JEE Straight Lines Questions

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.

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.

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

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 $$\alpha = \frac{13}{3} - 4 = \frac{1}{3}$$.

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

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 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.

Let the incident ray emit from the origin $$O(0, 0)$$ making an angle of $$30^\circ$$ with the positive $$x$$-axis. Its equation is given by:

$$y = \tan(30^\circ)x = \frac{1}{\sqrt{3}}x \implies x = \sqrt{3}y$$

The reflecting line acts as a plane mirror:

$$x + y = 1$$

According to the laws of reflection, the line containing the reflected ray will pass through the mirror image of the source point with respect to the line.

Step 1: Find the image of the origin $$O(0, 0)$$ in the line $$x + y - 1 = 0$$

Using the standard mirror image formula:

$$\frac{x_2 - 0}{1} = \frac{y_2 - 0}{1} = -2\frac{1(0) + 1(0) - 1}{1^2 + 1^2} = -2\left(\frac{-1}{2}\right) = 1$$

This gives $$x_2 = 1$$ and $$y_2 = 1$$. Thus, the image of the origin is the point $$M(1, 1)$$, and the line containing the reflected ray must pass through $$M$$.

Step 2: Find the point of incidence $$P$$

The incident ray intersects the line $$x + y = 1$$ at the point of incidence $$P$$. Substituting $$x = \sqrt{3}y$$ into the mirror line equation:

$$\sqrt{3}y + y = 1 \implies y(\sqrt{3} + 1) = 1 \implies y = \frac{1}{\sqrt{3} + 1}$$

Substituting $$y$$ back to find the $$x$$-coordinate of $$P$$:

$$x = \frac{\sqrt{3}}{\sqrt{3} + 1}$$

So, the point of incidence is $$P\left(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1}\right)$$.

Step 3: Find the slope of the reflected ray ($$PM$$)

The reflected ray passes through both the point of incidence $$P$$ and the image point $$M(1, 1)$$. The slope $$m$$ of the line $$PM$$ is:

$$m = \frac{1 - \frac{1}{\sqrt{3} + 1}}{1 - \frac{\sqrt{3}}{\sqrt{3} + 1}} = \frac{\frac{\sqrt{3} + 1 - 1}{\sqrt{3} + 1}}{\frac{\sqrt{3} + 1 - \sqrt{3}}{\sqrt{3} + 1}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Step 4: Find the equation of the reflected ray

Using the point-slope form with $$M(1, 1)$$ and slope $$m = \sqrt{3}$$:

$$y - 1 = \sqrt{3}(x - 1)$$

Step 5: Find the intersection point $$Q$$ on the $$x$$-axis

The ray intersects the $$x$$-axis at $$Q$$, where $$y = 0$$. Substituting $$y = 0$$ into the reflected ray equation:

$$0 - 1 = \sqrt{3}(x - 1)$$

$$-1 = \sqrt{3}x - \sqrt{3}$$

$$\sqrt{3}x = \sqrt{3} - 1$$

$$x = \frac{\sqrt{3} - 1}{\sqrt{3}} = 1 - \frac{1}{\sqrt{3}}$$

To match standard rationalised formats, multiplying the numerator and denominator by $$(\sqrt{3} + 1)$$:

$$x = \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3}(\sqrt{3} + 1)} = \frac{3 - 1}{3 + \sqrt{3}} = \frac{2}{3 + \sqrt{3}}$$

Conclusion:

The abscissa of Q is equal to $$\frac{2}{3 + \sqrt{3}}$$ .

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

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: $$y - x = 5$$.

Let $$B(\alpha, -2\alpha)$$ and $$C(\beta + 3, \beta)$$

$$\frac{\alpha + \beta + 3 + 1}{3} = 2$$

$$\alpha + \beta + 4 = 6 \implies \alpha + \beta = 2 \implies \beta = 2 - \alpha \quad \text{--- (Equation 1)}$$

$$\frac{-2\alpha - 2 + \beta}{3} = a$$

$$-2\alpha - 2 + \beta = 3a$$

$$-2\alpha - 2 + (2 - \alpha) = 3a$$

$$-3\alpha = 3a \implies \alpha = -a \quad \text{--- (Equation 2)}$$

Since $$B(\alpha, -2\alpha)$$ lies on the line $$x + py = 21a$$:

$$\alpha + p(-2\alpha) = 21a$$

$$\alpha(1 - 2p) = 21a \quad \text{--- (Equation 3)}$$

$$-a(1 - 2p) = 21a$$

$$-1(1 - 2p) = 21 \implies -1 + 2p = 21 \implies 2p = 22 \implies p = 11$$

Since $$C(\beta + 3, \beta)$$ also lies on the line $$x + 11y = 21a$$:

$$(\beta + 3) + 11\beta = 21a$$

$$12\beta + 3 = 21a$$

$$21\alpha + 12\beta + 3 = 0$$

$$21\alpha + 12(2 - \alpha) + 3 = 0$$

$$9\alpha + 27 = 0 \implies \alpha = -3$$

$$\beta = 2 - (-3) = 5$$

$$B(-3, 6) \quad \text{and} \quad C(5 + 3, 5) \implies C(8, 5)$$

$$(BC)^2 = (8 - (-3))^2 + (5 - 6)^2$$

$$(BC)^2 = (11)^2 + (-1)^2 = 121 + 1 = 122$$

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 $$348$$.

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.

The area of the triangle

$$ABC$$

is

$$\frac12\left|1(0-\alpha)+\alpha(\alpha-\alpha)+0(\alpha-0)\right|=4.$$

Thus,

$$\frac12|\,-\alpha\,|=4,$$

which gives

$$|\alpha|=8.$$

Hence

$$\alpha=\pm 8.$$

Now, the points

$$(\alpha,-\alpha)$$

and

$$(-\alpha,\alpha)$$

lie on the line

$$y=-x,$$

since

$$-\alpha=-\alpha$$

and

$$\alpha=-(-\alpha).$$

Therefore, for the three points to be collinear, the point

$$(\alpha^2,\beta)$$

must also lie on

$$y=-x.$$

Hence,

$$\beta=-\alpha^2.$$

Since

$$|\alpha|=8,$$

we have

$$\alpha^2=64.$$

Therefore,

$$\beta=-64.$$

Hence, $$\text{Answer}=-64$$.

Points $$A$$ and $$A'$$ lie on $$y = 2$$, so let $$A = (a, 2)$$ and $$A' = (a', 2)$$. Given $$B = (2, 3)$$.

The line segments $$OA$$ and $$OB$$ subtend angle $$\frac{\pi}{4}$$ at the origin, meaning $$\angle AOB = \frac{\pi}{4}$$.

Slope of $$OB = \frac{3}{2}$$ and slope of $$OA = \frac{2}{a}$$.

Using the tangent formula for angle between two lines:

$$\tan\frac{\pi}{4} = \left|\frac{\frac{2}{a} - \frac{3}{2}}{1 + \frac{2}{a} \cdot \frac{3}{2}}\right| = \left|\frac{\frac{4 - 3a}{2a}}{\frac{a + 3}{a}}\right| = \left|\frac{4 - 3a}{2(a + 3)}\right| = 1$$

Case 1: $$4 - 3a = 2(a + 3) = 2a + 6$$

$$4 - 3a = 2a + 6 \implies -5a = 2 \implies a = -\frac{2}{5}$$

Case 2: $$4 - 3a = -(2a + 6) = -2a - 6$$

$$4 - 3a = -2a - 6 \implies -a = -10 \implies a = 10$$

So the two points are $$A = \left(-\frac{2}{5}, 2\right)$$ and $$A' = (10, 2)$$.

The distance between them is:

$$|AA'| = \left|10 - \left(-\frac{2}{5}\right)\right| = 10 + \frac{2}{5} = \frac{52}{5}$$

The answer is Option C: $$\dfrac{52}{5}$$.

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: $$14$$.

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: $$p = 3$$

Given

$$y^5-9xy+2x=0$$

Rearrange:

$$x(2-9y)+y^5=0$$

$$x=\frac{y^5}{9y-2}$$

Differentiate implicitly:

$$5y^4-9\left(x+y\frac{dx}{dy}\right)+2\frac{dx}{dy}=0$$

or directly differentiate

$$x=\frac{y^5}{9y-2}$$

Using the quotient rule,

$$\frac{dx}{dy}=\frac{5y^4(9y-2)-9y^5}{(9y-2)^2}$$

$$=\frac{y^4(36y-10)}{(9y-2)^2}$$

### Tangents parallel to the x-axis

A tangent parallel to the

$$x$$

-axis means

$$\frac{dy}{dx}=0$$

which is equivalent to

$$\frac{dx}{dy}\neq0$$

and undefined reciprocal, so we set

$$\frac{dx}{dy}=\infty$$

This occurs when

$$(9y-2)^2=0$$

$$y=\frac29$$

Substituting into the curve,

$$x=\frac{(2/9)^5}{9(2/9)-2}$$

gives denominator

$$0$$

which is impossible.

Hence there is no point with

$$9y-2=0$$

on the curve.

Therefore we instead use

$$\frac{dy}{dx}=0$$

from implicit differentiation.

Differentiating:

$$5y^4y'-9(xy'+y)+2=0$$

$$(5y^4-9x)y'=9y-2$$

Thus

$$y'=\frac{9y-2}{5y^4-9x}$$

For tangents parallel to the

$$x$$

-axis,

$$y'=0$$

$$9y-2=0$$

$$y=\frac29$$

Substituting into the curve:

$$\left(\frac29\right)^5-9x\left(\frac29\right)+2x=0$$

The coefficient of

$$x$$

vanishes, leaving

$$\left(\frac29\right)^5=0$$

which is impossible.

Hence

$$M=0$$

### Tangents parallel to the y-axis

A tangent parallel to the

$$y$$

-axis requires

$$5y^4-9x=0$$

and

$$9y-2\neq0$$

From

$$5y^4-9x=0$$

$$x=\frac59y^4$$

Substitute into the curve:

$$y^5-9\left(\frac59y^4\right)y+2\left(\frac59y^4\right)=0$$

$$y^5-5y^5+\frac{10}{9}y^4=0$$

$$y^4\left(-4y+\frac{10}{9}\right)=0$$

Hence

$$y=0$$

or

$$y=\frac{5}{18}$$

These give two distinct points on the curve, and for both,

$$9y-2\neq0$$

Therefore

$$N=2$$

Thus

$$M+N=0+2$$

$$=2$$

Final Answer :

$$2$$

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).

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 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.

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 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 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 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.

First, we write the three given equations clearly:

$$\begin{aligned} x - 2y + kz &= 1 \quad\quad (1)\\[4pt] 2x + y + z &= 2 \quad\quad (2)\\[4pt] 3x - y - kz &= 3 \quad\quad (3) \end{aligned}$$

Because the question asks for the locus of $$(x,y)$$ when a solution with $$z\neq 0$$ exists, we shall eliminate both $$k$$ and $$z$$ from the system. That elimination will leave us with a relation involving only $$x$$ and $$y$$, which will be the required straight-line equation.

We begin by isolating the product $$kz$$ from the first and the third equations. From Equation (1), moving all other terms to the right:

$$kz = 1 - x + 2y \quad\quad (4)$$

Next, from Equation (3) we first move $$3x - y$$ to the right, remembering the sign in front of $$kz$$:

$$ -\,kz = 3 - 3x + y $$

Multiplying both sides by $$-1$$ gives

$$ kz = 3x - y - 3 \quad\quad (5)$$

Now, because both expressions (4) and (5) equal the same quantity $$kz$$, we set them equal to one another:

$$ 1 - x + 2y = 3x - y - 3 $$

We gather like terms on one side. First, move every term to the left:

$$ 1 - x + 2y - 3x + y + 3 = 0 $$

Combining individual groups:

$$ (1+3) + (-x-3x) + (2y + y) = 0 $$

$$ 4 - 4x + 3y = 0 $$

Finally, multiplying the entire equation by $$-1$$ (so that the coefficient of $$x$$ is positive) yields

$$ 4x - 3y - 4 = 0 $$

This linear relation between $$x$$ and $$y$$ must be satisfied by every triplet $$(x,y,z)$$ that solves the original system. Notice that we have not imposed any extra restriction on $$z$$ except that it must be non-zero; the relation we obtained is therefore the complete locus in the $$xy$$-plane.

Equation $$4x - 3y - 4 = 0$$ matches Option A in the list provided.

Hence, the correct answer is Option A.

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.

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 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

$$m_2 = \sqrt{3}.$$

Hence the slope of the required line $$L$$ is $$m_2 = \sqrt{3}.$$

Because $$L$$ passes through the point $$(3,-2)$$, we use the point-slope form $$y - y_1 = m(x - x_1).$$ Substituting $$x_1 = 3,\; y_1 = -2$$ and $$m = \sqrt{3}$$, we write

$$y - (-2) = \sqrt{3}\,(x - 3).$$

Removing the double negative:

$$y + 2 = \sqrt{3}\,x - 3\sqrt{3}.$$

Now we bring every term to the left to obtain the standard (general) form:

$$$ \begin{aligned} y + 2 - \sqrt{3}\,x + 3\sqrt{3} &= 0 \\ y - \sqrt{3}\,x + 2 + 3\sqrt{3} &= 0. \end{aligned} $$$

This is already one of the listed options. We may still verify that the line indeed intersects the X-axis. Setting $$y = 0$$ in the equation gives

$$0 - \sqrt{3}\,x + 2 + 3\sqrt{3} = 0 \;\Longrightarrow\; \sqrt{3}\,x = 2 + 3\sqrt{3} \;\Longrightarrow\; x = \dfrac{2 + 3\sqrt{3}}{\sqrt{3}},$$

which is a real number, so the line meets the X-axis exactly once. All conditions are satisfied.

The expression $$y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$$ appears as Option D.

Hence, the correct answer is Option D.

First, we need to find the slope of the given line $$5x - y = 1$$. To do this, we rearrange it into slope-intercept form $$y = mx + c$$.

