Chapter 9: Straight Lines

Let the slope of the other line be $$m$$. The slope of the first line is $$m_1 = \dfrac{1}{2}$$.

If $$\theta$$ is the acute angle between two lines having slopes $$m_1$$ and $$m_2$$, then

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

Here $$\theta = \dfrac{\pi}{4}$$, so $$\tan\theta = \tan\dfrac{\pi}{4} = 1$$. Therefore

$$1 = \left|\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m}\right|$$

This means $$\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = 1$$ or $$\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = -1$$.

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

$$m - \dfrac{1}{2} = 1 + \dfrac{1}{2}m \;\Rightarrow\; m - \dfrac{1}{2}m = 1 + \dfrac{1}{2} \;\Rightarrow\; \dfrac{1}{2}m = \dfrac{3}{2} \;\Rightarrow\; m = 3$$

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

$$m - \dfrac{1}{2} = -\left(1 + \dfrac{1}{2}m\right) \;\Rightarrow\; m + \dfrac{1}{2}m = -1 + \dfrac{1}{2} \;\Rightarrow\; \dfrac{3}{2}m = -\dfrac{1}{2} \;\Rightarrow\; m = -\dfrac{1}{3}$$

Hence the slope of the other line is $$3$$ or $$-\dfrac{1}{3}$$, depending on which side of the first line it is drawn.

Slope of the line through $$(-2, 6)$$ and $$(4, 8)$$:

$$m_1 = \dfrac{8 - 6}{4 - (-2)} = \dfrac{2}{6} = \dfrac{1}{3}$$

Slope of the line through $$(8, 12)$$ and $$(x, 24)$$:

$$m_2 = \dfrac{24 - 12}{x - 8} = \dfrac{12}{x - 8}$$

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

$$m_1 m_2 = -1$$

$$\dfrac{1}{3} \cdot \dfrac{12}{x - 8} = -1 \;\Rightarrow\; \dfrac{12}{3(x - 8)} = -1 \;\Rightarrow\; \dfrac{4}{x - 8} = -1$$

$$4 = -(x - 8) \;\Rightarrow\; 4 = 8 - x \;\Rightarrow\; x = 4$$

Let the vertices be $$A(-4, 5)$$, $$B(0, 7)$$, $$C(5, -5)$$ and $$D(-4, -2)$$. Plot the points and join them in order to obtain quadrilateral $$ABCD$$.

Drawing: Mark $$A(-4, 5)$$ and $$D(-4, -2)$$ on the vertical line $$x = -4$$, mark $$B(0, 7)$$ on the $$y$$-axis, and $$C(5, -5)$$ in the fourth quadrant. Join $$AB$$, $$BC$$, $$CD$$ and $$DA$$.

Draw the diagonal $$AC$$. It divides the quadrilateral into two triangles, $$\triangle ABC$$ and $$\triangle ACD$$.

The area of a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$ is

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

Area of $$\triangle ABC$$ with $$A(-4, 5)$$, $$B(0, 7)$$, $$C(5, -5)$$:

$$= \dfrac{1}{2}\left|(-4)(7 - (-5)) + 0(-5 - 5) + 5(5 - 7)\right|$$

$$= \dfrac{1}{2}\left|(-4)(12) + 0 + 5(-2)\right| = \dfrac{1}{2}\left|-48 - 10\right| = \dfrac{1}{2}(58) = 29$$

Area of $$\triangle ACD$$ with $$A(-4, 5)$$, $$C(5, -5)$$, $$D(-4, -2)$$:

$$= \dfrac{1}{2}\left|(-4)(-5 - (-2)) + 5(-2 - 5) + (-4)(5 - (-5))\right|$$

$$= \dfrac{1}{2}\left|(-4)(-3) + 5(-7) + (-4)(10)\right| = \dfrac{1}{2}\left|12 - 35 - 40\right| = \dfrac{1}{2}(63) = 31.5$$

Total area of the quadrilateral:

$$\text{Area} = 29 + 31.5 = 60.5 \text{ square units}$$

The base of the triangle lies along the $$y$$-axis, and its mid-point is the origin $$O(0, 0)$$. The base has length $$2a$$ (the side of the equilateral triangle).

Since the mid-point of the base is the origin, the two end-points of the base lie on the $$y$$-axis at distance $$a$$ on either side of $$O$$. So the base vertices are

$$B(0, a) \quad \text{and} \quad C(0, -a)$$

Let the third vertex be $$A(x, 0)$$. By symmetry it lies on the $$x$$-axis, which is the perpendicular bisector of the base.

Each side of the triangle is $$2a$$. Consider side $$AB$$, which joins $$A(x, 0)$$ and $$B(0, a)$$:

$$AB^2 = (x - 0)^2 + (0 - a)^2 = x^2 + a^2$$

Since $$AB = 2a$$, we get $$AB^2 = 4a^2$$, so

$$x^2 + a^2 = 4a^2 \;\Rightarrow\; x^2 = 3a^2 \;\Rightarrow\; x = \pm\sqrt{3}\,a$$

Hence the third vertex is $$(\sqrt{3}\,a, 0)$$ or $$(-\sqrt{3}\,a, 0)$$.

Let the required point on the $$x$$-axis be $$P(x, 0)$$, since every point on the $$x$$-axis has $$y$$-coordinate $$0$$.

$$P$$ is equidistant from $$A(7, 6)$$ and $$B(3, 4)$$, so $$PA = PB$$, and hence $$PA^2 = PB^2$$.

$$PA^2 = (x - 7)^2 + (0 - 6)^2 = (x - 7)^2 + 36$$

$$PB^2 = (x - 3)^2 + (0 - 4)^2 = (x - 3)^2 + 16$$

Equating:

$$(x - 7)^2 + 36 = (x - 3)^2 + 16$$

$$x^2 - 14x + 49 + 36 = x^2 - 6x + 9 + 16$$

$$-14x + 85 = -6x + 25$$

$$85 - 25 = -6x + 14x \;\Rightarrow\; 60 = 8x \;\Rightarrow\; x = \dfrac{15}{2}$$

Hence the required point is $$\left(\dfrac{15}{2}, 0\right)$$.

First find the mid-point of the segment joining $$P(0, -4)$$ and $$B(8, 0)$$.

$$\text{Mid-point} = \left(\dfrac{0 + 8}{2}, \dfrac{-4 + 0}{2}\right) = (4, -2)$$

The required line passes through the origin $$(0, 0)$$ and the mid-point $$(4, -2)$$. Its slope is

$$m = \dfrac{-2 - 0}{4 - 0} = \dfrac{-2}{4} = -\dfrac{1}{2}$$

Let the points be $$A(4, 4)$$, $$B(3, 5)$$ and $$C(-1, -1)$$. Instead of the Pythagoras theorem we use slopes: if two sides of the triangle are perpendicular, the triangle is right-angled.

