we multiply both sides by $$8$$:

$$\beta - 17 = 8 \left(-\frac{3}{2}\right).$$

Calculating the right-hand side:

$$\beta - 17 = -12.$$

Finally, adding $$17$$ to both sides yields

$$\beta = -12 + 17 = 5.$$

Therefore, $$\beta$$ = 5.$$

Hence, the correct answer is Option 3.

Let the point on the x-axis be $$P(a,0)$$ and the point on the y-axis be $$Q(0,b)$$, where $$a$$ and $$b$$ are non-zero integers (they may be positive or negative).

Because $$OP$$ lies on the x-axis and $$OQ$$ lies on the y-axis, the angle at the origin is a right angle. For a right-angled triangle whose perpendicular sides have lengths $$|a|$$ and $$|b|$$, we first state the familiar area formula:

$$\text{Area}=\dfrac12 \times (\text{base}) \times (\text{height}).$$

Using $$|a|$$ as the base and $$|b|$$ as the height, we have

$$\text{Area}=\dfrac12\,|a|\,|b|.$$

The question tells us this area equals $$50$$, so

$$\dfrac12\,|a|\,|b| = 50 \quad\Longrightarrow\quad |a|\,|b| = 100.$$

Now we must find all ordered pairs of positive integers $$(|a|,|b|)$$ whose product is $$100$$. We factorise $$100$$:

$$100 = 2^2 \cdot 5^2.$$

For a number $$n=p_1^{\alpha_1}\,p_2^{\alpha_2}\dots$$, the total number of positive divisors is $$(\alpha_1+1)(\alpha_2+1)\dots$$. Therefore $$100$$ has $$(2+1)(2+1)=9$$ positive divisors.

Choosing any positive divisor $$d$$ as $$|a|$$ determines $$|b|$$ uniquely as $$\dfrac{100}{d}$$, so there are exactly $$9$$ ordered pairs $$(|a|,|b|)$$ satisfying $$|a||b|=100$$.

Next we remember that $$a$$ and $$b$$ themselves can be either positive or negative independently. For each absolute-value pair we thus have the four possibilities

$$(|a|,|b|)\;\rightarrow\;(+|a|,+|b|),\;(+|a|,-|b|),\;(-|a|,+|b|),\;(-|a|,-|b|).$$

These four sign choices place the points $$P$$ and $$Q$$ on the positive or negative parts of their respective axes, giving four distinct triangles.

Therefore, the total number of triangles is

$$9 \times 4 = 36.$$

Hence, the correct answer is Option A.

We have two sides of the triangle given by the straight-line equations $$3x-2y+6=0 \quad\text{and}\quad 4x+5y-20=0.$$ Their point of intersection will be one vertex of the triangle. Let us find that vertex first.

Solving the simultaneous equations $$\begin{aligned} 3x-2y+6&=0,\\ 4x+5y-20&=0, \end{aligned}$$ we get $$\begin{aligned} 3x-2y&=-6,\\ 4x+5y&=20. \end{aligned}$$ Multiply the first by $$5$$ and the second by $$2$$: $$\begin{aligned} 15x-10y&=-30,\\ 8x+10y&=40. \end{aligned}$$ Adding, $$23x=10 \;\Longrightarrow\; x=\dfrac{10}{23}.$$ Putting this in $$3x-2y=-6$$, $$3\!\left(\dfrac{10}{23}\right)-2y=-6 \;\Longrightarrow\; \dfrac{30}{23}-2y=-6 \;\Longrightarrow\; -2y=-6-\dfrac{30}{23} =-\dfrac{168}{23} \;\Longrightarrow\; y=\dfrac{84}{23}.$$ Hence the vertex at the intersection of the two given sides is $$C\!\left(\dfrac{10}{23},\,\dfrac{84}{23}\right).$$

The orthocenter is given as $$H(1,1).$$ Recall that “the orthocenter is the common point of the three altitudes of a triangle.” Therefore, the altitude from the unknown vertex $$B$$ (lying on the first side) is perpendicular to the second side, and the altitude from the unknown vertex $$A$$ (lying on the second side) is perpendicular to the first side.

Altitude from $$B$$ to the side $$4x+5y-20=0$$ For the line $$4x+5y-20=0$$ the normal (perpendicular) vector is $$(4,5)$$. Hence a line that is perpendicular to this side must have the same direction vector $$(4,5)$$ and therefore slope $$m=\dfrac{5}{4}.$$ Because the altitude passes through the orthocenter $$H(1,1)$$ and the vertex $$B(x_b,y_b)$$, the slope condition gives $$\dfrac{y_b-1}{x_b-1}=\dfrac{5}{4}\;\Longrightarrow\;4(y_b-1)=5(x_b-1) \;\Longrightarrow\;5x_b-4y_b=1. \quad -(1)$$ At the same time, $$B$$ lies on the first side $$3x-2y+6=0$$: $$3x_b-2y_b+6=0 \;\Longrightarrow\; 3x_b-2y_b=-6. \quad -(2)$$ Solving (1) and (2): from (2) $$x_b=\dfrac{-6+2y_b}{3},$$ substitute into (1): $$5\!\left(\dfrac{-6+2y_b}{3}\right)-4y_b=1 \;\Longrightarrow\; \dfrac{-30+10y_b}{3}-4y_b=1 \;\Longrightarrow\; -30+10y_b-12y_b=3 \;\Longrightarrow\; -2y_b=33 \;\Longrightarrow\; y_b=-\dfrac{33}{2},$$ and hence $$x_b=\dfrac{-6+2\!\left(-\dfrac{33}{2}\right)}{3}=\dfrac{-6-33}{3}=-13.$$ Thus $$B(-13,-33/2).$$

Altitude from $$A$$ to the side $$3x-2y+6=0$$ For the line $$3x-2y+6=0$$ the normal vector is $$(3,-2)$$, so a perpendicular line has that direction and hence slope $$m=\dfrac{-2}{3}.$$ Because the altitude passes through $$H(1,1)$$ and $$A(x_a,y_a)$$, $$\dfrac{y_a-1}{x_a-1}=-\dfrac{2}{3} \;\Longrightarrow\;3(y_a-1)=-2(x_a-1) \;\Longrightarrow\;2x_a+3y_a=5. \quad -(3)$$ Also $$A$$ lies on the second side $$4x+5y-20=0$$: $$4x_a+5y_a-20=0 \;\Longrightarrow\;4x_a+5y_a=20. \quad -(4)$$ Solving (3) and (4): multiply (3) by $$2$$: $$4x_a+6y_a=10$$. Subtract this from (4): $$(4x_a+5y_a)-(4x_a+6y_a)=20-10 \;\Longrightarrow\;-y_a=10 \;\Longrightarrow\; y_a=-10,$$ then from (3) $$2x_a+3(-10)=5 \;\Longrightarrow\; 2x_a=35 \;\Longrightarrow\; x_a=\dfrac{35}{2}.$$ So $$A\!\left(\dfrac{35}{2},-10\right).$$

The third side is the straight line through $$A$$ and $$B$$. Using the two-point form, for points $$\bigl(x_1,y_1\bigr)=A$$ and $$\bigl(x_2,y_2\bigr)=B$$, $$(y-y_1)(x_2-x_1)=(y_2-y_1)(x-x_1).$$ Here $$x_2-x_1=-13-\dfrac{35}{2}=-\dfrac{61}{2},\qquad y_2-y_1=-\dfrac{33}{2}-(-10)=-\dfrac{13}{2}.$$ Substituting, $$(y+10)\!\left(-\dfrac{61}{2}\right)=\!\left(-\dfrac{13}{2}\right)\!\left(x-\dfrac{35}{2}\right).$$ Multiply by $$2$$ to clear the denominators: $$-61(y+10)=-13\!\left(x-\dfrac{35}{2}\right).$$ Expand both sides: $$-61y-610=-13x+\dfrac{455}{2}.$$ Multiply by $$2$$ once more to eliminate the remaining fraction: $$-122y-1220=-26x+455.$$ Bring every term to the left: $$26x-122y-1675=0.$$

This matches option B in the list. Hence, the correct answer is Option 2.