Starting with $$5x - y = 1$$:

Subtract $$5x$$ from both sides: $$-y = -5x + 1$$

Multiply both sides by $$-1$$: $$y = 5x - 1$$

So, the slope of this line is 5.

Since line L is perpendicular to this line, the product of their slopes must be $$-1$$. Let the slope of L be $$m$$. Then:

$$m \times 5 = -1$$

Solving for $$m$$: $$m = -\frac{1}{5}$$

Thus, the slope of line L is $$-\frac{1}{5}$$.

The equation of line L can be written as $$y = -\frac{1}{5}x + b$$, where $$b$$ is the y-intercept. To avoid fractions, multiply through by 5:

$$5y = -x + 5b$$

Rearrange to standard form: $$x + 5y = 5b$$

Let $$c = 5b$$, so the equation becomes:

$$x + 5y = c$$     (1)

Now, we find the intercepts of line L with the coordinate axes to use the area condition.

When $$x = 0$$: $$5y = c$$ $$\Rightarrow$$ $$y = \frac{c}{5}$$   (y-intercept)

When $$y = 0$$: $$x = c$$   (x-intercept)

The area of the triangle formed with the axes is given as 5 sq units. The area formula is $$\frac{1}{2} \times |\text{x-intercept}| \times |\text{y-intercept}|$$. Since area is positive, we use absolute values:

$$\frac{1}{2} \times |c| \times \left|\frac{c}{5}\right| = 5$$

Since $$\left|\frac{c}{5}\right| = \frac{|c|}{5}$$, this becomes:

$$\frac{1}{2} \times |c| \times \frac{|c|}{5} = 5$$

Simplify: $$\frac{1}{2} \times \frac{|c|^2}{5} = 5$$

Because $$|c|^2 = c^2$$, we have:

$$\frac{c^2}{10} = 5$$

Multiply both sides by 10: $$c^2 = 50$$

Take square roots: $$c = \sqrt{50}$$ or $$c = -\sqrt{50}$$

Simplify: $$c = 5\sqrt{2}$$ or $$c = -5\sqrt{2}$$

So, the possible equations for line L are:

$$x + 5y = 5\sqrt{2}$$     or     $$x + 5y = -5\sqrt{2}$$

Now, we need to find the distance from line L to the line $$x + 5y = 0$$.

Write both lines in the form $$ax + by + k = 0$$:

The reference line is $$x + 5y = 0$$, so $$x + 5y + 0 = 0$$.

For the first case, $$x + 5y - 5\sqrt{2} = 0$$.

For the second case, $$x + 5y + 5\sqrt{2} = 0$$ (since $$x + 5y = -5\sqrt{2}$$ implies $$x + 5y + 5\sqrt{2} = 0$$).

The distance $$d$$ between two parallel lines $$ax + by + c_1 = 0$$ and $$ax + by + c_2 = 0$$ is:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

Here, $$a = 1$$, $$b = 5$$, so $$\sqrt{a^2 + b^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$$.

For the first line L: $$x + 5y - 5\sqrt{2} = 0$$, so $$c_1 = -5\sqrt{2}$$. The reference line has $$c_2 = 0$$.

Distance: $$d_1 = \frac{|-5\sqrt{2} - 0|}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{26}}$$

For the second line L: $$x + 5y + 5\sqrt{2} = 0$$, so $$c_1 = 5\sqrt{2}$$.

Distance: $$d_2 = \frac{|5\sqrt{2} - 0|}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{26}}$$

Both distances are the same. Simplify:

$$\frac{5\sqrt{2}}{\sqrt{26}} = 5 \sqrt{\frac{2}{26}} = 5 \sqrt{\frac{1}{13}} = \frac{5}{\sqrt{13}}$$

Alternatively, $$\frac{5\sqrt{2}}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{2 \times 13}} = \frac{5\sqrt{2}}{\sqrt{2} \times \sqrt{13}} = \frac{5}{\sqrt{13}}$$

So, the distance is $$\frac{5}{\sqrt{13}}$$ units.

Comparing with the options:

A. $$\frac{7}{\sqrt{13}}$$

B. $$\frac{7}{\sqrt{5}}$$

C. $$\frac{5}{\sqrt{13}}$$

D. $$\frac{5}{\sqrt{7}}$$

Hence, the correct answer is Option C.

We are given that a line segment intercepted between the coordinate axes is trisected at the point A(4, 3), and A is nearer to the x-axis. We need to find the equation of this line.

Let the line intersect the x-axis at point P(a, 0) and the y-axis at point Q(0, b). The segment PQ is trisected, meaning it is divided into three equal parts by two points. Since A(4, 3) is nearer to the x-axis, it must be the trisection point closer to P (the x-intercept). Therefore, A divides PQ in the ratio 1:2, where PA:AQ = 1:2 (the part from P to A is one-third of PQ).

Using the section formula, if a point divides the line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n, its coordinates are:

$$ \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) $$

Here, P(a, 0) and Q(0, b), and A(4, 3) divides PQ in the ratio m:n = 1:2. Substituting:

For the x-coordinate:

$$ 4 = \frac{1 \cdot 0 + 2 \cdot a}{1 + 2} = \frac{2a}{3} $$

Solving for a:

$$ 4 = \frac{2a}{3} $$

Multiply both sides by 3:

$$ 12 = 2a $$

Divide both sides by 2:

$$ a = 6 $$

For the y-coordinate:

$$ 3 = \frac{1 \cdot b + 2 \cdot 0}{1 + 2} = \frac{b}{3} $$

Solving for b:

$$ 3 = \frac{b}{3} $$

Multiply both sides by 3:

$$ b = 9 $$

So, the x-intercept is 6 and the y-intercept is 9. The equation of a line in intercept form is:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

Substituting a = 6 and b = 9:

$$ \frac{x}{6} + \frac{y}{9} = 1 $$

To eliminate denominators, multiply both sides by the least common multiple of 6 and 9, which is 18:

$$ 18 \cdot \frac{x}{6} + 18 \cdot \frac{y}{9} = 18 \cdot 1 $$

Simplify each term:

$$ 3x + 2y = 18 $$

Thus, the equation is 3x + 2y = 18.

Now, comparing with the options:

A. 4x - 3y = 7

B. 3x + 2y = 18

C. 3x + 8y = 36

D. x + 3y = 13

Option B matches our equation. To verify, check if A(4, 3) lies on this line and if it trisects PQ with intercepts (6, 0) and (0, 9).

Substitute x = 4, y = 3 into 3x + 2y:

$$ 3(4) + 2(3) = 12 + 6 = 18 $$

This equals the right-hand side, so A lies on the line.

For trisection: the point dividing PQ from P(6, 0) to Q(0, 9) in the ratio 1:2 is:

$$ \left( \frac{1 \cdot 0 + 2 \cdot 6}{1 + 2}, \frac{1 \cdot 9 + 2 \cdot 0}{1 + 2} \right) = \left( \frac{12}{3}, \frac{9}{3} \right) = (4, 3) $$

This confirms A(4, 3) is the trisection point nearer to the x-axis (since the other trisection point at ratio 2:1 is (2, 6), which has a larger y-coordinate and is farther from the x-axis).

Hence, the correct answer is Option B.

We have the three vertices of the triangle as $$P(2,\,2),\;Q(6,\,-1)\; \text{and}\; R(7,\,3)$$. The median $$PS$$ is drawn from the vertex $$P$$ to the midpoint $$S$$ of the side $$QR$$, so we first find the coordinates of this midpoint.

The midpoint formula says that the midpoint of $$\bigl(x_1,\,y_1\bigr)$$ and $$\bigl(x_2,\,y_2\bigr)$$ is

$$\left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right).$$

Applying the formula to $$Q(6,\,-1)$$ and $$R(7,\,3)$$, we obtain

$$S\;=\;\left(\dfrac{6+7}{2},\;\dfrac{-1+3}{2}\right) \;=\;\left(\dfrac{13}{2},\;\dfrac{2}{2}\right) \;=\;\left(\dfrac{13}{2},\;1\right).$$

Now we determine the slope of the median $$PS$$. The slope formula for the line through points $$\bigl(x_1,\,y_1\bigr)$$ and $$\bigl(x_2,\,y_2\bigr)$$ is

$$m=\dfrac{y_2-y_1}{x_2-x_1}.$$

For the points $$P(2,\,2)$$ and $$S\!\left(\dfrac{13}{2},\,1\right)$$, we have

$$m_{PS} =\dfrac{1-2}{\dfrac{13}{2}-2} =\dfrac{-1}{\dfrac{13}{2}-\dfrac{4}{2}} =\dfrac{-1}{\dfrac{9}{2}} =-\dfrac{1}{9/2} =-\dfrac{2}{9}.$$

Hence the slope of any line parallel to $$PS$$ is also $$-\dfrac{2}{9}$$. We now need the equation of the line that passes through the point $$(1,\,-1)$$ and has this slope.

We use the point-slope form of a straight line, which is

$$y-y_1=m\bigl(x-x_1\bigr),$$

where $$\bigl(x_1,\,y_1\bigr)=(1,\,-1)$$ and $$m=-\dfrac{2}{9}$$. Substituting,

$$y-(-1)=-\dfrac{2}{9}\bigl(x-1\bigr).$$

Simplifying step by step:

$$y+1=-\dfrac{2}{9}x+\dfrac{2}{9}.$$

To clear the denominator, multiply every term by $$9$$:

$$9(y+1)=-2x+2.$$

Expanding the left side gives

$$9y+9=-2x+2.$$

Now bring every term to the same side to get the equation in standard form:

$$2x+9y+9-2=0 \;\Longrightarrow\; 2x+9y+7=0.$$

Thus the required equation is

$$2x+9y+7=0.$$

This matches Option D in the given list. Hence, the correct answer is Option D.

The base of the equilateral triangle lies along the line $$3x + 4y = 9$$, and one vertex is at $$(1, 2)$$. Since it is an equilateral triangle, the altitude from the vertex $$(1, 2)$$ to the base will be perpendicular to the base and will bisect the base. This altitude can be used to find the side length.

First, find the slope of the base line $$3x + 4y = 9$$. Rewrite the equation in slope-intercept form:

$$4y = -3x + 9$$

$$y = -\frac{3}{4}x + \frac{9}{4}$$

The slope of the base is $$-\frac{3}{4}$$. The slope of the altitude, which is perpendicular to the base, is the negative reciprocal. So, the slope of the altitude is:

$$m = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$$

The altitude passes through the vertex $$(1, 2)$$, so its equation is:

$$y - 2 = \frac{4}{3}(x - 1)$$

Simplify this equation:

$$y - 2 = \frac{4}{3}x - \frac{4}{3}$$

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

$$y = \frac{4}{3}x - \frac{4}{3} + \frac{6}{3}$$

$$y = \frac{4}{3}x + \frac{2}{3}$$

The foot of the perpendicular (denoted as point D) is the intersection of the altitude and the base line. Solve the system of equations:

Base: $$3x + 4y = 9$$

Altitude: $$y = \frac{4}{3}x + \frac{2}{3}$$

Substitute $$y$$ from the altitude equation into the base equation:

$$3x + 4\left(\frac{4}{3}x + \frac{2}{3}\right) = 9$$

$$3x + \frac{16}{3}x + \frac{8}{3} = 9$$

Multiply through by 3 to clear denominators:

$$9x + 16x + 8 = 27$$

$$25x + 8 = 27$$

$$25x = 19$$

$$x = \frac{19}{25}$$

Now substitute $$x = \frac{19}{25}$$ into the altitude equation to find $$y$$:

$$y = \frac{4}{3} \cdot \frac{19}{25} + \frac{2}{3}$$

$$y = \frac{76}{75} + \frac{2}{3}$$

Convert $$\frac{2}{3}$$ to a fraction with denominator 75:

$$\frac{2}{3} = \frac{2 \times 25}{3 \times 25} = \frac{50}{75}$$

$$y = \frac{76}{75} + \frac{50}{75} = \frac{126}{75}$$

Simplify $$\frac{126}{75}$$ by dividing numerator and denominator by 3:

$$y = \frac{42}{25}$$

So, point D is at $$\left(\frac{19}{25}, \frac{42}{25}\right)$$.

The distance between vertex A $$(1, 2)$$ and point D $$\left(\frac{19}{25}, \frac{42}{25}\right)$$ is the length of the altitude. Compute the differences:

$$\Delta x = \frac{19}{25} - 1 = \frac{19}{25} - \frac{25}{25} = -\frac{6}{25}$$

$$\Delta y = \frac{42}{25} - 2 = \frac{42}{25} - \frac{50}{25} = -\frac{8}{25}$$

The distance AD is:

$$\text{AD} = \sqrt{\left(-\frac{6}{25}\right)^2 + \left(-\frac{8}{25}\right)^2} = \sqrt{\frac{36}{625} + \frac{64}{625}} = \sqrt{\frac{100}{625}} = \frac{\sqrt{100}}{\sqrt{625}} = \frac{10}{25} = \frac{2}{5}$$

In an equilateral triangle with side length $$s$$, the height $$h$$ is given by $$h = \frac{\sqrt{3}}{2} s$$. Here, AD is the height, so:

$$\frac{2}{5} = \frac{\sqrt{3}}{2} s$$

Solve for $$s$$:

$$s = \frac{2}{5} \times \frac{2}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$$

Rationalize the denominator:

$$s = \frac{4}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{15}$$

Thus, the length of a side of the triangle is $$\frac{4\sqrt{3}}{15}$$. Comparing with the options, this matches option B.

Hence, the correct answer is Option B.

The circumcentre of the triangle is at the origin, denoted as $$O(0, 0)$$. The centroid $$G$$ is given as the midpoint of the line segment joining the points $$(a^2 + 1, a^2 + 1)$$ and $$(2a, -2a)$$, where $$a \neq 0$$.