Slope of $$AB = \dfrac{5 - 4}{3 - 4} = \dfrac{1}{-1} = -1$$

Slope of $$BC = \dfrac{-1 - 5}{-1 - 3} = \dfrac{-6}{-4} = \dfrac{3}{2}$$

Slope of $$CA = \dfrac{4 - (-1)}{4 - (-1)} = \dfrac{5}{5} = 1$$

Now consider the two sides $$AB$$ and $$CA$$ meeting at $$A$$:

$$(\text{slope of } AB) \times (\text{slope of } CA) = (-1)(1) = -1$$

Since the product of their slopes is $$-1$$, the sides $$AB$$ and $$CA$$ are perpendicular. Hence the angle at vertex $$A$$ is a right angle.

Therefore the points $$(4, 4)$$, $$(3, 5)$$ and $$(-1, -1)$$ are the vertices of a right-angled triangle, right-angled at $$(4, 4)$$.

The positive direction of the $$y$$-axis itself makes an angle of $$90^\circ$$ with the positive direction of the $$x$$-axis.

The required line is at $$30^\circ$$ from the positive $$y$$-axis, measured anticlockwise. So, measured anticlockwise from the positive $$x$$-axis, the inclination of the line is

$$\theta = 90^\circ + 30^\circ = 120^\circ$$

Hence the slope is

$$m = \tan\theta = \tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$$

Let the points be $$A(-2, -1)$$, $$B(4, 0)$$, $$C(3, 3)$$ and $$D(-3, 2)$$, taken in order. A quadrilateral is a parallelogram if both pairs of opposite sides are parallel, that is, have equal slopes.

Slope of $$AB = \dfrac{0 - (-1)}{4 - (-2)} = \dfrac{1}{6}$$

Slope of $$CD = \dfrac{2 - 3}{-3 - 3} = \dfrac{-1}{-6} = \dfrac{1}{6}$$

Since slope of $$AB$$ = slope of $$CD$$, the sides $$AB$$ and $$CD$$ are parallel.

Slope of $$BC = \dfrac{3 - 0}{3 - 4} = \dfrac{3}{-1} = -3$$

Slope of $$DA = \dfrac{-1 - 2}{-2 - (-3)} = \dfrac{-3}{1} = -3$$

Since slope of $$BC$$ = slope of $$DA$$, the sides $$BC$$ and $$DA$$ are parallel.

Both pairs of opposite sides are parallel, so $$ABCD$$ is a parallelogram.

First find the slope of the line joining $$(3, -1)$$ and $$(4, -2)$$:

$$m = \dfrac{-2 - (-1)}{4 - 3} = \dfrac{-1}{1} = -1$$

If $$\theta$$ is the angle that the line makes with the positive direction of the $$x$$-axis, then $$m = \tan\theta$$:

$$\tan\theta = -1$$

The angle $$\theta$$ in the range $$0^\circ \le \theta < 180^\circ$$ whose tangent is $$-1$$ is

$$\theta = 180^\circ - 45^\circ = 135^\circ$$

Hence the line makes an angle of $$135^\circ$$ with the $$x$$-axis.

Let the slope of one line be $$m$$. Then the slope of the other line, being double of it, is $$2m$$.

If $$\theta$$ is the angle between the lines, then $$\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right|$$. Taking $$m_1 = m$$, $$m_2 = 2m$$ and $$\tan\theta = \dfrac{1}{3}$$:

$$\dfrac{1}{3} = \left|\dfrac{2m - m}{1 + m \cdot 2m}\right| = \left|\dfrac{m}{1 + 2m^2}\right|$$

So $$\dfrac{m}{1 + 2m^2} = \dfrac{1}{3}$$ or $$\dfrac{m}{1 + 2m^2} = -\dfrac{1}{3}$$.

Case 1: $$3m = 1 + 2m^2$$

$$2m^2 - 3m + 1 = 0 \;\Rightarrow\; (2m - 1)(m - 1) = 0 \;\Rightarrow\; m = \dfrac{1}{2} \text{ or } m = 1$$

The pair of slopes $$(m, 2m)$$ is $$\left(\dfrac{1}{2}, 1\right)$$ or $$(1, 2)$$.

Case 2: $$3m = -(1 + 2m^2)$$

$$2m^2 + 3m + 1 = 0 \;\Rightarrow\; (2m + 1)(m + 1) = 0 \;\Rightarrow\; m = -\dfrac{1}{2} \text{ or } m = -1$$

The pair of slopes $$(m, 2m)$$ is $$\left(-\dfrac{1}{2}, -1\right)$$ or $$(-1, -2)$$.

Hence the slopes of the two lines are $$\dfrac{1}{2}$$ and $$1$$, or $$1$$ and $$2$$, or $$-\dfrac{1}{2}$$ and $$-1$$, or $$-1$$ and $$-2$$.

The line passes through the two points $$(x_1, y_1)$$ and $$(h, k)$$, and its slope is $$m$$.

By the definition of the slope of a line through two points,

$$m = \dfrac{\text{difference of } y\text{-coordinates}}{\text{difference of } x\text{-coordinates}} = \dfrac{k - y_1}{h - x_1}$$

(here $$h \neq x_1$$, otherwise the slope would be undefined). Multiplying both sides by $$(h - x_1)$$:

$$m(h - x_1) = k - y_1$$

That is, $$k - y_1 = m(h - x_1)$$, which is the required result.

We need the two lines through $$(-2, 3)$$ that are parallel to the coordinate axes.

Line parallel to the $$x$$-axis: Every point on a line parallel to the $$x$$-axis has the same $$y$$-coordinate. Since the line passes through $$(-2, 3)$$, that constant $$y$$-coordinate is $$3$$. Its equation is

$$y = 3$$

Line parallel to the $$y$$-axis: Every point on a line parallel to the $$y$$-axis has the same $$x$$-coordinate. Since the line passes through $$(-2, 3)$$, that constant $$x$$-coordinate is $$-2$$. Its equation is

$$x = -2$$

The equation of a line through the point $$(x_0, y_0)$$ with slope $$m$$ (point-slope form) is

$$y - y_0 = m(x - x_0)$$

Here $$(x_0, y_0) = (-2, 3)$$ and $$m = -4$$:

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

$$y - 3 = -4(x + 2)$$

$$y - 3 = -4x - 8$$

$$4x + y + 5 = 0$$

First find the slope of the line through $$(1, -1)$$ and $$(3, 5)$$:

$$m = \dfrac{5 - (-1)}{3 - 1} = \dfrac{6}{2} = 3$$

Now use the point-slope form $$y - y_0 = m(x - x_0)$$ with the point $$(1, -1)$$:

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

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

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

Using the intercept form, a line with $$x$$-intercept $$a$$ and $$y$$-intercept $$b$$ has the equation