Let the feet of the two vertical poles be denoted by $$A$$ and $$B$$. The pole at $$A$$ is $$20\text{ m}$$ high, while the pole at $$B$$ is $$80\text{ m}$$ high. Assume the (unknown) horizontal distance between the two feet to be $$L$$ metres. Because everything lies in one vertical plane, we introduce a convenient coordinate system:

• Choose the foot $$A$$ as the origin, giving $$A\equiv(0,0)$$, where the first coordinate is horizontal and the second is vertical (height). • Then the foot $$B$$ has coordinates $$B\equiv(L,0)$$. • The top of the shorter pole is $$A'\equiv(0,20)$$, and the top of the taller pole is $$B'\equiv(L,80)$$.

We draw two straight segments: 1. $$A'B$$ joins the top of the shorter pole to the foot of the taller pole. 2. $$B'A$$ joins the top of the taller pole to the foot of the shorter pole.

Let the two lines intersect at $$P(x_P,h)$$, where $$h$$ is the required height above the horizontal plane.

Parametric form of each line.

Along $$A'B$$, take a parameter $$\lambda\;(0\le\lambda\le1)$$ measured from $$A'$$ to $$B$$. Using the two-point formula, a general point on $$A'B$$ is

$$\begin{aligned} x &= 0 + \lambda(L-0) = \lambda L,\\[4pt] z &= 20 + \lambda(0-20) = 20\,(1-\lambda). \end{aligned}$$

Along $$B'A$$, take a parameter $$\mu\;(0\le\mu\le1)$$ measured from $$B'$$ to $$A$$. A general point on $$B'A$$ is

$$\begin{aligned} x &= L + \mu(0-L) = L\,(1-\mu),\\[4pt] z &= 80 + \mu(0-80) = 80\,(1-\mu). \end{aligned}$$

Equating the coordinates at the intersection $$P$$.

The same point $$P$$ must satisfy both sets of equations, so first equate the horizontal coordinates:

$$\lambda L = L(1-\mu)\;\Longrightarrow\;\lambda = 1-\mu.$$

Next, equate the vertical (height) coordinates:

$$20(1-\lambda) = 80(1-\mu).$$

Substituting $$\lambda = 1-\mu$$ into this equation, we obtain

$$20\bigl[1-(1-\mu)\bigr] = 80(1-\mu) \;\Longrightarrow\; 20\mu = 80(1-\mu).$$

Now solve step by step:

$$\begin{aligned} 20\mu &= 80 - 80\mu,\\[4pt] 20\mu + 80\mu &= 80,\\[4pt] 100\mu &= 80,\\[4pt] \mu &= \frac{80}{100} = 0.8. \end{aligned}$$

Since $$\lambda = 1 - \mu$$, we have $$\lambda = 1 - 0.8 = 0.2.$$ Height of the point $$P$$. Using either expression for $$z$$, say $$z = 80(1-\mu)$$, we get

$$h = 80(1 - 0.8) = 80 \times 0.2 = 16.$$

Checking with the other formula $$z = 20(1-\lambda)$$ gives the same value:

$$h = 20(1 - 0.2) = 20 \times 0.8 = 16,$$

confirming the consistency.

The point of intersection therefore lies $$16\text{ m}$$ above the horizontal plane.

Hence, the correct answer is Option A.

We are given two simultaneous inequalities $$|x-y|\le 2$$ and $$|x+y|\le 2.$$ To understand the common region, we first convert each absolute-value inequality into two separate linear inequalities.

For $$|x-y|\le 2$$ we write

$$-(2)\le x-y\le 2\; \Longrightarrow\; x-y=-2 \quad\text{and}\quad x-y=2$$

For $$|x+y|\le 2$$ we write

$$-(2)\le x+y\le 2\; \Longrightarrow\; x+y=-2 \quad\text{and}\quad x+y=2$$

Thus the boundary of the region consists of the four straight lines

$$x-y=2,\qquad x-y=-2,\qquad x+y=2,\qquad x+y=-2.$$

Next, we find the points where these lines intersect pairwise. Each intersection of one line from the first pair with one line from the second pair will be a vertex of the enclosed figure.

1. Taking $$x-y=2$$ and $$x+y=2:$$

Adding, we get $$2x=4\;\Longrightarrow\;x=2.$$ Substituting $$x=2$$ into $$x+y=2$$ gives $$y=0.$$ So one vertex is $$(2,0).$$

2. Taking $$x-y=2$$ and $$x+y=-2:$$

Adding, we get $$2x=0\;\Longrightarrow\;x=0.$$ Substituting into $$x-y=2$$ gives $$y=-2.$$ So another vertex is $$(0,-2).$$

3. Taking $$x-y=-2$$ and $$x+y=2:$$

Adding, we get $$2x=0\;\Longrightarrow\;x=0.$$ Substituting into $$x-y=-2$$ gives $$y=2.$$ So the third vertex is $$(0,2).$$

4. Taking $$x-y=-2$$ and $$x+y=-2:$$

Adding, we get $$2x=-4\;\Longrightarrow\;x=-2.$$ Substituting into $$x+y=-2$$ gives $$y=0.$$ So the fourth vertex is $$(-2,0).$$

The four vertices are therefore

$$(2,0),\;(0,2),\;(-2,0),\;(0,-2).$$

Because the vertices are equally spaced and each side joins consecutive vertices, we expect a quadrilateral. To see whether it is a square, we compute one side length using the distance formula.

Formula (distance between two points) $$\bigl(x_1,y_1\bigr),\bigl(x_2,y_2\bigr):\;d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$

Take consecutive vertices $$(2,0)$$ and $$(0,2)$$:

$$d=\sqrt{(0-2)^2+(2-0)^2}=\sqrt{(-2)^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}.$$

Thus each side has length $$2\sqrt{2}.$$ Since all four sides are equal and the adjacent sides are at right angles (you can verify the slopes are negative reciprocals: slope of $$x-y=2$$ is $$1$$, slope of $$x+y=2$$ is $$-1$$), the figure is a square.

Because the side length is $$2\sqrt{2},$$ the area would be $$(2\sqrt{2})^2=8,$$ but the option list asks only for the side length.

Therefore the region is bounded by a square whose side length is $$2\sqrt{2}$$ units.

Hence, the correct answer is Option D.

Let us denote the required straight line by the equation $$L$$. Because this line meets the $$x$$-axis at the point $$P(a,0)$$ and the $$y$$-axis at the point $$Q(0,b)$$, its equation can be written in the intercept form

$$\frac{x}{a}+\frac{y}{b}=1.$$

We are told that the line always passes through the fixed point $$(2,3)$$. Substituting $$x=2,\;y=3$$ in the above equation gives

$$\frac{2}{a}+\frac{3}{b}=1.$$

Next, we complete the rectangle $$OPRQ$$ with $$O(0,0)$$ at the origin, $$P(a,0)$$ on the $$x$$-axis, and $$Q(0,b)$$ on the $$y$$-axis. In such an axis-parallel rectangle, the fourth (opposite) vertex $$R$$ has the coordinates $$R(a,b)$$ because the sides $$OP$$ and $$OQ$$ are respectively horizontal and vertical.