Using the midpoint formula, the coordinates of the centroid $$G$$ are calculated as follows:

$$$ x_G = \frac{(a^2 + 1) + (2a)}{2} = \frac{a^2 + 2a + 1}{2} $$$

$$$ y_G = \frac{(a^2 + 1) + (-2a)}{2} = \frac{a^2 - 2a + 1}{2} $$$

Simplifying the expressions:

$$$ a^2 + 2a + 1 = (a + 1)^2 $$$

$$$ a^2 - 2a + 1 = (a - 1)^2 $$$

So,

$$$ x_G = \frac{(a + 1)^2}{2} $$$

$$$ y_G = \frac{(a - 1)^2}{2} $$$

In any triangle, the centroid divides the line segment joining the orthocentre $$H$$ and circumcentre $$O$$ in the ratio $$2:1$$, with the centroid closer to the orthocentre. The vector relation is:

$$$ \overrightarrow{G} = \frac{2\overrightarrow{O} + \overrightarrow{H}}{3} $$$

Since the circumcentre is at the origin, $$\overrightarrow{O} = \vec{0}$$, so:

$$$ \overrightarrow{G} = \frac{\overrightarrow{H}}{3} $$$

Thus,

$$$ \overrightarrow{H} = 3\overrightarrow{G} $$$

Therefore, the coordinates of the orthocentre $$H$$ are three times the coordinates of the centroid $$G$$:

$$$ x_H = 3 \times x_G = 3 \times \frac{(a + 1)^2}{2} = \frac{3(a + 1)^2}{2} $$$

$$$ y_H = 3 \times y_G = 3 \times \frac{(a - 1)^2}{2} = \frac{3(a - 1)^2}{2} $$$

So, $$H$$ is at $$\left( \frac{3(a + 1)^2}{2}, \frac{3(a - 1)^2}{2} \right)$$.

The problem requires finding the line on which $$H$$ lies for any $$a \neq 0$$. Substituting the coordinates of $$H$$ into each option will determine which equation is satisfied identically for all $$a$$.

Option A: $$y - (a^2 + 1)x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H - (a^2 + 1)x_H = \frac{3(a - 1)^2}{2} - (a^2 + 1) \cdot \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 - (a^2 + 1)(a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 = a^2 - 2a + 1 $$$

$$$ (a + 1)^2 = a^2 + 2a + 1 $$$

$$$ (a^2 + 1)(a + 1)^2 = (a^2 + 1)(a^2 + 2a + 1) = a^4 + 2a^3 + a^2 + a^2 + 2a + 1 = a^4 + 2a^3 + 2a^2 + 2a + 1 $$$

So,

$$$ (a - 1)^2 - (a^2 + 1)(a + 1)^2 = (a^2 - 2a + 1) - (a^4 + 2a^3 + 2a^2 + 2a + 1) = -a^4 - 2a^3 - a^2 - 4a $$$

This is not zero for all $$a$$, so Option A is incorrect.

Option B: $$y - 2a x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H - 2a x_H = \frac{3(a - 1)^2}{2} - 2a \cdot \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 - 2a (a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 = a^2 - 2a + 1 $$$

$$$ 2a (a + 1)^2 = 2a (a^2 + 2a + 1) = 2a^3 + 4a^2 + 2a $$$

So,

$$$ (a - 1)^2 - 2a (a + 1)^2 = (a^2 - 2a + 1) - (2a^3 + 4a^2 + 2a) = -2a^3 - 3a^2 - 4a + 1 $$$

This is not zero for all $$a$$, so Option B is incorrect.

Option C: $$y + x = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ y_H + x_H = \frac{3(a - 1)^2}{2} + \frac{3(a + 1)^2}{2} = \frac{3}{2} \left[ (a - 1)^2 + (a + 1)^2 \right] $$$

Expand:

$$$ (a - 1)^2 + (a + 1)^2 = (a^2 - 2a + 1) + (a^2 + 2a + 1) = 2a^2 + 2 $$$

So,

$$$ \frac{3}{2} \times (2a^2 + 2) = 3(a^2 + 1) $$$

This equals zero only if $$a^2 + 1 = 0$$, which is not true for real $$a$$, so Option C is incorrect.

Option D: $$(a - 1)^2 x - (a + 1)^2 y = 0$$

Substitute $$x_H$$ and $$y_H$$:

$$$ (a - 1)^2 x_H - (a + 1)^2 y_H = (a - 1)^2 \cdot \frac{3(a + 1)^2}{2} - (a + 1)^2 \cdot \frac{3(a - 1)^2}{2} $$$

Factor out $$\frac{3}{2}$$:

$$$ = \frac{3}{2} (a - 1)^2 (a + 1)^2 - \frac{3}{2} (a + 1)^2 (a - 1)^2 = 0 $$$

This is identically zero for all $$a \neq 0$$, so the orthocentre $$H$$ lies on this line for any $$a$$.

Hence, the correct answer is Option D.

The three given lines are:

1. $$x + 2ay + a = 0$$

2. $$x + 3by + b = 0$$

3. $$x + 4ay + a = 0$$

Since the lines are concurrent, they all pass through a common point. For three lines to be concurrent, the determinant of their coefficients must be zero. The general form of a line is $$px + qy + r = 0$$. For the given lines:

- Line 1: coefficients are $$p_1 = 1$$, $$q_1 = 2a$$, $$r_1 = a$$

- Line 2: coefficients are $$p_2 = 1$$, $$q_2 = 3b$$, $$r_2 = b$$

- Line 3: coefficients are $$p_3 = 1$$, $$q_3 = 4a$$, $$r_3 = a$$

The determinant condition is:

$$ \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4a & a \\ \end{vmatrix} = 0 $$

Expanding the determinant:

$$ = 1 \cdot \left( (3b)(a) - (b)(4a) \right) - 2a \cdot \left( (1)(a) - (b)(1) \right) + a \cdot \left( (1)(4a) - (3b)(1) \right) $$

Compute each part:

First part: $$ (3b)(a) - (b)(4a) = 3ab - 4ab = -ab $$

Second part: $$ -2a \cdot (a - b) = -2a(a - b) = -2a^2 + 2ab $$

Third part: $$ a \cdot (4a - 3b) = 4a^2 - 3ab $$

Summing them up:

$$ -ab + (-2a^2 + 2ab) + (4a^2 - 3ab) = -ab - 2a^2 + 2ab + 4a^2 - 3ab $$

Combine like terms:

For $$a^2$$ terms: $$ -2a^2 + 4a^2 = 2a^2 $$

For $$ab$$ terms: $$ -ab + 2ab - 3ab = (-1 + 2 - 3)ab = -2ab $$

So, the expression becomes:

$$ 2a^2 - 2ab = 0 $$

Factor out 2:

$$ 2(a^2 - ab) = 0 $$

Divide both sides by 2:

$$ a^2 - ab = 0 $$

Factor:

$$ a(a - b) = 0 $$

So, either $$a = 0$$ or $$a - b = 0$$ (i.e., $$a = b$$).

Now, check for distinct lines. If $$a = 0$$:

- Line 1: $$x + 2(0)y + 0 = 0$$ → $$x = 0$$

- Line 2: $$x + 3by + b = 0$$

- Line 3: $$x + 4(0)y + 0 = 0$$ → $$x = 0$$

Lines 1 and 3 are identical (both are $$x = 0$$), violating distinctness. Thus, $$a = 0$$ is invalid.

Therefore, $$a = b$$. The point $$(a, b)$$ satisfies $$b = a$$, meaning it lies on the straight line $$y = x$$.

Hence, the correct answer is Option C.

We have the two straight-line equations

$$4ax + 2ay + c = 0 \qquad\qquad (1)$$

$$5bx + 2by + d = 0 \qquad\qquad (2)$$

Because both $$a$$ and $$b$$ are non-zero, we can divide each equation by its respective non-zero constant so that the coefficients of $$x$$ and $$y$$ become simpler.

Dividing (1) by $$2a$$ gives

$$2x + y + \frac{c}{2a}=0 \; \; \Longrightarrow \; \; y = -2x - \frac{c}{2a} \qquad (3)$$

Dividing (2) by $$b$$ gives

$$5x + 2y + \frac{d}{b}=0 \; \; \Longrightarrow \; \; 2y = -5x - \frac{d}{b} \; \; \Longrightarrow \; \; y = -\frac52 x - \frac{d}{2b} \qquad (4)$$

The point of intersection satisfies both (3) and (4), so we equate the right-hand sides:

$$-2x - \frac{c}{2a} \;=\; -\frac52 x - \frac{d}{2b}$$

Multiplying every term by $$2$$ to clear the denominators, we obtain

$$-4x - \frac{c}{a} = -5x - \frac{d}{b}$$

Rearranging,

$$(-4x + 5x) = -\frac{d}{b} + \frac{c}{a} \; \; \Longrightarrow \; \; x = \frac{c}{a} - \frac{d}{b}$$

To combine the two fractions we take the common denominator $$ab$$:

$$x = \frac{cb - ad}{ab} \qquad (5)$$

Now we substitute this value of $$x$$ into equation (3) to obtain $$y$$.

Using (3):

$$y = -2x - \frac{c}{2a}$$

Substituting from (5):

$$\begin{aligned} y &= -2\left(\frac{cb - ad}{ab}\right) - \frac{c}{2a} \\ &= \frac{-2(cb - ad)}{ab} - \frac{c}{2a} \end{aligned}$$

To combine the two terms, we again choose the common denominator $$2ab$$:

$$\frac{-2(cb - ad)}{ab} = \frac{-4(cb - ad)}{2ab} = \frac{-4cb + 4ad}{2ab}$$

and

$$-\frac{c}{2a} = \frac{-cb}{2ab}$$

Adding these we get

$$y = \frac{-4cb + 4ad - cb}{2ab} = \frac{-5cb + 4ad}{2ab} \qquad (6)$$

The point lies in the fourth quadrant, so its $$x$$-coordinate is positive and its $$y$$-coordinate is negative. Moreover, the point is equidistant from the two coordinate axes. The distance of a point $$(x,\,y)$$ from the $$y$$-axis is $$|x|$$ and from the $$x$$-axis is $$|y|$$. Equidistance therefore gives the condition

$$|x| = |y|$$

Since the point is in the fourth quadrant, $$x > 0$$ and $$y < 0$$, so $$|x| = x$$ and $$|y| = -y$$. Hence

$$x = -y \qquad (7)$$

Substituting (5) and (6) into (7), we get

$$\frac{cb - ad}{ab} = -\left(\frac{-5cb + 4ad}{2ab}\right)$$

Multiplying both sides by $$2ab$$ to eliminate denominators yields

$$2(cb - ad) = 5cb - 4ad$$

Expanding the left side and then gathering like terms,

$$2cb - 2ad = 5cb - 4ad$$

Bringing all terms to one side,

$$2cb - 2ad - 5cb + 4ad = 0$$

$$(-3cb) + 2ad = 0$$

Multiplying through by $$-1$$ for convenience, we arrive at

$$3cb - 2ad = 0$$

Since multiplication is commutative, $$3cb = 3bc$$, so the final relation is

$$3bc - 2ad = 0$$

This matches Option A.

Hence, the correct answer is Option A.

We are given three points P, Q, and R. Point P is at (5, 3). Point R lies on the x-axis, so its y-coordinate is 0. Let the coordinates of R be (h, 0). The equation of line RQ is given as $$x - 2y = 2$$. Also, PQ is parallel to the x-axis. Since PQ is parallel to the x-axis and P has y-coordinate 3, the y-coordinate of Q must also be 3. Let the coordinates of Q be (a, 3).

Since R lies on the line RQ, we substitute R(h, 0) into the equation $$x - 2y = 2$$:

$$h - 2 \times 0 = 2$$

$$h = 2$$

So, R is at (2, 0).

Now, Q(a, 3) must also lie on the line RQ because RQ is the line joining R and Q. Substitute Q(a, 3) into the equation $$x - 2y = 2$$:

$$a - 2 \times 3 = 2$$

$$a - 6 = 2$$

$$a = 8$$

Therefore, Q is at (8, 3).

We now have the coordinates:

P(5, 3), Q(8, 3), R(2, 0)

The centroid of triangle PQR is found by averaging the x-coordinates and y-coordinates of the vertices. Let the centroid be G(x_g, y_g).

$$x_g = \frac{x_P + x_Q + x_R}{3} = \frac{5 + 8 + 2}{3} = \frac{15}{3} = 5$$

$$y_g = \frac{y_P + y_Q + y_R}{3} = \frac{3 + 3 + 0}{3} = \frac{6}{3} = 2$$

So, the centroid G is at (5, 2).

We need to determine which line this centroid lies on by substituting (5, 2) into each option.

Option A: $$x - 2y + 1 = 0$$

Substitute x = 5, y = 2:

$$5 - 2 \times 2 + 1 = 5 - 4 + 1 = 2 \neq 0$$

Not satisfied.

Option B: $$2x + y - 9 = 0$$

Substitute x = 5, y = 2:

$$2 \times 5 + 2 - 9 = 10 + 2 - 9 = 3 \neq 0$$

Not satisfied.

Option C: $$2x - 5y = 0$$

Substitute x = 5, y = 2:

$$2 \times 5 - 5 \times 2 = 10 - 10 = 0$$

Satisfied.

Option D: $$5x - 2y = 0$$

Substitute x = 5, y = 2:

$$5 \times 5 - 2 \times 2 = 25 - 4 = 21 \neq 0$$

Not satisfied.

Hence, the centroid lies on the line given in Option C: $$2x - 5y = 0$$. So, the answer is Option C.

A light ray starts at point P(1, 3) and reflects off the x-axis at point Q, then passes through R(6, 7). Since Q lies on the x-axis, its coordinates are (x, 0), and we need to find the abscissa x.

When a ray reflects off a surface, the angle of incidence equals the angle of reflection. For reflection over the x-axis, the mirror image of point R(6, 7) is R'(6, -7). The path from P to Q to R is equivalent to a straight line from P to R' because reflection preserves the path length and direction in the mirrored space. Therefore, Q is the point where the line segment joining P(1, 3) and R'(6, -7) intersects the x-axis.