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

Here $$a = -3$$ and $$b = 2$$:

$$\dfrac{x}{-3} + \dfrac{y}{2} = 1$$

Multiplying both sides by $$6$$:

$$-2x + 3y = 6$$

$$2x - 3y + 6 = 0$$

Equation of the $$x$$-axis: Every point lying on the $$x$$-axis has its $$y$$-coordinate equal to $$0$$, and this is the only condition that all such points satisfy. Hence the equation of the $$x$$-axis is

$$y = 0$$

Equation of the $$y$$-axis: Every point lying on the $$y$$-axis has its $$x$$-coordinate equal to $$0$$. Hence the equation of the $$y$$-axis is

$$x = 0$$

Use the point-slope form $$y - y_0 = m(x - x_0)$$ with the point $$(-4, 3)$$ and slope $$m = \dfrac{1}{2}$$:

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

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

Multiply both sides by $$2$$:

$$2(y - 3) = x + 4$$

$$2y - 6 = x + 4$$

$$x - 2y + 10 = 0$$

Use the point-slope form $$y - y_0 = m(x - x_0)$$ with the point $$(0, 0)$$ and slope $$m$$:

$$y - 0 = m(x - 0)$$

$$y = mx$$

This is the equation of the required line through the origin.

The slope of the line is $$m = \tan 75^\circ$$. Compute it using $$75^\circ = 45^\circ + 30^\circ$$:

$$\tan 75^\circ = \dfrac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \dfrac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Rationalise by multiplying numerator and denominator by $$(\sqrt{3} + 1)$$:

$$\tan 75^\circ = \dfrac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{3 + 2\sqrt{3} + 1}{3 - 1} = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

So $$m = 2 + \sqrt{3}$$. Use the point-slope form with the point $$(2, 2\sqrt{3})$$:

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

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

$$y - 2\sqrt{3} = (2 + \sqrt{3})x - 4 - 2\sqrt{3}$$

$$y = (2 + \sqrt{3})x - 4$$

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

The line meets the $$x$$-axis $$3$$ units to the left of the origin, so it passes through the point $$(-3, 0)$$.

Use the point-slope form $$y - y_0 = m(x - x_0)$$ with $$(x_0, y_0) = (-3, 0)$$ and $$m = -2$$:

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

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

$$y = -2x - 6$$

$$2x + y + 6 = 0$$

The line meets the $$y$$-axis $$2$$ units above the origin, so it passes through the point $$(0, 2)$$.

It makes an angle of $$30^\circ$$ with the positive direction of the $$x$$-axis, so its slope is

$$m = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$$

Use the point-slope form with the point $$(0, 2)$$:

$$y - 2 = \dfrac{1}{\sqrt{3}}(x - 0)$$

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

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

$$x - \sqrt{3}\,y + 2\sqrt{3} = 0$$

Slope of the line through $$(-1, 1)$$ and $$(2, -4)$$:

$$m = \dfrac{-4 - 1}{2 - (-1)} = \dfrac{-5}{3}$$

Use the point-slope form $$y - y_0 = m(x - x_0)$$ with the point $$(-1, 1)$$:

$$y - 1 = -\dfrac{5}{3}(x - (-1))$$

$$3(y - 1) = -5(x + 1)$$

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

$$5x + 3y + 2 = 0$$

The median through vertex $$R$$ joins $$R$$ to the mid-point of the opposite side $$PQ$$.

Mid-point of $$PQ$$, with $$P(2, 1)$$ and $$Q(-2, 3)$$:

$$M = \left(\dfrac{2 + (-2)}{2}, \dfrac{1 + 3}{2}\right) = (0, 2)$$

The median is the line joining $$R(4, 5)$$ and $$M(0, 2)$$. Its slope:

$$m = \dfrac{2 - 5}{0 - 4} = \dfrac{-3}{-4} = \dfrac{3}{4}$$

Equation through $$M(0, 2)$$ using the point-slope form:

$$y - 2 = \dfrac{3}{4}(x - 0)$$

$$4(y - 2) = 3x \;\Rightarrow\; 4y - 8 = 3x$$

$$3x - 4y + 8 = 0$$

Slope of the line through $$(2, 5)$$ and $$(-3, 6)$$:

$$m_1 = \dfrac{6 - 5}{-3 - 2} = \dfrac{1}{-5} = -\dfrac{1}{5}$$

The required line is perpendicular to this line, so its slope $$m$$ satisfies $$m \cdot m_1 = -1$$:

$$m = -\dfrac{1}{m_1} = -\dfrac{1}{-1/5} = 5$$

Use the point-slope form with the point $$(-3, 5)$$:

$$y - 5 = 5(x - (-3))$$

$$y - 5 = 5(x + 3)$$

$$y - 5 = 5x + 15$$

$$5x - y + 20 = 0$$

The required line is perpendicular to the segment joining $$A(1, 0)$$ and $$B(2, 3)$$, and it crosses this segment at the point dividing $$AB$$ in the ratio $$1 : n$$.

By the section formula, the point dividing $$A(1, 0)$$ and $$B(2, 3)$$ internally in the ratio $$1 : n$$ is

$$P = \left(\dfrac{1 \cdot 2 + n \cdot 1}{1 + n}, \dfrac{1 \cdot 3 + n \cdot 0}{1 + n}\right) = \left(\dfrac{n + 2}{n + 1}, \dfrac{3}{n + 1}\right)$$

Slope of $$AB$$:

$$m_{AB} = \dfrac{3 - 0}{2 - 1} = 3$$

The required line is perpendicular to $$AB$$, so its slope is $$m = -\dfrac{1}{3}$$.

Equation of the line through $$P$$ with slope $$-\dfrac{1}{3}$$:

$$y - \dfrac{3}{n + 1} = -\dfrac{1}{3}\left(x - \dfrac{n + 2}{n + 1}\right)$$

Multiply both sides by $$3$$:

$$3y - \dfrac{9}{n + 1} = -x + \dfrac{n + 2}{n + 1}$$

$$x + 3y = \dfrac{9}{n + 1} + \dfrac{n + 2}{n + 1} = \dfrac{9 + n + 2}{n + 1} = \dfrac{n + 11}{n + 1}$$

Multiplying both sides by $$(n + 1)$$:

$$(1 + n)\,x + 3(1 + n)\,y = n + 11$$

Let the line cut off equal intercepts $$a$$ on each axis (with $$a \neq 0$$). Using the intercept form with $$x$$-intercept $$a$$ and $$y$$-intercept $$a$$:

$$\dfrac{x}{a} + \dfrac{y}{a} = 1 \;\Rightarrow\; x + y = a$$

The line passes through $$(2, 3)$$, so substitute $$x = 2$$, $$y = 3$$:

$$2 + 3 = a \;\Rightarrow\; a = 5$$

Hence the equation of the line is

$$x + y = 5$$

Let the $$x$$-intercept be $$a$$ and the $$y$$-intercept be $$b$$. The intercept form of the line is

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

The sum of the intercepts is $$9$$, so $$a + b = 9$$, which gives $$b = 9 - a$$.