To describe the locus of $$R$$ we simply rename its coordinates:

$$x=a,\qquad y=b.$$

We now translate the earlier condition $$\dfrac{2}{a}+\dfrac{3}{b}=1$$ into $$x$$ and $$y$$. Replacing $$a$$ by $$x$$ and $$b$$ by $$y$$, we have

$$\frac{2}{x}+\frac{3}{y}=1.$$

To clear the denominators we multiply every term by $$xy$$, obtaining

$$2y+3x=xy.$$

Re-arranging the terms to match the standard presentation, we write

$$3x+2y=xy.$$

This equation represents the path traced by the point $$R$$ for every admissible position of the line through $$(2,3)$$ that meets both coordinate axes.

Hence, the correct answer is Option D.

We have a triangle whose vertex $$A$$ is fixed at the point $$(1,\,2)$$. The median drawn from vertex $$B$$ is the straight line whose equation is $$x + y = 5$$, while the median drawn from vertex $$C$$ is the vertical line $$x = 4$$. Using the definition of a median, each of these lines must pass through the corresponding vertex and also through the midpoint of the opposite side. We therefore assign

$$B\,(b_1,\,b_2), \qquad C\,(c_1,\,c_2).$$

First we work with the median through $$B$$. The midpoint of side $$AC$$ is obtained from the midpoint formula:

$$M_1\left(\frac{1 + c_1}{2},\; \frac{2 + c_2}{2}\right).$$

Because both $$B$$ and $$M_1$$ lie on the line $$x + y = 5$$, we substitute their coordinates into that equation one by one.

For the vertex $$B$$ we get

$$b_1 + b_2 = 5 \qquad\qquad (1).$$

For the midpoint $$M_1$$ we have

$$\frac{1 + c_1}{2} + \frac{2 + c_2}{2} = 5.$$

Simplifying the left‐hand side:

$$\frac{(1 + c_1) + (2 + c_2)}{2} = 5 \;\Longrightarrow\; \frac{3 + c_1 + c_2}{2} = 5.$$

Multiplying by two we obtain

$$3 + c_1 + c_2 = 10 \;\Longrightarrow\; c_1 + c_2 = 7 \qquad\qquad (2).$$

Now we use the information about the median through $$C$$. Since that median is the line $$x = 4$$, any point on it has an $$x$$-coordinate equal to $$4$$. Therefore

$$c_1 = 4 \qquad\qquad (3).$$

The same line also contains the midpoint of side $$AB$$. The midpoint of $$AB$$ is

$$M_2\left(\frac{1 + b_1}{2},\; \frac{2 + b_2}{2}\right).$$

Its $$x$$-coordinate must satisfy $$\dfrac{1 + b_1}{2} = 4$$. Multiplying by two gives

$$1 + b_1 = 8 \;\Longrightarrow\; b_1 = 7 \qquad\qquad (4).$$

We substitute $$b_1 = 7$$ into equation (1):

$$7 + b_2 = 5 \;\Longrightarrow\; b_2 = -2.$$

Next we substitute $$c_1 = 4$$ into equation (2):

$$4 + c_2 = 7 \;\Longrightarrow\; c_2 = 3.$$

Thus the three vertices are now completely known:

$$A(1,\,2), \quad B(7,\,-2), \quad C(4,\,3).$$

To find the area of $$\triangle ABC$$ we apply the determinant formula for the area of a triangle formed by points $$(x_1, y_1), (x_2, y_2), (x_3, y_3)$$:

$$\text{Area} \;=\; \frac12 \,\left|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\right|.$$

Assigning $$A(1, 2) \to (x_1, y_1),\; B(7, -2) \to (x_2, y_2),\; C(4, 3) \to (x_3, y_3)$$ we calculate each term carefully:

$$y_2 - y_3 = -2 - 3 = -5,$$

$$x_1(y_2 - y_3) = 1 \times (-5) = -5,$$

$$y_3 - y_1 = 3 - 2 = 1,$$

$$x_2(y_3 - y_1) = 7 \times 1 = 7,$$

$$y_1 - y_2 = 2 - (-2) = 4,$$

$$x_3(y_1 - y_2) = 4 \times 4 = 16.$$

Adding these three products we obtain

$$-5 + 7 + 16 = 18.$$

Taking the absolute value (already positive) and dividing by two gives the required area:

$$\text{Area} = \frac12 \times 18 = 9 \text{ square units}.$$

Hence, the correct answer is Option B.

The given line is $$3x + y = \lambda$$, where $$\lambda \neq 0$$.

The line meets the x-axis ($$y = 0$$) at $$A = \left(\frac{\lambda}{3}, 0\right)$$ and the y-axis ($$x = 0$$) at $$B = (0, \lambda)$$.

The foot of the perpendicular from the origin $$O(0,0)$$ to the line $$3x + y = \lambda$$ is point $$P$$.

The line $$3x + y = \lambda$$ can be written as $$3x + y - \lambda = 0$$. Using the formula for the foot of the perpendicular from $$(x_0, y_0)$$ to $$ax + by + c = 0$$:

$$\frac{x - x_0}{a} = \frac{y - y_0}{b} = -\frac{ax_0 + by_0 + c}{a^2 + b^2}$$

Here $$a = 3$$, $$b = 1$$, $$c = -\lambda$$, and $$(x_0, y_0) = (0, 0)$$:

$$\frac{x}{3} = \frac{y}{1} = -\frac{0 + 0 - \lambda}{9 + 1} = \frac{\lambda}{10}$$

So $$P = \left(\frac{3\lambda}{10}, \frac{\lambda}{10}\right)$$.

Now we find the distances $$BP$$ and $$PA$$.

$$BP^2 = \left(\frac{3\lambda}{10} - 0\right)^2 + \left(\frac{\lambda}{10} - \lambda\right)^2 = \frac{9\lambda^2}{100} + \frac{81\lambda^2}{100} = \frac{90\lambda^2}{100}$$

$$BP = \frac{3\lambda\sqrt{10}}{10}$$

$$PA^2 = \left(\frac{\lambda}{3} - \frac{3\lambda}{10}\right)^2 + \left(0 - \frac{\lambda}{10}\right)^2 = \left(\frac{10\lambda - 9\lambda}{30}\right)^2 + \frac{\lambda^2}{100} = \frac{\lambda^2}{900} + \frac{\lambda^2}{100} = \frac{\lambda^2 + 9\lambda^2}{900} = \frac{10\lambda^2}{900}$$

$$PA = \frac{\lambda\sqrt{10}}{30}$$

Therefore: $$\frac{BP}{PA} = \frac{\frac{3\lambda\sqrt{10}}{10}}{\frac{\lambda\sqrt{10}}{30}} = \frac{3}{10} \times 30 = 9$$

So $$BP : PA = 9 : 1$$.

The answer is Option A: 9 : 1.