To find the equation of the line passing through P(1, 3) and R'(6, -7), first calculate the slope m:

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 3}{6 - 1} = \frac{-10}{5} = -2 $$

Using the point-slope form with point P(1, 3):

$$ y - y_1 = m(x - x_1) $$

$$ y - 3 = -2(x - 1) $$

$$ y - 3 = -2x + 2 $$

$$ y = -2x + 2 + 3 $$

$$ y = -2x + 5 $$

This line intersects the x-axis where y = 0:

$$ 0 = -2x + 5 $$

$$ 2x = 5 $$

$$ x = \frac{5}{2} $$

Thus, the abscissa of Q is $$ \frac{5}{2} $$.

Alternatively, using the reflection property directly, let Q be (x, 0). The incident ray is from P(1, 3) to Q(x, 0), so its direction vector is (x - 1, -3). The reflected ray is from Q(x, 0) to R(6, 7), so its direction vector is (6 - x, 7). For reflection over the x-axis, the x-component of the direction remains the same, and the y-component reverses. Thus, the reflected direction should be proportional to (x - 1, 3) (since the incident y-component is -3, reversing gives +3). Therefore, the vectors (6 - x, 7) and (x - 1, 3) must be parallel, meaning their components are proportional:

$$ \frac{6 - x}{x - 1} = \frac{7}{3} $$

Cross-multiplying:

$$ 3(6 - x) = 7(x - 1) $$

$$ 18 - 3x = 7x - 7 $$

$$ 18 + 7 = 7x + 3x $$

$$ 25 = 10x $$

$$ x = \frac{25}{10} = \frac{5}{2} $$

Both methods confirm the abscissa of Q is $$ \frac{5}{2} $$.

Hence, the correct answer is Option D.

To solve this problem, we are given that the image of point P(2, 3) in a line L is Q(4, 5). We need to find the image of point R(0, 0) in the same line L. The image of a point in a line means that the line acts as the perpendicular bisector of the segment joining the point and its image. Therefore, for any point and its image, the line L must satisfy two conditions: it passes through the midpoint of the segment, and it is perpendicular to the segment.

First, we find the equation of line L using points P(2, 3) and Q(4, 5). The midpoint of segment PQ is calculated as follows:

Midpoint M = $$\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 4}{2}, \frac{3 + 5}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)$$.

Next, we find the slope of PQ. The slope is given by the change in y-coordinates divided by the change in x-coordinates:

Slope of PQ = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - 2} = \frac{2}{2} = 1$$.

Since line L is perpendicular to PQ, its slope is the negative reciprocal of the slope of PQ. Therefore, slope of L = $$-\frac{1}{1} = -1$$.

Now, line L passes through the midpoint M(3, 4) and has a slope of -1. Using the point-slope form of a line equation:

$$y - y_1 = m(x - x_1)$$

Substituting the values:

$$y - 4 = -1(x - 3)$$

Simplifying:

$$y - 4 = -x + 3$$

$$y = -x + 3 + 4$$

$$y = -x + 7$$

So, the equation of line L is $$y = -x + 7$$, which can be rewritten as $$x + y - 7 = 0$$.

Now, we need to find the image of point R(0, 0) in this line. Let the image be S(x, y). Since L is the perpendicular bisector of segment RS, the midpoint of RS must lie on L, and RS must be perpendicular to L.

First, the midpoint of RS (with R(0, 0) and S(x, y)) is:

Midpoint = $$\left( \frac{0 + x}{2}, \frac{0 + y}{2} \right) = \left( \frac{x}{2}, \frac{y}{2} \right)$$.

This midpoint must satisfy the equation of L: $$x + y - 7 = 0$$. Substituting the midpoint coordinates:

$$\frac{x}{2} + \frac{y}{2} - 7 = 0$$

Multiplying both sides by 2 to eliminate denominators:

$$x + y - 14 = 0 \quad \text{(Equation 1)}$$.

Next, RS must be perpendicular to L. The slope of L is -1 (from the equation $$y = -x + 7$$). The slope of RS is given by $$\frac{y - 0}{x - 0} = \frac{y}{x}$$.

Since the lines are perpendicular, the product of their slopes is -1:

$$(\text{slope of L}) \times (\text{slope of RS}) = -1$$

$$(-1) \times \left( \frac{y}{x} \right) = -1$$

$$-\frac{y}{x} = -1$$

Multiplying both sides by -1:

$$\frac{y}{x} = 1$$

$$y = x \quad \text{(Equation 2)}$$.

Now, we substitute Equation 2 into Equation 1:

$$x + y - 14 = 0$$

$$x + x - 14 = 0$$ (since y = x)

$$2x - 14 = 0$$

$$2x = 14$$

$$x = 7$$

Then, from Equation 2, y = x = 7.

Therefore, the image of R(0, 0) is (7, 7).

Comparing with the options:

A. (2, 2)

B. (4, 5)

C. (3, 4)

D. (7, 7)

The point (7, 7) corresponds to Option D.

Hence, the correct answer is Option D.

To solve this problem, we need to evaluate two statements about the angles between given lines. Recall that the angle between two lines depends only on their slopes, not on the constant terms in their equations. The slope of a line given by $$ax + by + c = 0$$ is $$m = -\frac{a}{b}$$.

First, consider the line $$2x + 3y + c_1 = 0$$. Its slope is $$m_1 = -\frac{2}{3}$$. Now, the line $$-x + 5y + c_2 = 0$$ has slope $$m_2 = -\frac{-1}{5} = \frac{1}{5}$$. Similarly, the line $$-x + 5y + c_3 = 0$$ has slope $$m_3 = -\frac{-1}{5} = \frac{1}{5}$$. Notice that both lines $$-x + 5y + c_2 = 0$$ and $$-x + 5y + c_3 = 0$$ have the same slope $$\frac{1}{5}$$, regardless of the values of $$c_2$$ and $$c_3$$. This is because the constants $$c_2$$ and $$c_3$$ only shift the lines parallel to themselves without changing their slopes.

The angle $$\theta_1$$ is between the lines $$2x + 3y + c_1 = 0$$ (slope $$m_1 = -\frac{2}{3}$$) and $$-x + 5y + c_2 = 0$$ (slope $$m_2 = \frac{1}{5}$$). The formula for the tangent of the angle $$\theta$$ between two lines with slopes $$m_1$$ and $$m_2$$ is:

$$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

Substituting the slopes:

$$\tan \theta_1 = \left| \frac{-\frac{2}{3} - \frac{1}{5}}{1 + \left(-\frac{2}{3}\right)\left(\frac{1}{5}\right)} \right| = \left| \frac{ -\frac{10}{15} - \frac{3}{15} }{1 - \frac{2}{15}} \right| = \left| \frac{ -\frac{13}{15} }{ \frac{15}{15} - \frac{2}{15} } \right| = \left| \frac{ -\frac{13}{15} }{ \frac{13}{15} } \right| = \left| -1 \right| = 1$$

Similarly, the angle $$\theta_2$$ is between the lines $$2x + 3y + c_1 = 0$$ (slope $$m_1 = -\frac{2}{3}$$) and $$-x + 5y + c_3 = 0$$ (slope $$m_3 = \frac{1}{5}$$). Since $$m_3 = m_2 = \frac{1}{5}$$, the calculation is identical:

$$\tan \theta_2 = \left| \frac{-\frac{2}{3} - \frac{1}{5}}{1 + \left(-\frac{2}{3}\right)\left(\frac{1}{5}\right)} \right| = \left| \frac{ -\frac{10}{15} - \frac{3}{15} }{1 - \frac{2}{15}} \right| = \left| \frac{ -\frac{13}{15} }{ \frac{13}{15} } \right| = \left| -1 \right| = 1$$

Thus, $$\tan \theta_1 = \tan \theta_2 = 1$$, so $$\theta_1 = \theta_2 = 45^\circ$$ (since the angle between lines is acute and $$\tan^{-1}(1) = 45^\circ$$). Importantly, this result holds for any real numbers $$c_1$$, $$c_2$$, and $$c_3$$ because the constants do not affect the slopes.

Now, evaluate Statement-2: $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$. As shown above, since the slopes are identical and independent of $$c_2$$ and $$c_3$$, $$\theta_1$$ and $$\theta_2$$ are always equal. Thus, Statement-2 is true.

Statement-1 states that if $$c_2$$ and $$c_3$$ are proportional (i.e., $$c_3 = k c_2$$ for some constant $$k$$), then $$\theta_1 = \theta_2$$. Since we have established that $$\theta_1 = \theta_2$$ for all $$c_2$$ and $$c_3$$, this includes the case where $$c_2$$ and $$c_3$$ are proportional. Therefore, Statement-1 is also true.

However, Statement-2 provides the complete reason: $$\theta_1 = \theta_2$$ because the slopes are fixed regardless of $$c_2$$ and $$c_3$$. Statement-1 only specifies a subset of cases (proportional constants), but the equality holds universally. Thus, Statement-2 explains why Statement-1 is true, as the condition in Statement-1 is sufficient but not necessary, and the underlying reason is the slope invariance captured by Statement-2.

Hence, both statements are true, and Statement-2 is a correct explanation for Statement-1.

So, the answer is Option A.

The incident ray is given by the straight‐line equation $$x + \sqrt{3}\,y = \sqrt{3}.$$

First we write this equation in the more familiar $$y = mx + c$$ form. We simply isolate $$y$$:

We have

$$x + \sqrt{3}\,y = \sqrt{3}.$$

Subtract $$x$$ from both sides:

$$\sqrt{3}\,y = \sqrt{3} - x.$$

Now divide every term by $$\sqrt{3}$$:

$$y = \frac{\sqrt{3} - x}{\sqrt{3}} = -\frac{1}{\sqrt{3}}\,x + 1.$$

So the slope of the incident ray is $$m_{\text{incident}} = -\dfrac{1}{\sqrt{3}}.$$ However, for reflection from the $$x$$-axis we do not need the slope directly; we simply need to apply the geometrical property of reflection in the $$x$$-axis.

A reflection in the $$x$$-axis changes every point $$(x,\,y)$$ to $$(x,\,-y)$$. In other words:

$$y \;\longrightarrow\; -y.$$ This single rule gives the equation of the image (reflected) line immediately.

Starting again with the incident equation

$$x + \sqrt{3}\,y = \sqrt{3},$$

we replace $$y$$ by $$-y$$ to obtain the reflected line:

$$x + \sqrt{3}(-y) = \sqrt{3}.$$

Simplify the product $$\sqrt{3}(-y):$$

$$x - \sqrt{3}\,y = \sqrt{3}.$$

To put this result in a slightly cleaner form, we again isolate $$y$$. First subtract $$x$$ from both sides:

$$-\sqrt{3}\,y = \sqrt{3} - x.$$

Now divide by $$-\sqrt{3}$$ (remembering that dividing by a negative flips the sign):

$$y = \frac{x - \sqrt{3}}{\sqrt{3}}.$$

Finally multiply every term by $$\sqrt{3}$$ to clear the denominator:

$$\sqrt{3}\,y = x - \sqrt{3}.$$

This is exactly one of the options provided. Comparing with the choices, we see it matches Option D.

Hence, the correct answer is Option D.

To solve this problem, we need to find the condition on $$a$$ such that the three given lines $$x - 3y = p$$, $$ax + 2y = q$$, and $$ax + y = r$$ form a right-angled triangle. For three lines to form a triangle, they must intersect pairwise and not be concurrent. Additionally, for the triangle to be right-angled, one pair of lines must be perpendicular, meaning the product of their slopes must be $$-1$$.

First, we find the slopes of each line by rewriting them in the slope-intercept form $$y = mx + c$$, where $$m$$ is the slope.

For the first line $$x - 3y = p$$:
Rearrange: $$x - p = 3y$$ → $$y = \frac{1}{3}x - \frac{p}{3}$$.
So, slope $$m_1 = \frac{1}{3}$$.

For the second line $$ax + 2y = q$$:
Rearrange: $$2y = -ax + q$$ → $$y = -\frac{a}{2}x + \frac{q}{2}$$.
So, slope $$m_2 = -\frac{a}{2}$$.

For the third line $$ax + y = r$$:
Rearrange: $$y = -ax + r$$.
So, slope $$m_3 = -a$$.

Since the triangle is right-angled, one pair of lines must be perpendicular. We check all possible pairs:

Case 1: Lines 1 and 2 are perpendicular.
Set $$m_1 \cdot m_2 = -1$$:
$$\frac{1}{3} \cdot \left(-\frac{a}{2}\right) = -1$$ → $$-\frac{a}{6} = -1$$.
Multiply both sides by $$-6$$: $$a = 6$$.

Case 2: Lines 1 and 3 are perpendicular.
Set $$m_1 \cdot m_3 = -1$$:
$$\frac{1}{3} \cdot (-a) = -1$$ → $$-\frac{a}{3} = -1$$.
Multiply both sides by $$-3$$: $$a = 3$$.

Case 3: Lines 2 and 3 are perpendicular.
Set $$m_2 \cdot m_3 = -1$$:
$$\left(-\frac{a}{2}\right) \cdot (-a) = -1$$ → $$\frac{a^2}{2} = -1$$.
But $$\frac{a^2}{2}$$ is always non-negative for real $$a$$, so it cannot equal $$-1$$. Thus, no solution.

So, possible values are $$a = 3$$ or $$a = 6$$. Now, we must ensure that for these values, the lines form a triangle (i.e., they are not concurrent). The lines are concurrent if the determinant of their coefficients is zero:

$$ \begin{vmatrix} 1 & -3 & -p \\ a & 2 & -q \\ a & 1 & -r \\ \end{vmatrix} = 0 $$

Expanding the determinant:
$$1 \cdot (2 \cdot (-r) - (-q) \cdot 1) - (-3) \cdot (a \cdot (-r) - (-q) \cdot a) + (-p) \cdot (a \cdot 1 - 2 \cdot a)$$
$$= 1 \cdot (-2r + q) + 3 \cdot (-a r + a q) - p \cdot (-a)$$
$$= -2r + q - 3a r + 3a q + a p$$
Set to zero: $$a p + 3a q - 3a r + q - 2r = 0$$.