The line passes through $$(2, 2)$$:

$$\dfrac{2}{a} + \dfrac{2}{9 - a} = 1$$

Multiply through by $$a(9 - a)$$:

$$2(9 - a) + 2a = a(9 - a)$$

$$18 - 2a + 2a = 9a - a^2$$

$$18 = 9a - a^2$$

$$a^2 - 9a + 18 = 0 \;\Rightarrow\; (a - 3)(a - 6) = 0$$

So $$a = 3$$ or $$a = 6$$.

If $$a = 3$$, then $$b = 9 - 3 = 6$$:

$$\dfrac{x}{3} + \dfrac{y}{6} = 1 \;\Rightarrow\; 2x + y = 6$$

If $$a = 6$$, then $$b = 9 - 6 = 3$$:

$$\dfrac{x}{6} + \dfrac{y}{3} = 1 \;\Rightarrow\; x + 2y = 6$$

The line makes an angle $$\dfrac{2\pi}{3}$$ with the positive direction of the $$x$$-axis, so its slope is

$$m = \tan\dfrac{2\pi}{3} = \tan 120^\circ = -\sqrt{3}$$

It passes through $$(0, 2)$$. Using the point-slope form:

$$y - 2 = -\sqrt{3}(x - 0)$$

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

$$\sqrt{3}\,x + y - 2 = 0$$

Parallel line: A line parallel to the one above has the same slope $$-\sqrt{3}$$. This line crosses the $$y$$-axis $$2$$ units below the origin, so it passes through $$(0, -2)$$:

$$y - (-2) = -\sqrt{3}(x - 0)$$

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

$$\sqrt{3}\,x + y + 2 = 0$$

The perpendicular from the origin $$O(0, 0)$$ meets the required line at $$P(-2, 9)$$. So the segment $$OP$$ is perpendicular to the line.

Slope of $$OP$$:

$$m_{OP} = \dfrac{9 - 0}{-2 - 0} = -\dfrac{9}{2}$$

Since the line is perpendicular to $$OP$$, its slope $$m$$ satisfies $$m \cdot m_{OP} = -1$$:

$$m = -\dfrac{1}{m_{OP}} = -\dfrac{1}{-9/2} = \dfrac{2}{9}$$

The line passes through $$P(-2, 9)$$. Using the point-slope form:

$$y - 9 = \dfrac{2}{9}(x - (-2))$$

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

$$9y - 81 = 2x + 4$$

$$2x - 9y + 85 = 0$$

Since $$L$$ is a linear function of $$C$$, the relation between them is the equation of a straight line in the $$C$$-$$L$$ plane. This line passes through the two given data points $$(C, L) = (20,\ 124.942)$$ and $$(110,\ 125.134)$$.

Slope of the line:

$$m = \dfrac{125.134 - 124.942}{110 - 20} = \dfrac{0.192}{90}$$

Using the point-slope form with the point $$(20,\ 124.942)$$:

$$L - 124.942 = \dfrac{0.192}{90}(C - 20)$$

$$L = \dfrac{0.192}{90}(C - 20) + 124.942$$

This expresses $$L$$ in terms of $$C$$.

Let the selling price (in Rs per litre) be $$x$$ and the quantity sold (in litres) be $$y$$. The relationship is linear, so the point $$(x, y)$$ lies on a straight line through the two data points $$(14,\ 980)$$ and $$(16,\ 1220)$$.

Slope of the line:

$$m = \dfrac{1220 - 980}{16 - 14} = \dfrac{240}{2} = 120$$

Equation of the line through $$(14,\ 980)$$:

$$y - 980 = 120(x - 14)$$

To find the quantity sold at Rs $$17$$/litre, put $$x = 17$$:

$$y - 980 = 120(17 - 14) = 120 \times 3 = 360$$

$$y = 980 + 360 = 1340$$

Hence he could sell $$1340$$ litres of milk weekly at Rs $$17$$/litre.

Let the line meet the $$x$$-axis at $$A(p, 0)$$ and the $$y$$-axis at $$B(0, q)$$. The segment between the axes is $$AB$$.

The mid-point of $$AB$$ is

$$\left(\dfrac{p + 0}{2}, \dfrac{0 + q}{2}\right) = \left(\dfrac{p}{2}, \dfrac{q}{2}\right)$$

This mid-point is given to be $$P(a, b)$$, so

$$\dfrac{p}{2} = a \;\Rightarrow\; p = 2a, \qquad \dfrac{q}{2} = b \;\Rightarrow\; q = 2b$$

Using the intercept form with $$x$$-intercept $$p = 2a$$ and $$y$$-intercept $$q = 2b$$:

$$\dfrac{x}{2a} + \dfrac{y}{2b} = 1$$

Multiplying both sides by $$2$$:

$$\dfrac{x}{a} + \dfrac{y}{b} = 2$$

which is the required equation.

Let the line meet the $$x$$-axis at $$A(a, 0)$$ and the $$y$$-axis at $$B(0, b)$$. The point $$R(h, k)$$ divides the segment $$AB$$ in the ratio $$1 : 2$$, measured from $$A$$ on the $$x$$-axis (so $$AR : RB = 1 : 2$$).

By the section formula, the point dividing $$A(a, 0)$$ and $$B(0, b)$$ internally in the ratio $$1 : 2$$ is

$$\left(\dfrac{1 \cdot 0 + 2 \cdot a}{1 + 2}, \dfrac{1 \cdot b + 2 \cdot 0}{1 + 2}\right) = \left(\dfrac{2a}{3}, \dfrac{b}{3}\right)$$

Equating this to $$R(h, k)$$:

$$\dfrac{2a}{3} = h \;\Rightarrow\; a = \dfrac{3h}{2}, \qquad \dfrac{b}{3} = k \;\Rightarrow\; b = 3k$$

Using the intercept form $$\dfrac{x}{a} + \dfrac{y}{b} = 1$$:

$$\dfrac{x}{3h/2} + \dfrac{y}{3k} = 1 \;\Rightarrow\; \dfrac{2x}{3h} + \dfrac{y}{3k} = 1$$

Multiplying both sides by $$3$$:

$$\dfrac{2x}{h} + \dfrac{y}{k} = 3$$

We first find the equation of the line passing through two of the points, say $$A(3, 0)$$ and $$B(-2, -2)$$, and then check whether the third point lies on it.

Slope of $$AB$$:

$$m = \dfrac{-2 - 0}{-2 - 3} = \dfrac{-2}{-5} = \dfrac{2}{5}$$

Equation of line $$AB$$ using the point-slope form through $$A(3, 0)$$:

$$y - 0 = \dfrac{2}{5}(x - 3)$$

$$5y = 2(x - 3) \;\Rightarrow\; 2x - 5y - 6 = 0$$

Now substitute the third point $$C(8, 2)$$ into the left-hand side:

$$2(8) - 5(2) - 6 = 16 - 10 - 6 = 0$$

Since $$C(8, 2)$$ satisfies the equation of line $$AB$$, the point $$C$$ lies on the line $$AB$$. Therefore the three points $$(3, 0)$$, $$(-2, -2)$$ and $$(8, 2)$$ are collinear.