Let us denote the three given vertices by

$$A(k,\,-3k),\qquad B(5,\,k),\qquad C(-k,\,2).$$

The area of a triangle with vertices $$(x_1,y_1),\,(x_2,y_2),\,(x_3,y_3)$$ is obtained from the determinant formula

$$\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$$

Substituting $$(x_1,y_1)=(k,-3k),\;(x_2,y_2)=(5,k),\;(x_3,y_3)=(-k,2)$$ we have

$$\begin{aligned} x_1(y_2-y_3)&=k\,(k-2)=k^2-2k,\\[2mm] x_2(y_3-y_1)&=5\bigl(2-(-3k)\bigr)=5(2+3k)=10+15k,\\[2mm] x_3(y_1-y_2)&=-k\bigl((-3k)-k\bigr)=-k(-4k)=4k^2. \end{aligned}$$

Adding these three results gives

$$k^2-2k+10+15k+4k^2=5k^2+13k+10.$$

Because the area is given as 28 square units, we must have

$$\dfrac12\bigl|5k^2+13k+10\bigr|=28\;\;\Longrightarrow\;\;\bigl|5k^2+13k+10\bigr|=56.$$

This absolute-value equation splits into two quadratic equations.

1. $$5k^2+13k+10=56\;\Longrightarrow\;5k^2+13k-46=0.$$

2. $$5k^2+13k+10=-56\;\Longrightarrow\;5k^2+13k+66=0.$$

The second quadratic has discriminant

$$\Delta_2=13^2-4\cdot5\cdot66=169-1320=-1151<0,$$

so it possesses no real roots and can be ignored. We therefore solve the first quadratic.

For $$5k^2+13k-46=0,$$ the discriminant is

$$\Delta_1=13^2-4\cdot5\cdot(-46)=169+920=1089=33^2.$$

Hence, using the quadratic formula

$$k=\dfrac{-13\pm33}{2\cdot5}=\dfrac{-13\pm33}{10}.$$

The two possible values are

$$k=\dfrac{20}{10}=2,\qquad k=\dfrac{-46}{10}=-\dfrac{23}{5}.$$

Because $$k$$ is required to be an integer, only $$k=2$$ is admissible.

Thus the triangle’s concrete vertices are

$$A(2,-6),\qquad B(5,2),\qquad C(-2,2).$$

To locate the orthocenter we intersect two of the altitudes of the triangle.

Altitude from vertex A:

The side $$BC$$ runs from $$B(5,2)$$ to $$C(-2,2).$$ Its slope is

$$m_{BC}=\dfrac{2-2}{5-(-2)}=0,$$

so $$BC$$ is the horizontal line $$y=2.$$ The altitude from $$A$$ is therefore perpendicular to a horizontal line, meaning it is vertical through $$x=x_A=2.$$ Hence the first altitude has equation

$$x=2.$$

Altitude from vertex B:

The side $$AC$$ joins $$A(2,-6)$$ and $$C(-2,2).$$ Its slope is

$$m_{AC}=\dfrac{2-(-6)}{-2-2}=\dfrac{8}{-4}=-2.$$

An altitude is perpendicular to the side, so the required slope is the negative reciprocal

$$m_{\perp B}=\dfrac12.$$

Passing through $$B(5,2)$$, the altitude from $$B$$ is written as

$$y-2=\dfrac12\,(x-5).$$

This rearranges to

$$y=\dfrac12x-\dfrac12.$$

Intersection of the two altitudes:

We simultaneously satisfy

$$x=2\qquad\text{and}\qquad y=\dfrac12x-\dfrac12.$$

Substituting $$x=2$$ into the second equation,

$$y=\dfrac12\cdot2-\dfrac12=1-\dfrac12=\dfrac12.$$

Hence the orthocenter is

$$H\bigl(2,\;\dfrac12\bigr).$$

The coordinates $$\left(2,\;\dfrac12\right)$$ correspond to Option D.

Hence, the correct answer is Option D.

We have a rhombus whose two adjacent sides lie on the straight lines

$$L_1:\;x-y+1=0 \qquad\text{and}\qquad L_2:\;7x-y-5=0.$$

For any straight line written as $$ax+by+c=0,$$ a direction vector is $$(b,\,-a).$$ Hence

$$\text{Direction along }L_1:\;(1,1),\qquad \text{Direction along }L_2:\;(1,7).$$

Let the side of the rhombus that is parallel to $$L_1$$ be represented by the vector

$$\vec v = a(1,1)=(a,a),$$

and the adjacent side parallel to $$L_2$$ be represented by

$$\vec w = b(1,7)=(b,7b).$$

Because every side of a rhombus has the same length, we impose

$$|\vec v| = |\vec w|.$$

First we write the magnitudes explicitly:

$$|\vec v|=\sqrt{a^{2}+a^{2}}=\sqrt{2\,a^{2}}=\sqrt2\,|a|,$$

$$|\vec w|=\sqrt{b^{2}+49\,b^{2}}=\sqrt{50\,b^{2}}=\sqrt{50}\,|b|=5\sqrt2\,|b|.$$

Equating the two gives

$$\sqrt2\,|a| = 5\sqrt2\,|b| \;\Longrightarrow\; |a| = 5|b|.$$

This can be written more simply as

$$a = 5b \quad\text{or}\quad a = -5b.$$

That is, the two side-vectors can point in the same sense ($$a=5b$$) or in opposite senses ($$a=-5b$$).

The diagonals of any parallelogram, and therefore of a rhombus, bisect each other. Hence if the point of intersection of the diagonals is $$O(-1,-2),$$ and if we take a vertex $$A$$ of the rhombus, then

$$\overrightarrow{OA} = -\dfrac{\vec v + \vec w}{2}.$$

We examine the two sign possibilities separately.

Case 1: $$a=5b$$ (both side vectors in the same general direction).

Here

$$\vec v = (5b,5b),\qquad \vec w = (b,7b).$$

Adding them,

$$\vec v+\vec w=(5b+b,\;5b+7b)=(6b,\;12b).$$

Therefore

$$\overrightarrow{OA}= -\dfrac{(6b,\,12b)}{2}=(-3b,\,-6b).$$

This vector has components in the ratio $$1:2.$$ None of the option vectors from $$O(-1,-2)$$ to the listed points is in that ratio, so this case produces no admissible vertex.

Case 2: $$a=-5b$$ (the two side vectors oppose each other).

Now

$$\vec v = (-5b,-5b),\qquad \vec w = (b,7b).$$

Add again:

$$\vec v+\vec w=(-5b+b,\,-5b+7b)=(-4b,\;2b).$$

Hence

$$\overrightarrow{OA}= -\dfrac{(-4b,\,2b)}{2}=(2b,\,-b).$$

This vector is proportional to $$(2,-1).$$ We now look for a choice of $$b$$ that places $$A$$ at one of the given options.

From each option we compute the displacement from $$O(-1,-2)$$.

Option A gives $$\left(\dfrac13,-\dfrac83\right)-(-1,-2)=\left(\dfrac13+1,\;-\dfrac83+2\right)=\left(\dfrac43,\;-\dfrac23\right).$$

This is precisely $$\left( \dfrac43,\;-\dfrac23 \right)=\left(2\cdot\dfrac23,\;-\dfrac23\right)=\bigl(2b,\,-b\bigr)$$ if we take

$$b=\dfrac23.$$

Because a consistent positive value of $$b$$ has been found, Option A indeed represents a vertex. (With $$b=\dfrac23$$ we have $$a=-5b=-\dfrac{10}{3},$$ which still satisfies $$|\vec v|=|\vec w|.$$)

No other option produces a vector proportional to $$(2,-1),$$ so no other choice can be a vertex.

The coordinate we obtained is therefore exactly the point in Option A, namely $$\left(\dfrac13,\,-\dfrac83\right).$$

Hence, the correct answer is Option A.