Since $$p$$, $$q$$, and $$r$$ are arbitrary constants, we can choose values such that this expression is not zero. For example:

• For $$a = 3$$, choose $$p = 1$$, $$q = 1$$, $$r = 1$$: Lines are $$x - 3y = 1$$, $$3x + 2y = 1$$, $$3x + y = 1$$. Solving the last two: $$3x + 2y = 1$$ and $$3x + y = 1$$ gives $$y = 0$$, $$x = \frac{1}{3}$$. Plugging into first: $$\frac{1}{3} - 0 = \frac{1}{3} \neq 1$$, so not concurrent.
• For $$a = 6$$, choose $$p = 1$$, $$q = 1$$, $$r = 1$$: Lines are $$x - 3y = 1$$, $$6x + 2y = 1$$, $$6x + y = 1$$. Solving the last two: $$6x + 2y = 1$$ and $$6x + y = 1$$ gives $$y = 0$$, $$x = \frac{1}{6}$$. Plugging into first: $$\frac{1}{6} - 0 = \frac{1}{6} \neq 1$$, so not concurrent.

Thus, for both $$a = 3$$ and $$a = 6$$, we can choose $$p$$, $$q$$, and $$r$$ such that the lines form a triangle. Since one pair is perpendicular in each case, the triangle is right-angled.

The values $$a = 3$$ and $$a = 6$$ must satisfy a quadratic equation. Checking the options:

• Option A: $$a^2 - 9a + 18 = 0$$ → $$(a - 3)(a - 6) = 0$$, roots $$a = 3, 6$$.
• Option B: $$a^2 - 6a - 12 = 0$$ → discriminant $$36 + 48 = 84$$, roots $$\frac{6 \pm \sqrt{84}}{2} = 3 \pm \sqrt{21}$$, not 3 or 6.
• Option C: $$a^2 - 6a - 18 = 0$$ → discriminant $$36 + 72 = 108$$, roots $$\frac{6 \pm \sqrt{108}}{2} = 3 \pm 3\sqrt{3}$$, not 3 or 6.
• Option D: $$a^2 - 9a + 12 = 0$$ → discriminant $$81 - 48 = 33$$, roots $$\frac{9 \pm \sqrt{33}}{2}$$, not 3 or 6.

Hence, the correct answer is Option A.

First, we need to find the x-intercept and y-intercept of the given reference line $$3x + 4y = 12$$.

To find the x-intercept, set $$y = 0$$ and solve for $$x$$:

$$$ 3x + 4(0) = 12 \implies 3x = 12 \implies x = \frac{12}{3} = 4 $$$

So, the x-intercept of the reference line is 4.

To find the y-intercept, set $$x = 0$$ and solve for $$y$$:

$$$ 3(0) + 4y = 12 \implies 4y = 12 \implies y = \frac{12}{4} = 3 $$$

So, the y-intercept of the reference line is 3.

Now, for line $$L$$, its x-intercept is double that of the reference line. Double of 4 is $$2 \times 4 = 8$$.

The y-intercept of $$L$$ is half that of the reference line. Half of 3 is $$\frac{1}{2} \times 3 = \frac{3}{2}$$.

The equation of a line in intercept form is $$\frac{x}{a} + \frac{y}{b} = 1$$, where $$a$$ is the x-intercept and $$b$$ is the y-intercept. For line $$L$$, $$a = 8$$ and $$b = \frac{3}{2}$$, so the equation is:

$$$ \frac{x}{8} + \frac{y}{\frac{3}{2}} = 1 $$$

Simplify $$\frac{y}{\frac{3}{2}}$$ by multiplying by the reciprocal: $$\frac{y}{\frac{3}{2}} = y \times \frac{2}{3} = \frac{2y}{3}$$. So the equation becomes:

$$$ \frac{x}{8} + \frac{2y}{3} = 1 $$$

To find the slope, we convert this to slope-intercept form $$y = mx + c$$, where $$m$$ is the slope. Solve for $$y$$:

Subtract $$\frac{x}{8}$$ from both sides:

$$$ \frac{2y}{3} = 1 - \frac{x}{8} $$$

Multiply both sides by 3 to eliminate the denominator:

$$$ 2y = 3 \left(1 - \frac{x}{8}\right) $$$

Distribute the 3:

$$$ 2y = 3 \times 1 - 3 \times \frac{x}{8} = 3 - \frac{3x}{8} $$$

Divide both sides by 2:

$$$ y = \frac{1}{2} \left(3 - \frac{3x}{8}\right) = \frac{3}{2} - \frac{3x}{16} $$$

Rewrite as:

$$$ y = -\frac{3}{16}x + \frac{3}{2} $$$

Comparing to $$y = mx + c$$, the slope $$m$$ is $$-\frac{3}{16}$$.

Looking at the options:

A. $$-3$$

B. $$-\frac{3}{8}$$

C. $$-\frac{3}{2}$$

D. $$-\frac{3}{16}$$

The slope $$-\frac{3}{16}$$ matches option D. Hence, the correct answer is Option D.

We are told that the three mid-points of the sides of some triangle are $$P(0,1),\;Q(1,1),\;R(1,0).$$ Our first task is to recover the actual vertices of the triangle from these mid-points.

For any triangle, if $$P,Q,R$$ are the mid-points of $$BC,CA,AB$$ respectively, then vectors give the relations

$$A = Q + R - P,\qquad B = P + R - Q,\qquad C = P + Q - R.$$

Substituting the given mid-points:

$$A = (1,1) + (1,0) - (0,1) = (2,0),$$

$$B = (0,1) + (1,0) - (1,1) = (0,0),$$

$$C = (0,1) + (1,1) - (1,0) = (0,2).$$

So the triangle’s vertices are $$A(2,0),\;B(0,0),\;C(0,2).$$

Next we need the lengths of the sides opposite these vertices. Using the distance formula $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$$ we compute

Side $$a$$ (opposite $$A$$) is $$BC$$: $$a = \sqrt{(0-0)^2+(0-2)^2}=2.$$

Side $$b$$ (opposite $$B$$) is $$CA$$: $$b = \sqrt{(0-2)^2+(2-0)^2}= \sqrt{4+4}=2\sqrt{2}.$$

Side $$c$$ (opposite $$C$$) is $$AB$$: $$c = \sqrt{(2-0)^2+(0-0)^2}=2.$$

For the coordinates of the incentre we use the well-known formula: if the vertices are $$A(x_1,y_1),\;B(x_2,y_2),\;C(x_3,y_3)$$ with opposite sides $$a,b,c,$$ then

$$I\left(\dfrac{ax_1+bx_2+cx_3}{a+b+c},\; \dfrac{ay_1+by_2+cy_3}{a+b+c}\right).$$

We need only the $$x$$-coordinate. Putting in the values we have just found:

Numerator:

$$ax_1 = 2\cdot 2 = 4,\qquad bx_2 = 2\sqrt{2}\cdot 0 = 0,\qquad cx_3 = 2\cdot 0 = 0.$$

Hence $$ax_1+bx_2+cx_3 = 4.$$

Denominator:

$$a+b+c = 2 + 2\sqrt{2} + 2 = 4 + 2\sqrt{2}.$$

Therefore

$$x_I = \dfrac{4}{4 + 2\sqrt{2}} = \dfrac{4}{2(2+\sqrt{2})} = \dfrac{2}{2+\sqrt{2}}.$$

To rationalise the denominator we multiply the numerator and the denominator by $$2-\sqrt{2}:$$

$$x_I = \dfrac{2}{2+\sqrt{2}}\times\dfrac{2-\sqrt{2}}{2-\sqrt{2}} = \dfrac{2(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} = \dfrac{2(2-\sqrt{2})}{4-2} = \dfrac{2(2-\sqrt{2})}{2} = 2-\sqrt{2}.$$

Thus the $$x$$-coordinate of the incentre is $$2-\sqrt{2}.$$

Hence, the correct answer is Option D.

We are given an isosceles triangle with the base endpoints at (2a, 0) and (0, a). One side is given by the equation x = 2a. Since the point (2a, 0) lies on this line, and the other endpoint (0, a) does not, the side x = 2a must be one of the legs, not the base. Therefore, the third vertex, let's call it C, must lie on the line x = 2a. So, the coordinates of C are (2a, k) for some k.

In an isosceles triangle with base AB (where A is (2a, 0) and B is (0, a)), the two legs AC and BC must be equal. So, we set AC = BC.

First, calculate AC. The distance between A(2a, 0) and C(2a, k) is the absolute difference in y-coordinates since x-coordinates are the same: AC = |k - 0| = |k|.

Next, calculate BC. The distance between B(0, a) and C(2a, k) is found using the distance formula: BC = √[(2a - 0)² + (k - a)²] = √[4a² + (k - a)²].

Set AC equal to BC:

$$ |k| = \sqrt{4a^2 + (k - a)^2} $$

Since distances are positive, we can square both sides to eliminate the square root:

$$ k^2 = 4a^2 + (k - a)^2 $$

Expand the right side:

$$ k^2 = 4a^2 + k^2 - 2ak + a^2 $$

Simplify:

$$ k^2 = k^2 + 5a^2 - 2ak $$

Subtract k² from both sides:

$$ 0 = 5a^2 - 2ak $$

Rearrange:

$$ 2ak = 5a^2 $$

Assuming a ≠ 0 (as a = 0 would make the triangle degenerate), divide both sides by a:

$$ 2k = 5a $$

Solve for k:

$$ k = \frac{5a}{2} $$

So, the coordinates of C are (2a, 5a/2).

Now, we have the three vertices: A(2a, 0), B(0, a), and C(2a, 5a/2). To find the area of the triangle, use the formula for the area given three points (x₁, y₁), (x₂, y₂), (x₃, y₃):

$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$

Assign A(2a, 0) as (x₁, y₁), B(0, a) as (x₂, y₂), and C(2a, 5a/2) as (x₃, y₃). Substitute the values:

$$ \text{Area} = \frac{1}{2} \left| (2a)(a - \frac{5a}{2}) + (0)(\frac{5a}{2} - 0) + (2a)(0 - a) \right| $$

Simplify each term inside the absolute value:

First term: (2a)(a - 5a/2) = (2a)(-3a/2) = -3a²

Second term: (0)(5a/2) = 0

Third term: (2a)(-a) = -2a²

Sum the terms:

$$ -3a^2 + 0 - 2a^2 = -5a^2 $$

Take the absolute value:

$$ |-5a^2| = 5a^2 $$

Now multiply by 1/2:

$$ \text{Area} = \frac{1}{2} \times 5a^2 = \frac{5a^2}{2} $$

So, the area of the triangle is (5/2)a² square units.

Comparing with the options:

A. (5/4)a²

B. (5/2)a²

C. (25/4)a²

D. 5a²

The area matches option B. Hence, the correct answer is Option B.

We are given vertices A(-3, 2) and B(-2, 1) of triangle ABC. The centroid of the triangle lies on the line $$3x + 4y + 2 = 0$$. We need to find the line on which vertex C(x, y) lies.

The centroid G of a triangle is the average of the coordinates of its vertices. So, the coordinates of G are:

$$ G_x = \frac{x_A + x_B + x_C}{3} = \frac{-3 + (-2) + x}{3} = \frac{-5 + x}{3} $$

$$ G_y = \frac{y_A + y_B + y_C}{3} = \frac{2 + 1 + y}{3} = \frac{3 + y}{3} $$

Since the centroid lies on the line $$3x + 4y + 2 = 0$$, we substitute the coordinates of G into this equation:

$$ 3 \left( \frac{-5 + x}{3} \right) + 4 \left( \frac{3 + y}{3} \right) + 2 = 0 $$

Simplify each term:

$$ 3 \times \frac{-5 + x}{3} = -5 + x $$

$$ 4 \times \frac{3 + y}{3} = \frac{4(3 + y)}{3} $$

So the equation becomes:

$$ (-5 + x) + \frac{4(3 + y)}{3} + 2 = 0 $$

To eliminate the fraction, multiply every term by 3:

$$ 3 \times (-5 + x) + 3 \times \frac{4(3 + y)}{3} + 3 \times 2 = 0 \times 3 $$

Simplify:

$$ 3(-5 + x) + 4(3 + y) + 6 = 0 $$

Expand the expressions:

$$ 3 \times (-5) + 3 \times x + 4 \times 3 + 4 \times y + 6 = 0 $$

$$ -15 + 3x + 12 + 4y + 6 = 0 $$

Combine the constant terms:

$$ -15 + 12 = -3 $$

$$ -3 + 6 = 3 $$

So the equation is:

$$ 3x + 4y + 3 = 0 $$

This is the equation that vertex C(x, y) must satisfy. Comparing with the options:

A. $$4x + 3y + 5 = 0$$

B. $$3x + 4y + 3 = 0$$

C. $$4x + 3y + 3 = 0$$

D. $$3x + 4y + 5 = 0$$

Option B matches the derived equation. Hence, the correct answer is Option B.

The two given lines are
  $$x + y = 1 \; \; -(1)$$
  $$2(x+y) = 3 \;\Longrightarrow\; x + y = \frac32 \; \; -(2)$$

Write each line in the form $$L=0$$:
  $$L_1 : x + y - 1 = 0$$
  $$L_2 : x + y - \frac32 = 0$$

A point $$P(x_0,y_0)$$ lies between two non-parallel lines $$L_1=0$$ and $$L_2=0$$ if the algebraic distances of the point from the lines have opposite signs, i.e.
  $$L_1(x_0,y_0)\;L_2(x_0,y_0) \lt 0.$$ This is because the product is negative only when the point is on opposite sides of the two lines.

For the point $$P(1,a)$$:
  $$L_1(1,a)=1+a-1=a$$
  $$L_2(1,a)=1+a-\frac32=a-\frac12$$

Condition for lying between the lines:
  $$a\left(a-\frac12\right) \lt 0$$

The product of two real numbers is negative when the numbers have opposite signs. Thus
  $$a \gt 0 \quad\text{and}\quad a-\frac12 \lt 0 \quad\Longrightarrow\quad 0 \lt a \lt \frac12$$

Hence $$a$$ must lie in the open interval $$\left(0,\frac12\right).$$

Option D which is: $$\left(0, \dfrac{1}{2}\right)$$

The three sides of the triangle are the pair-wise intersections of the given lines.