The distance of a point $$(x_1, y_1)$$ from the line $$Ax + By + C = 0$$ is

$$d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Here the line is $$3x - 4y - 26 = 0$$, so $$A = 3$$, $$B = -4$$, $$C = -26$$, and the point is $$(x_1, y_1) = (3, -5)$$:

$$d = \dfrac{|3(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}} = \dfrac{|9 + 20 - 26|}{\sqrt{9 + 16}}$$

$$d = \dfrac{|3|}{\sqrt{25}} = \dfrac{3}{5}$$

The distance between two parallel lines $$Ax + By + C_1 = 0$$ and $$Ax + By + C_2 = 0$$ is

$$d = \dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

Here the lines are $$3x - 4y + 7 = 0$$ and $$3x - 4y + 5 = 0$$, so $$A = 3$$, $$B = -4$$, $$C_1 = 7$$ and $$C_2 = 5$$:

$$d = \dfrac{|7 - 5|}{\sqrt{3^2 + (-4)^2}} = \dfrac{2}{\sqrt{25}} = \dfrac{2}{5}$$

First write the line in the general form $$Ax + By + C = 0$$:

$$12(x + 6) = 5(y - 2)$$

$$12x + 72 = 5y - 10$$

$$12x - 5y + 82 = 0$$

The distance of the point $$(-1, 1)$$ from this line is

$$d = \dfrac{|12(-1) - 5(1) + 82|}{\sqrt{12^2 + (-5)^2}} = \dfrac{|-12 - 5 + 82|}{\sqrt{144 + 25}}$$

$$d = \dfrac{|65|}{\sqrt{169}} = \dfrac{65}{13} = 5$$

First write the line in general form. From $$\dfrac{x}{3} + \dfrac{y}{4} = 1$$, multiply both sides by $$12$$:

$$4x + 3y = 12 \;\Rightarrow\; 4x + 3y - 12 = 0$$

Let a required point on the $$x$$-axis be $$(a, 0)$$. Its distance from the line is $$4$$:

$$\dfrac{|4a + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = 4$$

$$\dfrac{|4a - 12|}{5} = 4 \;\Rightarrow\; |4a - 12| = 20$$

So $$4a - 12 = 20$$ or $$4a - 12 = -20$$.

If $$4a - 12 = 20$$: $$4a = 32 \;\Rightarrow\; a = 8$$.

If $$4a - 12 = -20$$: $$4a = -8 \;\Rightarrow\; a = -2$$.

Hence the required points are $$(8, 0)$$ and $$(-2, 0)$$.

Any line parallel to $$3x - 4y + 2 = 0$$ has the same coefficients of $$x$$ and $$y$$, so it can be written as

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

for some constant $$k$$. This line passes through $$(-2, 3)$$, so substitute $$x = -2$$, $$y = 3$$:

$$3(-2) - 4(3) + k = 0$$

$$-6 - 12 + k = 0 \;\Rightarrow\; k = 18$$

Hence the required line is

$$3x - 4y + 18 = 0$$

Find the slope of $$x - 7y + 5 = 0$$ by writing it as $$y = \dfrac{x + 5}{7}$$, which gives slope $$\dfrac{1}{7}$$.

The required line is perpendicular to it, so its slope is

$$m = -\dfrac{1}{1/7} = -7$$

The $$x$$-intercept is $$3$$, so the line passes through the point $$(3, 0)$$. Using the point-slope form:

$$y - 0 = -7(x - 3)$$

$$y = -7x + 21$$

$$7x + y - 21 = 0$$

Find the slopes of the two lines.

$$\sqrt{3}\,x + y = 1 \;\Rightarrow\; y = -\sqrt{3}\,x + 1$$, so $$m_1 = -\sqrt{3}$$.

$$x + \sqrt{3}\,y = 1 \;\Rightarrow\; y = \dfrac{1 - x}{\sqrt{3}}$$, so $$m_2 = -\dfrac{1}{\sqrt{3}}$$.

If $$\theta$$ is the angle between the lines,

$$\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right| = \left|\dfrac{-\frac{1}{\sqrt{3}} - (-\sqrt{3})}{1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right|$$

Numerator: $$-\dfrac{1}{\sqrt{3}} + \sqrt{3} = \dfrac{-1 + 3}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}$$.   Denominator: $$1 + 1 = 2$$.

$$\tan\theta = \left|\dfrac{2/\sqrt{3}}{2}\right| = \dfrac{1}{\sqrt{3}}$$

So $$\theta = 30^\circ$$. The other angle between the lines is its supplement, $$180^\circ - 30^\circ = 150^\circ$$.

Slope of the line joining $$(h, 3)$$ and $$(4, 1)$$:

$$m_1 = \dfrac{1 - 3}{4 - h} = \dfrac{-2}{4 - h}$$

Slope of the line $$7x - 9y - 19 = 0$$: writing it as $$y = \dfrac{7x - 19}{9}$$ gives $$m_2 = \dfrac{7}{9}$$.

The two lines intersect at right angles, so $$m_1 m_2 = -1$$:

$$\dfrac{-2}{4 - h} \cdot \dfrac{7}{9} = -1$$

$$\dfrac{-14}{9(4 - h)} = -1$$

$$-14 = -9(4 - h) \;\Rightarrow\; 14 = 9(4 - h)$$

$$14 = 36 - 9h \;\Rightarrow\; 9h = 22 \;\Rightarrow\; h = \dfrac{22}{9}$$

The slope of the line $$Ax + By + C = 0$$ (with $$B \neq 0$$) is found by writing $$y = -\dfrac{A}{B}x - \dfrac{C}{B}$$, which gives slope $$-\dfrac{A}{B}$$.

The required line is parallel to it, so it has the same slope $$-\dfrac{A}{B}$$, and it passes through $$(x_1, y_1)$$. By the point-slope form:

$$y - y_1 = -\dfrac{A}{B}(x - x_1)$$

Multiplying both sides by $$B$$:

$$B(y - y_1) = -A(x - x_1)$$

$$A(x - x_1) + B(y - y_1) = 0$$

which is the required equation. (If $$B = 0$$, the line $$Ax + C = 0$$ is vertical; the line through $$(x_1, y_1)$$ parallel to it is $$x = x_1$$, i.e. $$A(x - x_1) = 0$$, which is again the stated form.)