Let the required straight line passing through the origin be written in the slope-intercept form

$$y = mx,$$

where $$m$$ is an arbitrary real number (the slope). Because this line contains the origin $$O(0,0),$$ every point on it can be written as $$\bigl(t,\;mt\bigr)$$ for some real parameter $$t$$. Our task is to find the points $$A$$ and $$B$$ at which this line meets the two given lines and then compare the distances $$OA$$ and $$OB$$ along the same straight line.

Intersection with the first given line. The first line is

$$3y = 10 - 4x.$$

Substituting $$y = mx$$ from the line through the origin, we get

$$3(mx) \;=\; 10 - 4x.$$

Simplifying,

$$3mx + 4x \;=\; 10,$$

$$x\,(3m + 4) \;=\; 10,$$

$$x_A \;=\; \dfrac{10}{\,3m + 4\,}.$$

Since $$y = mx,$$ the corresponding ordinate is

$$y_A \;=\; m\,x_A \;=\; \dfrac{10m}{\,3m + 4\,}.$$

Thus

$$A\Bigl(\dfrac{10}{3m + 4},\;\dfrac{10m}{3m + 4}\Bigr).$$

Intersection with the second given line. The second line is

$$8x + 6y + 5 = 0.$$

Again putting $$y = mx, $$ we obtain

$$8x + 6(mx) + 5 \;=\; 0,$$

$$x\,(8 + 6m) \;=\; -5,$$

$$x_B \;=\; -\dfrac{5}{\,8 + 6m\,}.$$

Hence

$$y_B \;=\; m\,x_B \;=\; -\dfrac{5m}{\,8 + 6m\,},$$

and

$$B\Bigl(-\dfrac{5}{8 + 6m},\;-\dfrac{5m}{8 + 6m}\Bigr).$$

Expressing the distances $$OA$$ and $$OB$$. Because $$A,\,O,\,B$$ all lie on the same line $$y = mx,$$ the vector from the origin to any point on this line is a scalar multiple of the direction vector $$(1,m).$$ If a point has coordinates $$(t,mt),$$ its distance from the origin is

$$\sqrt{t^{2} + (mt)^{2}} \;=\; |t|\sqrt{\,1 + m^{2}}.$$

Thus the distance from the origin is directly proportional to the absolute value of the $$x$$-coordinate. Therefore,

$$\dfrac{OA}{OB} \;=\; \dfrac{|x_A|}{|x_B|}.$$

Calculating the ratio.

$$\dfrac{OA}{OB} \;=\; \dfrac{\Bigl|\dfrac{10}{\,3m + 4\,}\Bigr|}{\Bigl|-\dfrac{5}{\,8 + 6m\,}\Bigr|}$$

$$\;\;=\; \dfrac{10}{5}\;\cdot\;\dfrac{|8 + 6m|}{|3m + 4|}$$

$$\;\;=\; 2\,\cdot\,\dfrac{|8 + 6m|}{|3m + 4|}.$$

Notice that the two factors $$8 + 6m$$ and $$3m + 4$$ always have the same sign (they are both positive or both negative for every real $$m$$), because

$$8 + 6m \;=\; 2\,(4 + 3m)$$

and

$$3m + 4 \;=\; 3m + 4.$$

The linear expressions differ only by the constant factor 2, so their signs coincide; hence the absolute values cancel the need for the sign, and we simply have

$$|8 + 6m| \;=\; 2\,|3m + 4|.$$

Substituting this into the ratio,

$$\dfrac{OA}{OB} \;=\; 2 \cdot \dfrac{2\,|3m + 4|}{|3m + 4|} \;=\; 2 \cdot 2 \;=\; 4.$$

Thus $$OA : OB = 4 : 1.$$ The origin $$O$$ therefore divides the segment $$AB$$ internally in the ratio $$4 : 1$$.

Hence, the correct answer is Option C.

To solve this problem, we need to find the locus of the midpoint of the segment AB, where AB is part of a variable line passing through the intersection of the given lines $$\frac{x}{3} + \frac{y}{4} = 1$$ and $$\frac{x}{4} + \frac{y}{3} = 1$$, and meeting the coordinate axes at points A and B (with A ≠ B).

First, we find the point of intersection of the two given lines. The equations are:

$$ \frac{x}{3} + \frac{y}{4} = 1 \quad \text{(Equation 1)} $$

$$ \frac{x}{4} + \frac{y}{3} = 1 \quad \text{(Equation 2)} $$

To eliminate fractions, multiply both equations by 12 (the LCM of 3 and 4):

For Equation 1: $$ 12 \times \frac{x}{3} + 12 \times \frac{y}{4} = 12 \times 1 $$ gives $$ 4x + 3y = 12 $$ \quad $$\text{(Equation 1a)}$$

For Equation 2: $$ 12 \times \frac{x}{4} + 12 \times \frac{y}{3} = 12 \times 1 $$ gives $$ 3x + 4y = 12 $$ \quad $$\text{(Equation 2a)}$$

Now solve the system:

$$ 4x + 3y = 12 \quad \text{(1a)} $$

$$ 3x + 4y = 12 \quad \text{(2a)} $$

Multiply Equation 1a by 4 and Equation 2a by 3:

$$ 4 \times (4x + 3y) = 4 \times 12 \implies 16x + 12y = 48 \quad \text{(Equation 1b)} $$

$$ 3 \times (3x + 4y) = 3 \times 12 \implies 9x + 12y = 36 \quad \text{(Equation 2b)} $$

Subtract Equation 2b from Equation 1b:

$$ (16x + 12y) - (9x + 12y) = 48 - 36 \implies 16x + 12y - 9x - 12y = 12 \implies 7x = 12 \implies x = \frac{12}{7} $$

Substitute $$ x = \frac{12}{7} $$ into Equation 1a:

$$ 4 \times \frac{12}{7} + 3y = 12 \implies \frac{48}{7} + 3y = 12 \implies 3y = 12 - \frac{48}{7} = \frac{84}{7} - \frac{48}{7} = \frac{36}{7} \implies y = \frac{36}{7} \times \frac{1}{3} = \frac{12}{7} $$

So the intersection point is $$ \left( \frac{12}{7}, \frac{12}{7} \right) $$.

The variable line passes through this point and meets the coordinate axes at A(a, 0) and B(0, b), where a and b are the x-intercept and y-intercept, respectively. The equation of this line in intercept form is:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

Since the line passes through $$ \left( \frac{12}{7}, \frac{12}{7} \right) $$, substitute these coordinates:

$$ \frac{\frac{12}{7}}{a} + \frac{\frac{12}{7}}{b} = 1 \implies \frac{12}{7a} + \frac{12}{7b} = 1 $$

Factor out $$ \frac{12}{7} $$:

$$ \frac{12}{7} \left( \frac{1}{a} + \frac{1}{b} \right) = 1 \implies \frac{12}{7} \left( \frac{a + b}{ab} \right) = 1 \implies \frac{12(a + b)}{7ab} = 1 \implies 12(a + b) = 7ab \quad \text{(Equation 3)} $$

The midpoint of AB, where A is (a, 0) and B is (0, b), is $$ M \left( \frac{a}{2}, \frac{b}{2} \right) $$. Denote the midpoint as (h, k), so:

$$ h = \frac{a}{2}, \quad k = \frac{b}{2} \implies a = 2h, \quad b = 2k $$

Substitute these into Equation 3:

$$ 12(2h + 2k) = 7 \times (2h) \times (2k) \implies 12 \times 2(h + k) = 7 \times 4hk \implies 24(h + k) = 28hk $$

Divide both sides by 4:

$$ 6(h + k) = 7hk \implies 7hk - 6h - 6k = 0 $$

This equation relates h and k. For the locus, replace h and k with x and y:

$$ 7xy - 6x - 6y = 0 \implies 7xy = 6x + 6y \implies 7xy = 6(x + y) $$

Comparing with the options, this matches option A.