Vertex A: intersection of $$x+3y=4$$ and $$3x+y=4$$.
Solve simultaneously: $$x+3y=4 \quad -(1)$$ $$3x+y=4 \quad -(2)$$ Multiply (1) by 3 and subtract (2): $$8y=8 \Rightarrow y=1$$. Insert in (2): $$3x+1=4 \Rightarrow x=1$$.
Hence $$A(1,\,1)$$.

Vertex B: intersection of $$3x+y=4$$ and $$x+y=0$$. Subtract: $$2x=4 \Rightarrow x=2$$, then $$y=-x=-2$$. So $$B(2,\,-2)$$.

Vertex C: intersection of $$x+3y=4$$ and $$x+y=0$$. Subtract: $$2y=4 \Rightarrow y=2$$, then $$x=-y=-2$$. Thus $$C(-2,\,2)$$.

Now compute the side lengths using the distance formula $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.

$$AB=\sqrt{(2-1)^2+(-2-1)^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}$$

$$AC=\sqrt{(-2-1)^2+(2-1)^2}=\sqrt{(-3)^2+1^2}=\sqrt{10}$$

$$BC=\sqrt{(-2-2)^2+(2+2)^2}=\sqrt{(-4)^2+4^2}=\sqrt{16+16}=4\sqrt{2}$$

Two sides are equal: $$AB=AC=\sqrt{10}$$, while $$BC=4\sqrt{2}$$ is different. Hence the triangle is isosceles.

To check for a right angle, note that the slopes of $$AB$$ and $$AC$$ are $$-3$$ and $$-\frac13$$, whose product is $$1 \neq -1$$. Therefore no angle is $$90^{\circ}$$, so the triangle is not right-angled.

Thus the triangle is isosceles.

Option C which is: isosceles

Let the given vertices be $$A(5,-1)$$ and $$B(-2,3)$$, and let the unknown third vertex be $$C(x,y)$$. The orthocentre is $$H(0,0)$$, i.e. the common intersection point of the three altitudes of the triangle.

Step 1: Slope of altitude through A
The altitude from $$A$$ passes through $$A(5,-1)$$ and $$H(0,0)$$.
Slope of this altitude, $$m_{AH}=\dfrac{-1-0}{5-0}=-\dfrac15$$.

This altitude is perpendicular to side $$BC$$, so
$$m_{BC}\;=\;-\dfrac1{m_{AH}}=5.$$

Therefore, for points $$B(-2,3)$$ and $$C(x,y)$$, we have
$$m_{BC}=\dfrac{y-3}{\,x+2\,}=5\quad\Longrightarrow\quad y-3=5(x+2).$$ Hence

$$y = 5x + 13 \quad -(1)$$

Step 2: Slope of altitude through B
The altitude from $$B$$ passes through $$B(-2,3)$$ and $$H(0,0)$$.
Slope of this altitude, $$m_{BH}=\dfrac{3-0}{-2-0}=-\dfrac32$$.

This altitude is perpendicular to side $$AC$$, so
$$m_{AC}\;=\;-\dfrac1{m_{BH}}=\dfrac23.$$

Therefore, for points $$A(5,-1)$$ and $$C(x,y)$$, we have
$$m_{AC}=\dfrac{y+1}{\,x-5\,}=\dfrac23\quad\Longrightarrow\quad 3(y+1)=2(x-5).$$ Hence

$$3y + 3 = 2x - 10 \;\Longrightarrow\; 3y = 2x - 13 \;\Longrightarrow\; y=\dfrac{2x-13}{3}\quad -(2)$$

Step 3: Solve simultaneous equations for C(x, y)
Equate the two expressions for $$y$$ from (1) and (2):

$$5x + 13 = \dfrac{2x-13}{3}.$$

Multiply by 3:
$$15x + 39 = 2x - 13.$$

$$13x = -52 \;\Longrightarrow\; x = -4.$$

Substitute $$x=-4$$ in (1):
$$y = 5(-4) + 13 = -20 + 13 = -7.$$

Thus the third vertex is $$C(-4,-7)$$.

Option B which is: $$(-4,\,-7)$$

The required point is the internal division point of $$(1,1)$$ and $$(2,4)$$ in the ratio $$3:2$$ (first point’s weight 3, second point’s weight 2).

Section-formula for internal division:
If $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are divided in ratio $$m:n$$, the coordinates are
$$\left(\frac{mx_2+nx_1}{m+n},\; \frac{my_2+ny_1}{m+n}\right).$$

Here $$m=3,\; n=2,\;(x_1,y_1)=(1,1),\;(x_2,y_2)=(2,4).$$
Hence the coordinates become
$$x=\frac{3\cdot2+2\cdot1}{3+2}= \frac{6+2}{5}= \frac{8}{5},$$
$$y=\frac{3\cdot4+2\cdot1}{3+2}= \frac{12+2}{5}= \frac{14}{5}.$$

This point $$(\frac{8}{5},\frac{14}{5})$$ lies on the line $$2x+y=k$$, so substitute:

$$2\left(\frac{8}{5}\right)+\frac{14}{5}=k \; \Longrightarrow \; \frac{16}{5}+\frac{14}{5}=k \; \Longrightarrow \; \frac{30}{5}=k.$$

Thus $$k=6$$.

Option C which is: $$6$$

The required line is parallel to the $$x$$-axis, so we only need the $$y$$-coordinate of the point of intersection of the two given lines.

Lines:
$$ax + 2by + 3b = 0$$ $$-(1)$$
$$bx - 2ay - 3a = 0$$ $$-(2)$$

Rewrite each in the form $$px + qy = r$$:

From $$(1): \; ax + 2by = -3b$$ $$-(3)$$
From $$(2): \; bx - 2ay = 3a$$ $$-(4)$$

Multiply $$(3)$$ by $$b$$ and $$(4)$$ by $$a$$ so that the $$x$$ terms become equal:

$$abx + 2b^{2}y = -3b^{2}$$ $$-(5)$$
$$abx - 2a^{2}y = 3a^{2}$$ $$-(6)$$

Subtract $$(6)$$ from $$(5)$$ to eliminate $$x$$:

$$(abx + 2b^{2}y) - (abx - 2a^{2}y) = -3b^{2} - 3a^{2}$$
$$2b^{2}y + 2a^{2}y = -3(b^{2} + a^{2})$$

Factor out $$2(a^{2}+b^{2})$$ on the left (note that $$(a,b)\neq(0,0)\Rightarrow a^{2}+b^{2}\neq0$$):

$$2(a^{2}+b^{2})y = -3(a^{2}+b^{2})$$

Divide both sides by $$2(a^{2}+b^{2})$$:

$$y = -\frac{3}{2}$$

Thus the intersection point has $$y$$-coordinate $$-\dfrac{3}{2}$$, independent of $$a$$ and $$b$$. A line parallel to the $$x$$-axis through this point is

$$y = -\frac{3}{2}$$

The constant $$y$$ is negative, so the line lies below the $$x$$-axis. Its perpendicular distance from the $$x$$-axis is $$\left|\,-\dfrac{3}{2}\,\right| = \dfrac{3}{2}$$.

Therefore, the correct description is: below $$x$$-axis at a distance $$3/2$$ from it.
Option C which is: below $$x$$-axis at a distance $$3/2$$ from it.

Let the required line meet the $$x$$-axis at $$P(a,0)$$ and the $$y$$-axis at $$Q(0,b)$$, where $$a \gt 0,\, b \gt 0$$.
The equation of the line in intercept form is

$$\frac{x}{a}+\frac{y}{b}=1 \qquad -(1)$$

Because the line passes through the given point $$(1,2)$$, substitute $$x = 1,\; y = 2$$ in $$-(1)$$:

$$\frac{1}{a}+\frac{2}{b}=1 \qquad -(2)$$

The area of the triangle $$OPQ$$ formed with the coordinate axes is

$$A=\frac{1}{2}\,|a|\,|b|=\frac{1}{2}\,ab \qquad -(3)$$

We have to minimise $$A$$ subject to the condition $$-(2)$$.

Solve $$-(2)$$ for $$a$$:

$$\frac{1}{a}=1-\frac{2}{b}=\frac{b-2}{b}\;\; \Longrightarrow\;\; a=\frac{b}{\,b-2\,}, \quad b \gt 2$$

Substitute this in $$-(3)$$ to express $$A$$ only in terms of $$b$$:

$$A(b)=\frac{1}{2}\,\frac{b}{\,b-2\,}\,b=\frac{1}{2}\,\frac{b^{2}}{\,b-2\,} \qquad -(4)$$

To find the minimum, differentiate:

$$\frac{dA}{db}=\frac{1}{2}\,\frac{(2b)(b-2)-b^{2}}{(b-2)^{2}}=\frac{1}{2}\,\frac{2b(b-2)-b^{2}}{(b-2)^{2}}=\frac{1}{2}\,\frac{b(b-4)}{(b-2)^{2}}$$

Set $$\dfrac{dA}{db}=0$$:

$$b(b-4)=0 \;\;\Longrightarrow\;\; b=0 \text{ or } b=4$$

The admissible range is $$b \gt 2$$, hence $$b=4$$.

With $$b=4$$, equation $$a=\dfrac{b}{b-2}$$ gives

$$a=\frac{4}{4-2}=2$$

Slope of the line $$PQ$$:

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

Thus, the line for which the triangle $$OPQ$$ has the least area has slope $$-2$$.

Option C which is: $$-2$$

First, write each line in the slope-intercept form $$y = mx + c$$ to read its slope quickly.

$$L_1:\;x - y = 1 \;\Longrightarrow\; y = x - 1 \;\; \Rightarrow\; m_1 = 1$$

$$L_2:\;x + y = 1 \;\Longrightarrow\; y = -x + 1 \;\Rightarrow\; m_2 = -1$$

$$L_3:\;2x + 2y = 5 \;\Longrightarrow\; x + y = \tfrac{5}{2} \;\Longrightarrow\; y = -x + \tfrac{5}{2} \;\Rightarrow\; m_3 = -1$$

$$L_4:\;2x - 2y = 7 \;\Longrightarrow\; x - y = \tfrac{7}{2} \;\Longrightarrow\; y = x - \tfrac{7}{2} \;\Rightarrow\; m_4 = 1$$

Now apply the two standard tests.

1. Parallel lines have equal slopes: $$m_p = m_q$$.
2. Perpendicular lines satisfy $$m_p \, m_q = -1$$.

$$m_1 \cdot m_2 = 1 \times (-1) = -1 \;\Rightarrow\; L_1 \perp L_2$$

$$m_1 \cdot m_3 = 1 \times (-1) = -1 \;\Rightarrow\; L_1 \perp L_3$$

$$m_2 = m_3 = -1 \;\Rightarrow\; L_2 \parallel L_3$$

$$m_1 = m_4 = 1 \;\Rightarrow\; L_1 \parallel L_4$$

$$m_2 \neq m_4 \quad$$ and $$\;m_2 \, m_4 = (-1)\times 1 = -1 \neq 1$$, so the slopes are different; therefore the two lines are not parallel, hence they meet. (Their product being $$-1$$ also shows they are perpendicular.) Thus $$L_2$$ intersects $$L_4$$.

Summarising:

• $$L_1 \perp L_2$$
• $$L_1 \perp L_3$$
• $$L_2$$ intersects $$L_4$$ (they are perpendicular).
• Any other claimed relations in the options are incorrect (e.g., $$L_1$$ does not intersect $$L_4$$ because they are parallel).

Only Option D lists all three correct relations.

Final Answer: Option D which is: $$L_1 \perp L_2, \; L_1 \perp L_3, \; L_2$$ intersect $$L_4$$.

Let the feet of the two poles be $$A$$ and $$B$$ on a horizontal line AB.
Height of pole at $$A$$ = $$20 \text{ m}$$, so its top is $$P(0,\,20)$$.
Height of pole at $$B$$ = $$80 \text{ m}$$, so its top is $$Q(d,\,80)$$, where $$d=AB$$ (the horizontal separation, whose value is not needed).

Choose a coordinate system with $$A(0,\,0)$$ and the positive $$x$$-axis along AB. Then:
• $$P(0,\,20)$$ is the top of the shorter pole.
• $$Q(d,\,80)$$ is the top of the taller pole.

The two required lines are:
1. $$PQ_1$$ joining $$P(0,\,20)$$ to the foot of the tall pole $$B(d,\,0)$$.
2. $$QP_1$$ joining $$Q(d,\,80)$$ to the foot of the short pole $$A(0,\,0)$$.

Equation of $$PB$$ (using two-point form):
$$\frac{y-20}{0-20}=\frac{x-0}{d-0} \;\Longrightarrow\; y = 20\!\left(1-\frac{x}{d}\right)$$.

Equation of $$QA$$ (straight line through the origin):
$$\frac{y-0}{80-0}=\frac{x-0}{d-0} \;\Longrightarrow\; y = 80\,\frac{x}{d}$$.

Let the intersection point of these two lines be $$R(x_0,\,y_0)$$. Equate the $$y$$-coordinates:

$$20\!\left(1-\frac{x_0}{d}\right)=80\,\frac{x_0}{d}$$
$$\Longrightarrow\; 20-20\,\frac{x_0}{d}=80\,\frac{x_0}{d}$$
$$\Longrightarrow\; 20 = 100\,\frac{x_0}{d}$$
$$\Longrightarrow\; \frac{x_0}{d}=0.2$$.

Substitute this ratio into $$y = 80\,\dfrac{x}{d}$$ to get the height of $$R$$:

$$y_0 = 80 \times 0.2 = 16$$.

Thus the two lines intersect at a height of $$16 \text{ m}$$ above the ground, independent of the horizontal separation of the poles.

Option A which is: 16

The line $$L$$ is given in intercept form as $$\frac{x}{5} + \frac{y}{b} = 1$$ and it passes through $$(13,\,32)$$.