Let the slope of the other line be $$m$$. One line has slope $$2$$, and the angle between the two lines is $$60^\circ$$.

$$\tan 60^\circ = \left|\dfrac{m - 2}{1 + 2m}\right| \;\Rightarrow\; \sqrt{3} = \left|\dfrac{m - 2}{1 + 2m}\right|$$

So $$\dfrac{m - 2}{1 + 2m} = \sqrt{3}$$ or $$\dfrac{m - 2}{1 + 2m} = -\sqrt{3}$$.

Case 1: $$m - 2 = \sqrt{3}(1 + 2m)$$

$$m - 2 = \sqrt{3} + 2\sqrt{3}\,m \;\Rightarrow\; m - 2\sqrt{3}\,m = 2 + \sqrt{3} \;\Rightarrow\; m(1 - 2\sqrt{3}) = 2 + \sqrt{3}$$

$$m = \dfrac{2 + \sqrt{3}}{1 - 2\sqrt{3}}$$

Rationalising (multiply numerator and denominator by $$1 + 2\sqrt{3}$$): the denominator is $$1 - 12 = -11$$ and the numerator is $$(2 + \sqrt{3})(1 + 2\sqrt{3}) = 8 + 5\sqrt{3}$$, so $$m = -\dfrac{8 + 5\sqrt{3}}{11}$$.

Case 2: $$m - 2 = -\sqrt{3}(1 + 2m)$$

$$m - 2 = -\sqrt{3} - 2\sqrt{3}\,m \;\Rightarrow\; m + 2\sqrt{3}\,m = 2 - \sqrt{3} \;\Rightarrow\; m(1 + 2\sqrt{3}) = 2 - \sqrt{3}$$

$$m = \dfrac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$$

Rationalising similarly gives $$m = \dfrac{5\sqrt{3} - 8}{11}$$.

Both lines pass through $$(2, 3)$$, so by the point-slope form the other line is

$$y - 3 = \dfrac{2 + \sqrt{3}}{1 - 2\sqrt{3}}(x - 2) \qquad \text{or} \qquad y - 3 = \dfrac{2 - \sqrt{3}}{1 + 2\sqrt{3}}(x - 2)$$

The right bisector (perpendicular bisector) of a segment passes through its mid-point and is perpendicular to the segment.

Mid-point of the segment joining $$(3, 4)$$ and $$(-1, 2)$$:

$$M = \left(\dfrac{3 + (-1)}{2}, \dfrac{4 + 2}{2}\right) = (1, 3)$$

Slope of the segment:

$$m_1 = \dfrac{2 - 4}{-1 - 3} = \dfrac{-2}{-4} = \dfrac{1}{2}$$

The right bisector is perpendicular to the segment, so its slope is

$$m = -\dfrac{1}{1/2} = -2$$

Equation of the right bisector through $$M(1, 3)$$:

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

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

$$2x + y - 5 = 0$$

Let the foot of the perpendicular be $$P(a, b)$$. Since $$P$$ lies on the given line,

$$3a - 4b - 16 = 0 \qquad \text{...(1)}$$

The slope of the line $$3x - 4y - 16 = 0$$ is $$\dfrac{3}{4}$$. The segment joining $$(-1, 3)$$ and $$P(a, b)$$ is perpendicular to the line, so its slope is $$-\dfrac{4}{3}$$:

$$\dfrac{b - 3}{a - (-1)} = -\dfrac{4}{3}$$

$$3(b - 3) = -4(a + 1) \;\Rightarrow\; 3b - 9 = -4a - 4 \;\Rightarrow\; 4a + 3b - 5 = 0 \qquad \text{...(2)}$$

Now solve (1) and (2). From (1): $$3a - 4b = 16$$. From (2): $$4a + 3b = 5$$.

Multiply (1) by $$3$$ and (2) by $$4$$:

$$9a - 12b = 48, \qquad 16a + 12b = 20$$

Adding these: $$25a = 68 \;\Rightarrow\; a = \dfrac{68}{25}$$.

Substitute into (2): $$3b = 5 - 4a = 5 - \dfrac{272}{25} = \dfrac{125 - 272}{25} = -\dfrac{147}{25}$$, so $$b = -\dfrac{49}{25}$$.

Hence the foot of the perpendicular is $$\left(\dfrac{68}{25}, -\dfrac{49}{25}\right)$$.

The point $$(-1, 2)$$ lies on the line $$y = mx + c$$, so

$$2 = m(-1) + c \;\Rightarrow\; c = m + 2 \qquad \text{...(1)}$$

The perpendicular from the origin $$O(0, 0)$$ meets the line at $$(-1, 2)$$, so the segment from $$O$$ to $$(-1, 2)$$ is perpendicular to the line.

Slope of this segment:

$$\dfrac{2 - 0}{-1 - 0} = -2$$

The line has slope $$m$$ and is perpendicular to this segment, so the product of slopes is $$-1$$:

$$m \times (-2) = -1 \;\Rightarrow\; m = \dfrac{1}{2}$$

From (1): $$c = \dfrac{1}{2} + 2 = \dfrac{5}{2}$$.

Length $$p$$. Write the first line as $$x\cos\theta - y\sin\theta - k\cos 2\theta = 0$$. The perpendicular distance from the origin is

$$p = \dfrac{|0 \cdot \cos\theta - 0 \cdot \sin\theta - k\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} = \dfrac{|k\cos 2\theta|}{1} = |k\cos 2\theta|$$

So $$p^2 = k^2\cos^2 2\theta$$.

Length $$q$$. Write the second line as $$x\sec\theta + y\csc\theta - k = 0$$. The perpendicular distance from the origin is

$$q = \dfrac{|-k|}{\sqrt{\sec^2\theta + \csc^2\theta}}$$

Now

$$\sec^2\theta + \csc^2\theta = \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}$$

Therefore

$$q = |k| \cdot |\sin\theta\cos\theta| = |k| \cdot \dfrac{|\sin 2\theta|}{2}$$

using the identity $$2\sin\theta\cos\theta = \sin 2\theta$$. Squaring, $$q^2 = \dfrac{k^2\sin^2 2\theta}{4}$$, so $$4q^2 = k^2\sin^2 2\theta$$.

Adding the two results:

$$p^2 + 4q^2 = k^2\cos^2 2\theta + k^2\sin^2 2\theta = k^2\left(\cos^2 2\theta + \sin^2 2\theta\right) = k^2$$

Hence $$p^2 + 4q^2 = k^2$$, as required.

The altitude from vertex $$A$$ is the line through $$A$$ perpendicular to the opposite side $$BC$$.

Slope of $$BC$$, with $$B(4, -1)$$ and $$C(1, 2)$$:

$$m_{BC} = \dfrac{2 - (-1)}{1 - 4} = \dfrac{3}{-3} = -1$$

The altitude is perpendicular to $$BC$$, so its slope is $$m = -\dfrac{1}{-1} = 1$$.

Equation of the altitude through $$A(2, 3)$$ with slope $$1$$:

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

Length of the altitude equals the perpendicular distance from $$A(2, 3)$$ to the line $$BC$$.