Hence, the correct answer is Option A.

We have the mirror (reflecting surface) given by the straight-line equation $$7x - y + 1 = 0$$. Its normal vector is obtained directly from its coefficients, so we take $$\vec n = (7,\,-1)$$. A direction vector along the mirror (i.e. tangent to the surface) must be perpendicular to this normal. One convenient choice is obtained by swapping the components and changing one sign; hence we select $$\vec t = (1,\,7)$$ because $$\vec t \cdot \vec n = 1\cdot 7 + 7\cdot (-1)=0$$, confirming orthogonality.

The ray after reflection travels along the line $$y + 2x = 1 \; \Longrightarrow \; 2x + y - 1 = 0.$$ Writing it in slope form, $$y = 1 - 2x,$$ the slope is $$-2,$$ so a convenient direction vector for the reflected ray is $$\vec r = (1,\,-2).$$ Both the incident and reflected rays meet the mirror at the point $$P(0,1),$$ so all vectors can be regarded with origin at that point.

Law of reflection (in vector form): decompose the incident vector $$\vec i$$ into two perpendicular components, one parallel to the mirror ($$\vec t$$ direction) and one normal to the mirror ($$\vec n$$ direction). The parallel component remains unchanged on reflection, while the normal component reverses its sign. Symbolically, if

$$\vec i = \alpha\,\vec t + \beta\,\vec n,$$ then after reflection we get $$\vec r = \alpha\,\vec t - \beta\,\vec n.$$

Our task is therefore to (i) express the known reflected vector $$\vec r$$ in terms of $$\vec t$$ and $$\vec n,$$ find $$\alpha$$ and $$\beta,$$ and then (ii) construct the incident vector $$\vec i$$ by flipping the sign of the normal component.

Step (i): Decompose $$\vec r$$ Write $$\vec r = \alpha\,\vec t + \beta\,\vec n$$ or in component form $$(1,\,-2)=\alpha(1,7)+\beta(7,\,-1).$$ Equating components:

For the $$x$$-coordinate: $$1 = \alpha + 7\beta \qquad (1)$$ For the $$y$$-coordinate: $$-2 = 7\alpha - \beta \qquad (2)$$

From (1) we obtain $$\alpha = 1 - 7\beta.$$ Substituting in (2):

$$-2 = 7(1 - 7\beta) - \beta = 7 - 49\beta - \beta = 7 - 50\beta.$$ Thus $$-50\beta = -9 \;\Longrightarrow\; \beta = \dfrac{9}{50}.$$

Then $$\alpha = 1 - 7\left(\dfrac{9}{50}\right)=1-\dfrac{63}{50}= \dfrac{50 - 63}{50}= -\dfrac{13}{50}.$$

Hence the reflected vector can indeed be written as $$\vec r = -\dfrac{13}{50}\,\vec t \;+\; \dfrac{9}{50}\,\vec n.$$

Step (ii): Construct the incident vector $$\vec i$$ According to the reflection rule, just reverse the sign of the normal component:

$$\vec i = \alpha\,\vec t \;-\; \beta\,\vec n = -\dfrac{13}{50}\,\vec t \;-\; \dfrac{9}{50}\,\vec n.$$

Insert $$\vec t = (1,7)$$ and $$\vec n = (7,-1):$$

$$\vec i = -\dfrac{13}{50}(1,7)\;-\;\dfrac{9}{50}(7,-1) = \left(-\dfrac{13}{50},\,-\dfrac{91}{50}\right) + \left(-\dfrac{63}{50},\,\dfrac{9}{50}\right) = \left(-\dfrac{76}{50},\, -\dfrac{82}{50}\right).$$

Simplify the common factor $$\dfrac{1}{50}:$$

$$\vec i = \left(-\dfrac{38}{25},\, -\dfrac{41}{25}\right).$$

Removing the denominator (multiply by $$25$$) gives the direction vector $$\vec i = (-38,\,-41).$$ The negative sign merely indicates direction; any non-zero scalar multiple represents the same straight line. For convenience we take the positive multiple $$\vec d = (38,\,41).$$

Equation of the incident line
The required line must pass through $$P(0,1)$$ and have direction vector $$\vec d = (38,41).$$ Slope $$m$$ of this vector is $$m = \dfrac{41}{38}.$$ Using the point-slope form,

$$y - 1 = m(x - 0) \;\Longrightarrow\; y - 1 = \dfrac{41}{38}\,x.$$

Eliminate the fraction by multiplying by $$38$$:

$$38(y - 1) = 41x \;\Longrightarrow\; 38y - 38 = 41x.$$

Rearrange to standard form:

$$41x - 38y + 38 = 0.$$

This matches Option C among the given choices.

Hence, the correct answer is Option C.

The given point is (2, 1). We need to translate this point parallel to the line L: $$x - y = 4$$ by a distance of $$2\sqrt{3}$$ units. The new point Q lies in the third quadrant, and we must find the equation of the line passing through Q and perpendicular to L.

First, find the direction vector of line L. The line $$x - y = 4$$ can be rewritten as $$y = x - 4$$, so its slope is 1. Therefore, the direction vector is proportional to (1, 1). The magnitude of this vector is $$\sqrt{1^2 + 1^2} = \sqrt{2}$$. The unit vector in the direction of (1, 1) is $$\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$.

The translation distance is $$2\sqrt{3}$$ units, so the displacement vector can be either $$2\sqrt{3} \times \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$$ or $$2\sqrt{3} \times \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$$. Simplifying:

$$2\sqrt{3} \times \frac{1}{\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{4 \times 3}{2}} = \sqrt{6}$$

So the displacement vectors are $$(\sqrt{6}, \sqrt{6})$$ and $$(-\sqrt{6}, -\sqrt{6})$$.

The two possible new points are:

Option 1: $$(2 + \sqrt{6}, 1 + \sqrt{6})$$

Option 2: $$(2 - \sqrt{6}, 1 - \sqrt{6})$$

Since Q lies in the third quadrant, both coordinates must be negative. Approximating $$\sqrt{6} \approx 2.45$$:

For Option 1: $$2 + 2.45 = 4.45 > 0$$ and $$1 + 2.45 = 3.45 > 0$$, so it is in the first quadrant.

For Option 2: $$2 - 2.45 = -0.45 < 0$$ and $$1 - 2.45 = -1.45 < 0$$, so it is in the third quadrant.

Thus, Q is $$(2 - \sqrt{6}, 1 - \sqrt{6})$$.

Now, find the line perpendicular to L passing through Q. The slope of L is 1, so the slope of a perpendicular line is the negative reciprocal, which is $$-1$$.