Substituting the point to find $$b$$:
$$\frac{13}{5} + \frac{32}{b} = 1 \implies \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$
$$\therefore\; b = \frac{32 \times 5}{-8} = -20$$

Hence the equation of $$L$$ is $$\frac{x}{5} - \frac{y}{20} = 1$$.
Multiplying by $$20$$ gives the general (ax + by + c = 0) form
$$4x - y - 20 = 0$$ $$-(1)$$

The parallel line $$K$$ is given as $$\frac{x}{c} + \frac{y}{3} = 1$$.
Convert to general form to compare slopes:
Multiply by $$3c$$: $$3x + cy = 3c$$
$$\Rightarrow y = -\frac{3}{c}x + 3$$

The slope of $$L$$ from $$-(1)$$ is $$4$$ (since $$y = 4x - 20$$).
For lines to be parallel, their slopes must be equal, so
$$-\frac{3}{c} = 4 \;\; \Longrightarrow \;\; c = -\frac{3}{4}$$

Substituting $$c = -\frac{3}{4}$$ back, the equation of $$K$$ becomes
$$-4x + y = 3 \;\; \text{or} \;\; 4x - y + 3 = 0$$ $$-(2)$$

The distance between two parallel lines $$ax + by + c_1 = 0$$ and $$ax + by + c_2 = 0$$ is
$$\displaystyle D = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$$

Here $$a = 4, \; b = -1, \; c_1 = -20 \;(\text{from } L), \; c_2 = 3 \;(\text{from } K).$$
$$D = \frac{|3 - (-20)|}{\sqrt{4^2 + (-1)^2}} = \frac{23}{\sqrt{16 + 1}} = \frac{23}{\sqrt{17}}$$

Therefore, the required distance is $$\frac{23}{\sqrt{17}}$$.
Option C which is: $$\frac{23}{\sqrt{17}}$$

The general form of a straight line is $$Ax + By + C = 0$$, whose slope is $$m = -\dfrac{A}{B}$$ (provided $$B \neq 0$$).

For the first line
$$p\bigl(p^{2}+1\bigr)x - y + q = 0$$
we have $$A_{1} = p(p^{2}+1)$$ and $$B_{1} = -1$$, so its slope is
$$m_{1} = -\dfrac{A_{1}}{B_{1}} = -\dfrac{p(p^{2}+1)}{-1} = p(p^{2}+1).$$

For the second line
$$(p^{2}+1)^{2}x + (p^{2}+1)\,y + 2q = 0$$
we have $$A_{2} = (p^{2}+1)^{2}$$ and $$B_{2} = (p^{2}+1)$$, so its slope is
$$m_{2} = -\dfrac{A_{2}}{B_{2}} = -\dfrac{(p^{2}+1)^{2}}{(p^{2}+1)} = -(p^{2}+1).$$

Let a third line with slope $$m$$ be perpendicular to both given lines. For perpendicularity we need

$$m\,m_{1} = -1 \quad\text{and}\quad m\,m_{2} = -1.$$ Cancelling $$m$$ gives $$m_{1} = m_{2}.$$

Therefore the two given lines themselves must be parallel:

$$p(p^{2}+1) = -(p^{2}+1).$$

Factorising,

$$(p^{2}+1)(p+1) = 0.$$

Since $$p^{2}+1 \gt 0$$ for all real $$p$$, the only real solution is

$$p = -1.$$

For $$p = -1$$ both lines have finite, equal slopes, so a unique line with slope $$m = -\dfrac{1}{m_{1}}$$ is perpendicular to them.

Hence there is exactly one real value of $$p$$ for which the two given lines are perpendicular to a common line.

Option B which is: exactly one value of $$p$$

The line segment $$PQ$$ joins the points $$P(1,4)$$ and $$Q(k,3)$$.

1. Mid-point of $$PQ$$
$$M\Bigl(\frac{1+k}{2},\frac{4+3}{2}\Bigr)=\Bigl(\frac{1+k}{2},\frac{7}{2}\Bigr).$$

2. Slope of $$PQ$$
$$m_{PQ}=\frac{3-4}{k-1}=\frac{-1}{k-1}.$$

3. Slope of the perpendicular bisector
If two lines are perpendicular, the product of their slopes is $$-1$$. Hence the slope of the perpendicular bisector is the negative reciprocal:
$$m_{\perp}=-(1/m_{PQ})=k-1.$$

4. Equation of the perpendicular bisector
Using point-slope form with point $$M$$ and slope $$m_{\perp}$$,
$$y-\frac{7}{2}=(k-1)\Bigl(x-\frac{1+k}{2}\Bigr).$$

5. Imposing the given y-intercept
The y-intercept is obtained by setting $$x=0$$.
Substituting $$x=0$$ in the above equation:
$$y=\frac{7}{2}+(k-1)\Bigl(-\frac{1+k}{2}\Bigr)$$ This y-value must be $$-4$$:
$$\frac{7}{2}-(k-1)\frac{1+k}{2}=-4.$$

Multiply by 2 to clear the denominator:
$$7-(k-1)(k+1)=-8.$$

Simplify:
$$7+8=(k-1)(k+1)$$ $$15=k^{2}-1$$ $$k^{2}=16$$ $$k=\pm4.$$

6. Selecting from the options
Both $$k=4$$ and $$k=-4$$ satisfy the condition, but only $$k=-4$$ appears among the given options.

Hence a possible value of $$k$$ is
Option D which is: $$-4$$.

Since $$AC$$ is the hypotenuse, the right angle must be at vertex $$B(1,1)$$. Hence the two sides through $$B$$, viz. $$\overline{AB}$$ and $$\overline{BC}$$, are perpendicular.

The slope of $$BC$$ is
$$m_{BC} = \frac{1-1}{2-1} = 0$$

A line perpendicular to a horizontal line (slope 0) is a vertical line whose slope is undefined. Therefore $$\overline{AB}$$ must be vertical, i.e. its $$x$$-coordinate is constant:
$$h = 1$$

Thus $$A$$ has coordinates $$(1,k)$$ and the two perpendicular legs of the right-angled triangle are
$$AB = |k-1|,\qquad BC = |2-1| = 1$$

The area of a right triangle is $$\dfrac12 \times(\text{product of the perpendicular legs})$$. Given that the area equals $$1$$, we have
$$\frac12 \times |k-1| \times 1 = 1$$
$$\Rightarrow\; |k-1| = 2$$

Solving the modulus equation:
$$k - 1 = 2 \;\;\text{or}\;\; k - 1 = -2$$
$$\Rightarrow\; k = 3 \;\;\text{or}\;\; k = -1$$

Hence the possible values of $$k$$ are $$\{-1,\,3\}$$.

Option C which is: $$\{-1, 3\}$$

For an angle bisector we need the two rays that form the angle and the vertex where they meet.
Here the vertex is $$Q(0,0)$$, the two rays are $$QP$$ and $$QR$$.

Step 1: Direction of $$QP$$
$$P(-1,0)$$ and $$Q(0,0)$$ give the vector $$\overrightarrow{QP} = (-1,0)$$.
This lies along the negative $$x$$-axis, so its direction makes an angle $$180^{\circ}$$ with the positive $$x$$-axis.
Slope of $$QP$$: $$m_1 = 0$$.

Step 2: Direction of $$QR$$
$$R(3,3\sqrt 3)$$ and $$Q(0,0)$$ give the vector $$\overrightarrow{QR} = (3,3\sqrt 3)$$.
Slope of $$QR$$: $$m_2 = \frac{3\sqrt 3}{3} = \sqrt 3$$.
The corresponding angle with the positive $$x$$-axis is $$\tan^{-1}(\sqrt 3)=60^{\circ}$$.

Step 3: Identify the interior angle at $$Q$$
Angles of the two rays measured from the positive $$x$$-axis are $$60^{\circ}$$ (for $$QR$$) and $$180^{\circ}$$ (for $$QP$$).
The smaller angle between them is $$180^{\circ}-60^{\circ}=120^{\circ}$$.
Hence the interior bisector will make half of this, i.e. $$\tfrac{120^{\circ}}{2}=60^{\circ}$$, with each ray.
Starting from $$60^{\circ}$$ and moving halfway towards $$180^{\circ}$$ takes us to
$$60^{\circ}+60^{\circ}=120^{\circ}$$.

Step 4: Slope of the interior bisector
Angle $$120^{\circ}$$ has slope $$\tan(120^{\circ}) = -\sqrt 3$$.
Therefore the bisector through the origin is
$$y = -\sqrt 3\,x$$.

Step 5: Write in general form
$$\sqrt 3\,x + y = 0$$.

Thus the equation of the bisector of $$\angle PQR$$ is
Option A which is: $$\sqrt{3}\,x + y = 0$$.

The coordinate axes are represented by the pair of lines $$xy = 0$$, i.e. $$x = 0$$ and $$y = 0$$. The bisectors of the angle between these two perpendicular lines are well known:

• the line $$y = x$$ (the internal bisector),
• the line $$y = -x$$ (the external bisector).

Hence, if one of the factors of the second-degree curve
$$my^2 + (1 - m^2)xy - mx^2 = 0$$
is a bisector, then either $$y = x$$ or $$y = -x$$ must satisfy the equation identically.

Let $$y = x$$. Substitute $$y = x$$ (so $$y^2 = x^2$$ and $$xy = x^2$$) into the given expression:

$$m\,x^2 + (1 - m^2)\,x^2 - m\,x^2 = 0$$
$$\Longrightarrow \bigl[m + (1 - m^2) - m\bigr]x^2 = 0$$
$$\Longrightarrow (1 - m^2)x^2 = 0.$$

Since $$x \neq 0$$ for a non-trivial line, we need
$$1 - m^2 = 0 \; \Longrightarrow \; m^2 = 1.$$

Let $$y = -x$$. Substitute $$y = -x$$ (so $$y^2 = x^2$$ and $$xy = -x^2$$) into the curve equation:

$$m\,x^2 + (1 - m^2)(-x^2) - m\,x^2 = 0$$
$$\Longrightarrow \bigl[m - (1 - m^2) - m\bigr]x^2 = 0$$
$$\Longrightarrow (m^2 - 1)x^2 = 0.$$

Again, for a genuine line $$x \neq 0$$, so we obtain
$$m^2 - 1 = 0 \; \Longrightarrow \; m^2 = 1.$$

Both cases lead to the single condition $$m^2 = 1$$, i.e. $$m = \pm1$$. Among the answer choices given, only $$m = 1$$ appears.

Therefore, the required value is
Option C which is: 1.

Any straight line that cuts the coordinate axes at points $$P(a,0)$$ on the $$x$$-axis and $$Q(0,b)$$ on the $$y$$-axis can be written in intercept form as
$$\frac{x}{a}+\frac{y}{b}=1$$

The segment $$PQ$$ is called the intercept of the line between the axes. Its midpoint is obtained by averaging the coordinates of $$P$$ and $$Q$$:
midpoint $$M=\left(\frac{a+0}{2},\,\frac{0+b}{2}\right)=\left(\frac{a}{2},\,\frac{b}{2}\right)$$.

According to the question, this midpoint coincides with the given point $$A(3,4)$$. Therefore
$$\frac{a}{2}=3 \quad\text{and}\quad \frac{b}{2}=4$$

Solving these gives the actual intercepts:
$$a=6,\; b=8$$

Substitute $$a$$ and $$b$$ back into the intercept form:
$$\frac{x}{6}+\frac{y}{8}=1$$

Eliminate denominators by multiplying throughout by the least common multiple $$24$$:
$$24\left(\frac{x}{6}+\frac{y}{8}\right)=24(1)$$
$$4x+3y=24$$

Thus the required equation of the line is $$4x+3y=24$$.

Option C which is: $$4x + 3y = 24$$

The point is $$P(a,\,a^{2})$$.

The two boundary lines are
  $$y=\tfrac{x}{2} \quad\text{and}\quad y=3x,$$
with the condition $$x\gt 0$$ (first-quadrant rays).

For any point whose abscissa is $$x=a\gt 0$$, the ordinates of the two boundary lines are
  Lower ray: $$y_L=\tfrac{a}{2},$$
  Upper ray: $$y_U=3a.$$

For the point $$P(a,a^{2})$$ to lie strictly inside the angle, its ordinate must satisfy
$$y_L \lt a^{2} \lt y_U.$$

That gives the double inequality
$$\frac{a}{2} \lt a^{2} \lt 3a.$$

Because $$a\gt 0$$, divide the two parts separately by $$a$$ (which keeps the inequality signs unchanged):

Left part: $$\frac{a^{2}}{a} \gt \frac{a}{2a}\; \Longrightarrow\; a \gt \frac{1}{2}.$$

Right part: $$\frac{a^{2}}{a} \lt \frac{3a}{a}\; \Longrightarrow\; a \lt 3.$$

Combining, we obtain
$$\frac{1}{2} \lt a \lt 3.$$

Hence $$a$$ lies in the open interval $$\left(\dfrac{1}{2},\,3\right).$$

Option C which is: $$\left(\dfrac{1}{2}, 3\right)$$

Let the three vertices of the triangle be $$A(1,1)$$ (given), $$B(x_B,y_B)$$ and $$C(x_C,y_C)$$.

The mid-point formula states that the mid-point $$M$$ of the segment joining $$P(x_1,y_1)$$ and $$Q(x_2,y_2)$$ is $$\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$$.