Equation of $$BC$$: it passes through $$B(4, -1)$$ with slope $$-1$$:

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

Distance from $$A(2, 3)$$ to this line:

$$d = \dfrac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$$

The line with intercepts $$a$$ and $$b$$ on the axes has equation

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

Multiplying both sides by $$ab$$ to get the general form:

$$bx + ay - ab = 0$$

The perpendicular distance from the origin $$(0, 0)$$ to this line is

$$p = \dfrac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \dfrac{|ab|}{\sqrt{a^2 + b^2}}$$

Squaring both sides:

$$p^2 = \dfrac{a^2 b^2}{a^2 + b^2}$$

Taking the reciprocal:

$$\dfrac{1}{p^2} = \dfrac{a^2 + b^2}{a^2 b^2} = \dfrac{a^2}{a^2 b^2} + \dfrac{b^2}{a^2 b^2} = \dfrac{1}{b^2} + \dfrac{1}{a^2}$$

Hence $$\dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$, as required.

For three lines to be concurrent, they must all pass through one common point.

Find the point of intersection of the first and third lines, $$2x + y - 3 = 0$$ and $$3x - y - 2 = 0$$. Adding them eliminates $$y$$:

$$(2x + y - 3) + (3x - y - 2) = 0 \;\Rightarrow\; 5x - 5 = 0 \;\Rightarrow\; x = 1$$

From $$2x + y - 3 = 0$$: $$y = 3 - 2x = 3 - 2 = 1$$. So the first and third lines meet at $$(1, 1)$$.

For concurrency, the second line $$5x + ky - 3 = 0$$ must also pass through $$(1, 1)$$:

$$5(1) + k(1) - 3 = 0 \;\Rightarrow\; 2 + k = 0 \;\Rightarrow\; k = -2$$

Through $$P(4, 1)$$ draw a line making an angle $$135^\circ$$ with the positive $$x$$-axis. Any point on this line at a distance $$r$$ from $$P$$ has coordinates

$$x = 4 + r\cos 135^\circ, \qquad y = 1 + r\sin 135^\circ$$

Since $$\cos 135^\circ = -\dfrac{1}{\sqrt{2}}$$ and $$\sin 135^\circ = \dfrac{1}{\sqrt{2}}$$,

$$x = 4 - \dfrac{r}{\sqrt{2}}, \qquad y = 1 + \dfrac{r}{\sqrt{2}}$$

The required distance is the value of $$r$$ for which this point lies on the line $$4x - y = 0$$. Substitute:

$$4\left(4 - \dfrac{r}{\sqrt{2}}\right) - \left(1 + \dfrac{r}{\sqrt{2}}\right) = 0$$

$$16 - \dfrac{4r}{\sqrt{2}} - 1 - \dfrac{r}{\sqrt{2}} = 0$$

$$15 - \dfrac{5r}{\sqrt{2}} = 0 \;\Rightarrow\; \dfrac{5r}{\sqrt{2}} = 15 \;\Rightarrow\; r = 3\sqrt{2}$$

Hence the required distance is $$3\sqrt{2}$$ units.

Let the image of $$Q(1, 2)$$ in the line be $$Q'(h, k)$$. Since the line acts as a mirror, it is the perpendicular bisector of the segment $$QQ'$$.

Condition 1 — mid-point of $$QQ'$$ lies on the line. The mid-point is $$\left(\dfrac{1 + h}{2}, \dfrac{2 + k}{2}\right)$$. Putting it in $$x - 3y + 4 = 0$$:

$$\dfrac{1 + h}{2} - 3 \cdot \dfrac{2 + k}{2} + 4 = 0$$

Multiply by $$2$$: $$(1 + h) - 3(2 + k) + 8 = 0 \;\Rightarrow\; h - 3k + 3 = 0 \qquad \text{...(1)}$$

Condition 2 — $$QQ'$$ is perpendicular to the line. The slope of $$x - 3y + 4 = 0$$ is $$\dfrac{1}{3}$$, so the slope of $$QQ'$$ must be $$-3$$:

$$\dfrac{k - 2}{h - 1} = -3 \;\Rightarrow\; k - 2 = -3(h - 1) \;\Rightarrow\; 3h + k - 5 = 0 \qquad \text{...(2)}$$

From (1), $$h = 3k - 3$$. Substitute into (2):

$$3(3k - 3) + k - 5 = 0 \;\Rightarrow\; 9k - 9 + k - 5 = 0 \;\Rightarrow\; 10k = 14 \;\Rightarrow\; k = \dfrac{7}{5}$$

Then $$h = 3k - 3 = \dfrac{21}{5} - 3 = \dfrac{6}{5}$$.

Hence the image of $$(1, 2)$$ is $$\left(\dfrac{6}{5}, \dfrac{7}{5}\right)$$.

Find the three vertices of the triangle as the pairwise intersections of the three lines.

Vertex $$A$$ — line 1 with $$x = 0$$: putting $$x = 0$$ in $$y = m_1 x + c_1$$ gives $$y = c_1$$, so $$A = (0, c_1)$$.

Vertex $$B$$ — line 2 with $$x = 0$$: putting $$x = 0$$ in $$y = m_2 x + c_2$$ gives $$y = c_2$$, so $$B = (0, c_2)$$.

Vertex $$C$$ — lines 1 and 2: setting $$m_1 x + c_1 = m_2 x + c_2$$,

$$x(m_1 - m_2) = c_2 - c_1 \;\Rightarrow\; x_C = \dfrac{c_2 - c_1}{m_1 - m_2}$$

The side $$AB$$ lies along the $$y$$-axis (the line $$x = 0$$), so take it as the base. Its length is

$$AB = |c_1 - c_2|$$

The height is the perpendicular distance of $$C$$ from the $$y$$-axis, which is the absolute value of the $$x$$-coordinate of $$C$$:

$$\text{height} = |x_C| = \left|\dfrac{c_2 - c_1}{m_1 - m_2}\right|$$

Therefore

$$\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \cdot |c_1 - c_2| \cdot \left|\dfrac{c_2 - c_1}{m_1 - m_2}\right|$$

$$= \dfrac{1}{2} \cdot \dfrac{|c_1 - c_2|^2}{|m_1 - m_2|} = \dfrac{(c_1 - c_2)^2}{2|m_1 - m_2|}$$

which is the required result.

Let the required line meet $$5x - y + 4 = 0$$ at $$A$$ and $$3x + 4y - 4 = 0$$ at $$B$$, with $$(1, 5)$$ the mid-point of $$AB$$.

Since $$A$$ lies on $$5x - y + 4 = 0$$, we may write $$A = (\alpha,\ 5\alpha + 4)$$.

Since $$B$$ lies on $$3x + 4y - 4 = 0$$, we may write $$B = \left(\beta,\ \dfrac{4 - 3\beta}{4}\right)$$.