The equation of a line with slope $$-1$$ passing through $$(x_0, y_0)$$ is $$y - y_0 = m(x - x_0)$$. Substituting $$m = -1$$, $$x_0 = 2 - \sqrt{6}$$, and $$y_0 = 1 - \sqrt{6}$$:

$$y - (1 - \sqrt{6}) = -1 \times (x - (2 - \sqrt{6}))$$

Simplify the right side:

$$y - 1 + \sqrt{6} = - (x - 2 + \sqrt{6})$$

$$y - 1 + \sqrt{6} = -x + 2 - \sqrt{6}$$

Bring all terms to the left side:

$$y - 1 + \sqrt{6} + x - 2 + \sqrt{6} = 0$$

Combine like terms:

$$x + y + (-\1 - 2) + (\sqrt{6} + \sqrt{6}) = 0$$

$$x + y - 3 + 2\sqrt{6} = 0$$

Rearrange to standard form:

$$x + y = 3 - 2\sqrt{6}$$

Comparing with the options:

A. $$x + y = 2 - \sqrt{6}$$

B. $$2x + 2y = 1 - \sqrt{6}$$ which simplifies to $$x + y = \frac{1 - \sqrt{6}}{2}$$

C. $$x + y = 3 - 3\sqrt{6}$$

D. $$x + y = 3 - 2\sqrt{6}$$

The equation $$x + y = 3 - 2\sqrt{6}$$ matches option D.

Hence, the correct answer is Option D.

We are asked to count all lattice points (points whose coordinates are both integers) that lie strictly inside the triangle whose vertices are $$A(0,0),\;B(0,41),\;C(41,0).$$

The most direct way is to use Pick’s Theorem, which applies to any simple polygon whose vertices have integer coordinates. The theorem states

$$\text{Area}=I+\frac{B}{2}-1,$$

where

$$I=\text{number of interior lattice points},\qquad B=\text{number of lattice points on the boundary}.$$

We first evaluate the area of the triangle. Since the triangle is right-angled at the origin, the area is half the product of the perpendicular sides:

$$\text{Area}=\frac12\times 41\times 41=\frac{1681}{2}=840.5.$$

Next we must count $$B,$$ the lattice points lying on the three sides.

1. On the vertical side $$AB$$ from $$(0,0)$$ to $$(0,41)$$, the $$x$$-coordinate is always $$0,$$ while $$y$$ runs through all integers from $$0$$ to $$41.$$ Thus the number of lattice points on $$AB$$ is

$$41-0+1=42.$$

2. On the horizontal side $$AC$$ from $$(0,0)$$ to $$(41,0),$$ the $$y$$-coordinate is always $$0,$$ while $$x$$ runs from $$0$$ to $$41.$$ Therefore the side $$AC$$ also contributes

$$41-0+1=42$$

lattice points.

3. For the hypotenuse $$BC,$$ join $$(0,41)$$ to $$(41,0).$$ Its equation is $$x+y=41.$$ We need all integer solutions with $$x\ge 0$$ and $$y\ge 0.$$ If $$x$$ takes the values $$0,1,2,\dots,41,$$ then $$y$$ is automatically $$41-x,$$ remaining an integer in the permitted range. Hence this side also possesses

$$41-0+1=42$$

lattice points.

Adding the counts of all three sides gives $$42+42+42=126,$$ but each of the three vertices $$(0,0),\;(0,41),\;(41,0)$$ has just been tallied twice, once for each adjoining side. To correct this double counting we subtract the number of vertices once:

$$B=126-3=123.$$

Now substitute the known area and boundary count into Pick’s formula:

$$840.5=I+\frac{123}{2}-1.$$

Simplifying the right-hand side step by step, we have

$$\frac{123}{2}=61.5,$$

so

$$I+61.5-1=I+60.5.$$

Equating the two sides gives

$$I+60.5=840.5.$$

Subtract $$60.5$$ from both sides:

$$I=840.5-60.5=780.$$

Thus exactly $$780$$ lattice points lie strictly inside the given triangle.

Hence, the correct answer is Option A.

We are told that the required line $$L$$ passes through the fixed point $$P(1,2)$$ and that the segment cut by this line on the co-ordinate axes is bisected at $$P$$.

To express an ordinary straight line in terms of its intercepts, we use the axis-intercept form (state the formula first):

$$\frac{x}{a} + \frac{y}{b} = 1,$$

where $$a$$ is the $$x$$-intercept (the point $$(a,0)$$ on the $$x$$-axis) and $$b$$ is the $$y$$-intercept (the point $$(0,b)$$ on the $$y$$-axis).

The midpoint of this intercepted segment with endpoints $$(a,0)$$ and $$(0,b)$$ is obtained from the midpoint formula $$\bigl(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\bigr).$$

So the midpoint is

$$\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right).$$

But we are told that this midpoint is exactly the point $$P(1,2).$$ Hence we can write two separate equalities, equating corresponding co-ordinates:

$$\frac{a}{2}=1 \quad\text{and}\quad \frac{b}{2}=2.$$

From the first equality we immediately get

$$a = 2,$$

and from the second equality we get

$$b = 4.$$

Now substitute $$a=2$$ and $$b=4$$ back into the intercept form of the line:

$$\frac{x}{2} + \frac{y}{4} = 1.$$

To put this in a simpler linear form, multiply every term by $$4$$:

$$4\left(\frac{x}{2}\right) + 4\left(\frac{y}{4}\right) = 4(1).$$

This simplifies step by step to

$$2x + y = 4.$$

So the explicit equation of line $$L$$ is

$$2x + y = 4,$$

which can also be rearranged as

$$y = 4 - 2x.$$

From $$y = 4 - 2x$$ we identify the slope of $$L$$. Writing the general slope-intercept form $$y = mx + c,$$ we compare and see that

$$m_1 = -2.$$

Next, we are asked for a line $$L_1$$ that is perpendicular to $$L$$ and that also passes through the point $$(-2,1).$$

If two lines are perpendicular, the product of their slopes equals $$-1$$. Hence, denoting by $$m_2$$ the slope of $$L_1$$, we use the perpendicular-slopes condition:

$$m_1 \, m_2 = -1.$$

Substituting $$m_1 = -2,$$ we obtain

$$(-2)\, m_2 = -1.$$

Dividing both sides by $$-2$$, we get

$$m_2 = \frac{1}{2}.$$

Thus the slope of $$L_1$$ is $$\frac12$$, and we know that the line must pass through the given point $$(-2,1).$$ We now write the equation of $$L_1$$ in point-slope form (state the formula):

$$y - y_0 = m(x - x_0),$$

where $$(x_0,y_0)=(-2,1)$$. Substituting these values and $$m = \tfrac12$$, we get

$$y - 1 = \frac12 \bigl(x - (-2)\bigr).$$

Since $$x - (-2) = x + 2,$$ the equation becomes

$$y - 1 = \frac12 (x + 2).$$

To make later substitution easy, expand the right-hand side:

$$y - 1 = \frac{x}{2} + 1.$$

Now add $$1$$ to both sides:

$$y = \frac{x}{2} + 2.$$

So the explicit equation of $$L_1$$ is

$$y = \frac{x}{2} + 2.$$

We must now find the point of intersection of the two lines. This point must satisfy both

$$y = 4 - 2x \quad\text{and}\quad y = \frac{x}{2} + 2.$$

We therefore set the two right-hand sides equal:

$$4 - 2x = \frac{x}{2} + 2.$$

To clear the fraction and make the algebra straightforward, multiply every term by $$2$$:

$$2(4) - 2(2x) = 2\left(\frac{x}{2}\right) + 2(2).$$

This simplifies to

$$8 - 4x = x + 4.$$

Now gather all the $$x$$-terms on one side and the constants on the other side. First add $$4x$$ to both sides:

$$8 = x + 4 + 4x.$$

That gives

$$8 = 5x + 4.$$

Next, subtract $$4$$ from both sides to isolate the term with $$x$$:

$$8 - 4 = 5x.$$

Hence

$$4 = 5x.$$

Divide by $$5$$ to obtain $$x$$:

$$x = \frac{4}{5}.$$

With this value of $$x$$, we substitute back into either of the two line equations to get $$y$$. Choosing the simpler $$y = 4 - 2x$$, we write

$$y = 4 - 2\left(\frac{4}{5}\right).$$

Compute the product inside the parentheses first:

$$2\left(\frac{4}{5}\right) = \frac{8}{5}.$$

Thus

$$y = 4 - \frac{8}{5}.$$

Rewrite $$4$$ as $$\frac{20}{5}$$ so we share a common denominator, and subtract:

$$y = \frac{20}{5} - \frac{8}{5} = \frac{12}{5}.$$

We have therefore obtained the intersection point

$$\left(\frac{4}{5},\frac{12}{5}\right).$$

This matches option B in the list provided. Hence, the correct answer is Option B.