Given mid-point $$M_1(-1,2)$$ lies on $$AB$$, so

$$\left(\dfrac{x_B+1}{2},\dfrac{y_B+1}{2}\right)=(-1,2).$$

Equating coordinates: $$\dfrac{x_B+1}{2}=-1 \;\Rightarrow\; x_B+1=-2 \;\Rightarrow\; x_B=-3,$$ $$\dfrac{y_B+1}{2}=2 \;\Rightarrow\; y_B+1=4 \;\Rightarrow\; y_B=3.$$

Thus $$B(-3,3).$$

Given mid-point $$M_2(3,2)$$ lies on $$AC$$, so

$$\left(\dfrac{x_C+1}{2},\dfrac{y_C+1}{2}\right)=(3,2).$$

Equating coordinates: $$\dfrac{x_C+1}{2}=3 \;\Rightarrow\; x_C+1=6 \;\Rightarrow\; x_C=5,$$ $$\dfrac{y_C+1}{2}=2 \;\Rightarrow\; y_C+1=4 \;\Rightarrow\; y_C=3.$$

Thus $$C(5,3).$$

The centroid $$G(x_G,y_G)$$ of a triangle with vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ is

$$\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right).$$

Substituting $$A(1,1), B(-3,3), C(5,3):$$

$$x_G=\dfrac{1+(-3)+5}{3}=\dfrac{3}{3}=1,$$ $$y_G=\dfrac{1+3+3}{3}=\dfrac{7}{3}.$$

Therefore, the centroid is $$\left(1,\dfrac{7}{3}\right).$$

Option C which is: $$\left(1, \dfrac{7}{3}\right).$$

The equation $$ax^2 + 2(a+b)xy + by^2 = 0$$ represents two straight lines through the origin (centre of the circle).
For a homogeneous second-degree expression $$ax^2 + 2hxy + by^2 = 0$$ the two lines make an angle $$\theta$$ given by

$$\tan\theta = \left|\dfrac{2\sqrt{\,h^2-ab\,}}{a+b}\right| \; -(1)$$

Here $$h = a + b$$, so

$$\tan\theta = \left|\dfrac{2\sqrt{(a+b)^2 - ab}}{a + b}\right| = \left|\dfrac{2\sqrt{a^2 + ab + b^2}}{a + b}\right| \; -(2)$$

Because the two lines are diameters, they divide the circle into four sectors whose central angles are the angles between the lines and their supplements. Let the smaller angle between the lines be $$\theta$$; the opposite angle is also $$\theta$$, while the other two are $$\pi - \theta$$ each.

One sector has area thrice that of another, so the corresponding central angles are in the ratio $$3:1$$.
Let the smaller sector have angle $$\theta$$ and the larger sector have angle $$\pi - \theta$$. Then

$$(\pi - \theta) = 3\theta \;\;\Longrightarrow\;\; \pi = 4\theta \;\;\Longrightarrow\;\; \theta = \dfrac{\pi}{4}$$

Hence $$\theta = 45^\circ$$ and therefore $$\tan\theta = 1$$.

Substituting $$\tan\theta = 1$$ in equation $$-(2)$$:

$$\dfrac{2\sqrt{a^2 + ab + b^2}}{|a + b|} = 1$$

Square both sides:

$$4\left(a^2 + ab + b^2\right) = (a + b)^2$$

Expand and collect terms:

$$4a^2 + 4ab + 4b^2 - (a^2 + 2ab + b^2) = 0$$
$$3a^2 + 2ab + 3b^2 = 0$$

This is exactly the condition given in Option D.

Final Answer: Option D which is: $$3a^2 + 2ab + 3b^2 = 0$$

The two given lines are
$$a x + 2 b y + 3 b = 0 \qquad -(1)$$
$$b x - 2 a y - 3 a = 0 \qquad -(2)$$
with $$(a,b)\neq(0,0)$$.

Let the point of intersection of these lines be $$P(x_0,\,y_0)$$. To find $$P$$ we solve the simultaneous equations $$-(1)$$ and $$-(2)$$.

Write the system in matrix form:
$$\begin{bmatrix} a & 2b \\[2pt] b & -2a \end{bmatrix} \begin{bmatrix} x_0 \\[2pt] y_0 \end{bmatrix} = \begin{bmatrix} -3b \\[2pt] 3a \end{bmatrix}.$$ The determinant of the coefficient matrix is
$$D = a(-2a) - b(2b) = -2(a^2 + b^2) \neq 0.$$ Hence a unique intersection point exists.

Finding $$x_0$$:
Replace the $$x$$-column by the constants and evaluate the determinant: $$D_x = \begin{vmatrix} -3b & 2b \\[2pt] 3a & -2a \end{vmatrix} = (-3b)(-2a) - (2b)(3a) = 6ab - 6ab = 0.$$ Therefore $$x_0 = \frac{D_x}{D} = \frac{0}{-2(a^2 + b^2)} = 0.$$

Finding $$y_0$$:
Replace the $$y$$-column by the constants: $$D_y = \begin{vmatrix} a & -3b \\[2pt] b & 3a \end{vmatrix} = a(3a) - b(-3b) = 3a^2 + 3b^2 = 3(a^2 + b^2).$$ Hence $$y_0 = \frac{D_y}{D} = \frac{3(a^2 + b^2)}{-2(a^2 + b^2)} = -\frac{3}{2}.$$

Thus the intersection point is $$P\bigl(0,\,-\frac{3}{2}\bigr).$$ A line parallel to the $$x$$-axis through this point has equation $$y = -\frac{3}{2}.$$

The ordinate is negative, so the line lies below the $$x$$-axis, and its perpendicular distance from the $$x$$-axis is $$\left|-\dfrac{3}{2}\right| = \dfrac{3}{2}$$.

Hence the required description is:
Option A — below the $$x$$-axis at a distance of $$\dfrac{3}{2}$$ from it.

Let the coordinates of the third vertex C of the triangle be given by:

$$ (h, k) $$

The coordinates of the given vertices are A(2, -3) and B(-2, 1). Let the coordinates of the centroid of the triangle ABC be denoted as G(X, Y).

The formula for the coordinates of the centroid of a triangle is:

$$ X = \frac{x_1 + x_2 + x_3}{3} $$

$$ Y = \frac{y_1 + y_2 + y_3}{3} $$

Substitute the given values into the centroid formula:

$$ X = \frac{2 + (-2) + h}{3} $$

$$ Y = \frac{-3 + 1 + k}{3} $$

Simplify the numerator expressions for both coordinates:

$$ X = \frac{h}{3} $$

$$ Y = \frac{k - 2}{3} $$

The problem states that the centroid G(X, Y) moves along a specific straight line:

$$ 2x + 3y = 1 $$

Since the point G(X, Y) lies on this line, its coordinates must satisfy the line equation:

$$ 2X + 3Y = 1 $$

Substitute the expressions for X and Y in terms of h and k back into this equation where $$ X = \frac{h}{3} $$ and $$ Y = \frac{k - 2}{3} $$:

$$ 2\left(\frac{h}{3}\right) + 3\left(\frac{k - 2}{3}\right) = 1 $$

Multiply the entire equation by 3 to clear the denominators:

$$ 2h + 3(k - 2) = 3 $$

Expand the terms inside the parentheses:

$$ 2h + 3k - 6 = 3 $$

Isolate the constant terms on the right side of the equation:

$$ 2h + 3k = 3 + 6 $$

$$ 2h + 3k = 9 $$

To find the locus of the moving vertex C, replace the coordinates (h, k) with the general coordinate variables (x, y):

$$ 2x + 3y = 9 $$

Final Answer:

The locus of the vertex C is the line $$ 2x + 3y = 9 $$.

Let the equation of the straight line in intercept form be:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

In this form, a represents the x-intercept and b represents the y-intercept. The problem states that the sum of these intercepts is -1:

$$ a + b = -1 $$

Isolate the variable b in terms of a:

$$ b = -1 - a $$

Substitute this expression for b back into the intercept form of the line:

$$ \frac{x}{a} + \frac{y}{-1 - a} = 1 $$

$$ \frac{x}{a} - \frac{y}{a + 1} = 1 $$

The line passes through the point (4, 3). Therefore, substituting $$ x = 4 $$ and $$ y = 3 $$ into the equation must yield a true statement:

$$ \frac{4}{a} - \frac{3}{a + 1} = 1 $$

Find a common denominator on the left side to merge the fractions:

$$ \frac{4(a + 1) - 3a}{a(a + 1)} = 1 $$

Expand the numerator and simplify the expression:

$$ \frac{4a + 4 - 3a}{a^2 + a} = 1 $$

$$ \frac{a + 4}{a^2 + a} = 1 $$

Cross-multiply to eliminate the fraction and form a quadratic equation:

$$ a + 4 = a^2 + a $$

Subtract a from both sides of the equation:

$$ 4 = a^2 $$

$$ a^2 = 4 $$

Taking the square root gives two possible values for a:

$$ a = 2 $$

$$ a = -2 $$

Now calculate the corresponding value of b for each value of a using the relationship $$ b = -1 - a $$.

Case 1: When $$ a = 2 $$, the value of b is:

$$ b = -1 - 2 $$

$$ b = -3 $$

Substitute $$ a = 2 $$ and $$ b = -3 $$ back into the intercept equation:

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

$$\implies \frac{x}{2} - \frac{y}{3} = 1 $$

Case 2: When $$ a = -2 $$, the value of b is:

$$ b = -1 - (-2) $$

$$ b = -1 + 2 $$

$$ b = 1 $$

Substitute $$ a = -2 $$ and $$ b = 1 $$ back into the intercept equation:

$$ \frac{x}{-2} + \frac{y}{1} = 1 $$

Final Answer:

The equations of the straight lines are $$\frac{x}{2} - \frac{y}{3} = 1 $$ and $$ \frac{x}{-2} + \frac{y}{1} = 1 $$.

Let the given pair of straight lines be represented by the equation:

$$ 6x^2 - xy + 4cy^2 = 0 $$

One of the lines given by this equation is:

$$ 3x + 4y = 0 $$

From this line equation, we can express $$ x $$ in terms of $$ y $$:

$$ 3x = -4y $$

$$ x = -\frac{4y}{3} $$

Since this line is part of the pair of lines, any point on it must satisfy the joint equation. Substitute this expression for $$ x $$ into the joint equation:

$$ 6\left(-\frac{4y}{3}\right)^2 - \left(-\frac{4y}{3}\right)y + 4cy^2 = 0 $$

Expand the squared term and simplify the multiplication:

$$ 6\left(\frac{16y^2}{9}\right) + \frac{4y^2}{3} + 4cy^2 = 0 $$

Reduce the fraction in the first term by dividing $$ 6 $$ and $$ 9 $$ by their common factor $$ 3 $$:

$$ 2\left(\frac{16y^2}{3}\right) + \frac{4y^2}{3} + 4cy^2 = 0 $$

$$ \frac{32y^2}{3} + \frac{4y^2}{3} + 4cy^2 = 0 $$

Combine the first two terms since they have a common denominator:

$$ \frac{36y^2}{3} + 4cy^2 = 0 $$

$$ 12y^2 + 4cy^2 = 0 $$

Factor out the common term $$ y^2 $$ from the equation:

$$ y^2(12 + 4c) = 0 $$

Since this equation holds true for all points along the line where $$ y \neq 0 $$, the expression inside the parentheses must be equal to $$ 0 $$:

$$ 12 + 4c = 0 $$

Solve for $$ c $$ by isolating the variable:

$$ 4c = -12 $$

$$ c = -3 $$

Final Answer:

The value of $$ c $$ is $$ -3 $$.

Let the given pair of straight lines passing through the origin be represented by the homogeneous quadratic equation:

$$ x^2 - 2cxy - 7y^2 = 0 $$

We can rewrite this equation in the standard form of a pair of straight lines:

$$ ax^2 + 2hxy + by^2 = 0 $$

By comparing the coefficients of the given equation with the standard form, we can identify the following values:

$$ a = 1 $$

$$ 2h = -2c \implies h = -c $$

$$ b = -7 $$

Let the individual slopes of the two lines represented by the joint equation be denoted as m1 and m2. According to the theory of pairs of straight lines, the formulas for the sum and product of the slopes are:

$$ m_1 + m_2 = -\frac{2h}{b} $$

$$ m_1 \cdot m_2 = \frac{a}{b} $$

Substitute the specific values of a, h, and b into these two relationships:

$$ m_1 + m_2 = -\frac{-2c}{-7} = -\frac{2c}{7} $$

$$ m_1 \cdot m_2 = \frac{1}{-7} = -\frac{1}{7} $$

The problem states a specific condition relating the sum and the product of the slopes, which is that the sum of the slopes is four times their product:

$$ m_1 + m_2 = 4(m_1 \cdot m_2) $$

Substitute the fractional expressions for the sum and product into this given condition:

$$ -\frac{2c}{7} = 4\left(-\frac{1}{7}\right) $$

$$ -\frac{2c}{7} = -\frac{4}{7} $$

Since both sides have a common denominator of 7, we can multiply the entire equation by -7 to isolate the variable term:

$$ 2c = 4 $$

Divide both sides by 2 to find the final value of c:

$$ c = 2 $$

Final Answer:

The value of c is 2.

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 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)

Lines AB: $$3y - x = 2$$ and AC: $$x + y = 2$$, with B and C on the x-axis.

Point B lies on AB and the x-axis ($$y = 0$$): $$-x = 2$$, so $$B = (-2, 0)$$.

Point C lies on AC and the x-axis ($$y = 0$$): $$x = 2$$, so $$C = (2, 0)$$.

Point A is the intersection of AB and AC: substituting $$x = 2 - y$$ into $$3y - x = 2$$:

$$3y - (2-y) = 2$$, giving $$4y = 4$$, so $$y = 1$$ and $$x = 1$$. Thus $$A = (1, 1)$$.

Finding the orthocentre P of triangle ABC:

Since BC lies on the x-axis (horizontal), the altitude from A to BC is vertical: $$x = 1$$.

The slope of AC is $$-1$$, so the altitude from B to AC has slope $$1$$. Through $$B(-2, 0)$$: $$y = x + 2$$.

Intersection of $$x = 1$$ and $$y = x + 2$$: $$P = (1, 3)$$.

Area of triangle PBC with vertices $$P(1,3)$$, $$B(-2,0)$$, $$C(2,0)$$:

Base $$BC = 4$$, height = y-coordinate of P = 3.

Area $$= \frac{1}{2} \times 4 \times 3 = 6$$

Hence, the correct answer is Option D.

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 $$\boxed{3}$$.

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 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.

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.

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).