The given equation is $$(2x - 3y + 4) + k(x - 2y + 3) = 0$$. This represents a family of lines passing through the intersection of:

$$L_1: 2x - 3y + 4 = 0$$ and $$L_2: x - 2y + 3 = 0$$

To find the intersection point ($$P$$), solve the two equations

$$\implies \mathbf{y = 2}$$, $$\mathbf{x = 1}$$.

Thus, all lines in the family pass through the fixed point $$P(1, 2)$$.

Let $$A(2, 3)$$ be the given point and $$A'$$ be its image in any line $$L$$ from the family.

By the definition of a reflection, the line $$L$$ is the perpendicular bisector of the segment $$AA'$$.

Since the fixed point $$P(1, 2)$$ lies on every such line $$L$$, it must be equidistant from the point $$A$$ and its image $$A'$$.

$$\text{Therefore, } PA = PA'$$

$$PA = \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Since $$PA' = PA = \sqrt{2}$$, the image $$A'$$ always remains at a constant distance of $$\sqrt{2}$$ from the fixed point $$P(1, 2)$$.

The locus of a point moving at a constant distance from a fixed point is a circle, 

with Center: $$(1, 2)$$, and Radius: $$\sqrt{2}$$

To determine the nature of the points $$\left(0, \frac{8}{3}\right)$$, $$(1, 3)$$, and $$(82, 30)$$, we first check if they are collinear, meaning they lie on a straight line. Three points are collinear if the area of the triangle they form is zero. The area formula for a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, and $$(x_3, y_3)$$ is:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$

Substituting the given points:

$$x_1 = 0$$, $$y_1 = \frac{8}{3}$$

$$x_2 = 1$$, $$y_2 = 3$$

$$x_3 = 82$$, $$y_3 = 30$$

Compute the differences:

$$y_2 - y_3 = 3 - 30 = -27$$

$$y_3 - y_1 = 30 - \frac{8}{3} = \frac{90}{3} - \frac{8}{3} = \frac{82}{3}$$

$$y_1 - y_2 = \frac{8}{3} - 3 = \frac{8}{3} - \frac{9}{3} = -\frac{1}{3}$$

Now plug into the area formula:

$$\text{Area} = \frac{1}{2} \left| 0 \cdot (-27) + 1 \cdot \left(\frac{82}{3}\right) + 82 \cdot \left(-\frac{1}{3}\right) \right|$$

Simplify inside the absolute value:

$$0 + \frac{82}{3} - \frac{82}{3} = 0$$

So,

$$\text{Area} = \frac{1}{2} \left| 0 \right| = 0$$

Since the area is zero, the points are collinear and lie on a straight line.

Alternatively, we can verify by checking the slopes between each pair of points. The slope between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is given by:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Slope between $$\left(0, \frac{8}{3}\right)$$ and $$(1, 3)$$:

$$m_1 = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9}{3} - \frac{8}{3}}{1} = \frac{\frac{1}{3}}{1} = \frac{1}{3}$$

Slope between $$(1, 3)$$ and $$(82, 30)$$:

$$m_2 = \frac{30 - 3}{82 - 1} = \frac{27}{81} = \frac{1}{3}$$

Slope between $$\left(0, \frac{8}{3}\right)$$ and $$(82, 30)$$:

$$m_3 = \frac{30 - \frac{8}{3}}{82 - 0} = \frac{\frac{90}{3} - \frac{8}{3}}{82} = \frac{\frac{82}{3}}{82} = \frac{82}{3} \times \frac{1}{82} = \frac{1}{3}$$

All slopes are equal to $$\frac{1}{3}$$, confirming that the points are collinear.

Since the points lie on a straight line, they do not form a triangle. Therefore, options A, B, and D, which describe types of triangles, are incorrect.

Hence, the correct answer is Option C.

We begin by examining the line already given in the question. The equation $$\sqrt{3}x + y = 1$$ can be rewritten in the slope-intercept form $$y = mx + c$$ in order to read its slope directly. Subtracting $$\sqrt{3}x$$ from both sides we get

$$y = -\sqrt{3}\,x + 1.$$

So, the slope of this reference line is $$m_1 = -\sqrt{3}.$$

The new line $$L$$ must make an angle of $$60^\circ$$ with this reference line. For two lines whose slopes are $$m_1$$ and $$m_2$$, the magnitude of the angle $$\theta$$ between them is given by the standard formula

$$\tan\theta \;=\;\left|\dfrac{m_2 - m_1}{1 + m_1\,m_2}\right|.$$

Here $$\theta = 60^\circ$$ and thus $$\tan 60^\circ = \sqrt{3}.$$ Putting these values in, we have

$$\Bigl|\dfrac{m_2 - (-\sqrt{3})}{1 + (-\sqrt{3})\,m_2}\Bigr| \;=\; \sqrt{3}.$$

Simplifying the numerator inside the absolute value, we obtain

$$\Bigl|\dfrac{m_2 + \sqrt{3}}{\,1 - \sqrt{3}\,m_2\,}\Bigr| \;=\; \sqrt{3}.$$

This absolute-value equation splits into two linear equations:

1. $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = \sqrt{3},$$ 2. $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = -\sqrt{3}.$$

First possibility. Take $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = \sqrt{3}.$$ Cross-multiplying:

$$(m_2 + \sqrt{3}) = \sqrt{3}\,(1 - \sqrt{3}\,m_2).$$

Expanding the right side gives

$$m_2 + \sqrt{3} = \sqrt{3} - 3\,m_2.$$

Bringing all terms to the left:

$$m_2 + 3\,m_2 + \sqrt{3} - \sqrt{3} = 0 \;\Longrightarrow\; 4\,m_2 = 0 \;\Longrightarrow\; m_2 = 0.$$

A slope of zero corresponds to a horizontal line $$y = \text{constant}.$$ Passing this horizontal line through the given point $$(3,-2)$$ would give $$y = -2,$$ which never meets the X-axis (where $$y = 0$$). Therefore this solution is inadmissible.

Second possibility. Now take $$\dfrac{m_2 + \sqrt{3}}{1 - \sqrt{3}\,m_2} = -\sqrt{3}.$$ Cross-multiplying again:

$$(m_2 + \sqrt{3}) = -\sqrt{3}\,(1 - \sqrt{3}\,m_2).$$

Expanding the right side gives

$$m_2 + \sqrt{3} = -\sqrt{3} + 3\,m_2.$$

Collecting like terms on the left:

$$$ \begin{aligned} m_2 - 3\,m_2 + \sqrt{3} + \sqrt{3} &= 0 \\ -2\,m_2 + 2\sqrt{3} &= 0. \end{aligned} $$$

Dividing by $$-2$$ we get

$$m_2 = \sqrt{3}.$$