The mid-point of $$AB$$ is $$(1, 5)$$. Equating $$x$$-coordinates:

$$\dfrac{\alpha + \beta}{2} = 1 \;\Rightarrow\; \alpha + \beta = 2 \qquad \text{...(1)}$$

Equating $$y$$-coordinates:

$$\dfrac{(5\alpha + 4) + \dfrac{4 - 3\beta}{4}}{2} = 5 \;\Rightarrow\; (5\alpha + 4) + \dfrac{4 - 3\beta}{4} = 10$$

Multiply by $$4$$: $$4(5\alpha + 4) + (4 - 3\beta) = 40 \;\Rightarrow\; 20\alpha + 16 + 4 - 3\beta = 40$$

$$20\alpha - 3\beta = 20 \qquad \text{...(2)}$$

From (1), $$\beta = 2 - \alpha$$. Substitute into (2):

$$20\alpha - 3(2 - \alpha) = 20 \;\Rightarrow\; 20\alpha - 6 + 3\alpha = 20 \;\Rightarrow\; 23\alpha = 26 \;\Rightarrow\; \alpha = \dfrac{26}{23}$$

So $$A = \left(\dfrac{26}{23},\ 5 \cdot \dfrac{26}{23} + 4\right) = \left(\dfrac{26}{23},\ \dfrac{222}{23}\right)$$.

The required line passes through $$A$$ and $$(1, 5)$$. Its slope:

$$m = \dfrac{\dfrac{222}{23} - 5}{\dfrac{26}{23} - 1} = \dfrac{\dfrac{222 - 115}{23}}{\dfrac{26 - 23}{23}} = \dfrac{107}{3}$$

Equation through $$(1, 5)$$:

$$y - 5 = \dfrac{107}{3}(x - 1) \;\Rightarrow\; 3(y - 5) = 107(x - 1)$$

$$3y - 15 = 107x - 107 \;\Rightarrow\; 107x - 3y - 92 = 0$$

Let the moving point be $$P(h, k)$$. Its distances from the two lines $$3x - 2y - 5 = 0$$ and $$3x + 2y - 5 = 0$$ are equal:

$$\dfrac{|3h - 2k - 5|}{\sqrt{3^2 + (-2)^2}} = \dfrac{|3h + 2k - 5|}{\sqrt{3^2 + 2^2}}$$

Both denominators equal $$\sqrt{13}$$, so they cancel:

$$|3h - 2k - 5| = |3h + 2k - 5|$$

This gives $$3h - 2k - 5 = \pm(3h + 2k - 5)$$.

Case 1 (taking $$+$$): $$3h - 2k - 5 = 3h + 2k - 5 \;\Rightarrow\; -2k = 2k \;\Rightarrow\; k = 0$$.

Case 2 (taking $$-$$): $$3h - 2k - 5 = -(3h + 2k - 5) \;\Rightarrow\; 3h - 5 = -3h + 5 \;\Rightarrow\; 6h = 10 \;\Rightarrow\; h = \dfrac{5}{3}$$.

Replacing $$(h, k)$$ by $$(x, y)$$, the point lies on $$y = 0$$ or on $$x = \dfrac{5}{3}$$. Each of these is the equation of a straight line.

Hence the path of the moving point is a straight line.

Let the intercepts on the axes be $$a$$ and $$b$$. We are given

$$a + b = 1, \qquad ab = -6$$

So $$a$$ and $$b$$ are the roots of the quadratic equation $$t^2 - (a + b)t + ab = 0$$:

$$t^2 - t - 6 = 0 \;\Rightarrow\; (t - 3)(t + 2) = 0 \;\Rightarrow\; t = 3 \text{ or } t = -2$$

So either $$(a, b) = (3, -2)$$ or $$(a, b) = (-2, 3)$$.

Using the intercept form $$\dfrac{x}{a} + \dfrac{y}{b} = 1$$:

If $$a = 3$$, $$b = -2$$: $$\dfrac{x}{3} + \dfrac{y}{-2} = 1$$. Multiplying by $$6$$: $$2x - 3y = 6$$, i.e. $$2x - 3y - 6 = 0$$.

If $$a = -2$$, $$b = 3$$: $$\dfrac{x}{-2} + \dfrac{y}{3} = 1$$. Multiplying by $$6$$: $$-3x + 2y = 6$$, i.e. $$3x - 2y + 6 = 0$$.

First write the line in general form. From $$\dfrac{x}{3} + \dfrac{y}{4} = 1$$, multiply both sides by $$12$$:

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

Let a required point on the $$y$$-axis be $$(0, b)$$. Its distance from the line is $$4$$:

$$\dfrac{|4(0) + 3b - 12|}{\sqrt{4^2 + 3^2}} = 4$$

$$\dfrac{|3b - 12|}{5} = 4 \;\Rightarrow\; |3b - 12| = 20$$

So $$3b - 12 = 20$$ or $$3b - 12 = -20$$.

If $$3b - 12 = 20$$: $$3b = 32 \;\Rightarrow\; b = \dfrac{32}{3}$$.

If $$3b - 12 = -20$$: $$3b = -8 \;\Rightarrow\; b = -\dfrac{8}{3}$$.

Hence the required points are $$\left(0, \dfrac{32}{3}\right)$$ and $$\left(0, -\dfrac{8}{3}\right)$$.

Let the two points be $$A(\cos\theta, \sin\theta)$$ and $$B(\cos\phi, \sin\phi)$$.

Slope of $$AB$$:

$$m = \dfrac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta}$$

Use the identities $$\sin\phi - \sin\theta = 2\cos\dfrac{\theta + \phi}{2}\sin\dfrac{\phi - \theta}{2}$$ and $$\cos\phi - \cos\theta = -2\sin\dfrac{\theta + \phi}{2}\sin\dfrac{\phi - \theta}{2}$$:

$$m = \dfrac{2\cos\frac{\theta + \phi}{2}\sin\frac{\phi - \theta}{2}}{-2\sin\frac{\theta + \phi}{2}\sin\frac{\phi - \theta}{2}} = -\cot\dfrac{\theta + \phi}{2}$$

Write $$\alpha = \dfrac{\theta + \phi}{2}$$, so the slope is $$-\dfrac{\cos\alpha}{\sin\alpha}$$. The equation of line $$AB$$ through $$A(\cos\theta, \sin\theta)$$ is

$$y - \sin\theta = -\dfrac{\cos\alpha}{\sin\alpha}(x - \cos\theta)$$

$$\sin\alpha\,(y - \sin\theta) = -\cos\alpha\,(x - \cos\theta)$$

$$x\cos\alpha + y\sin\alpha = \cos\theta\cos\alpha + \sin\theta\sin\alpha = \cos(\alpha - \theta)$$

Since $$\alpha - \theta = \dfrac{\theta + \phi}{2} - \theta = \dfrac{\phi - \theta}{2}$$, the line is

$$x\cos\alpha + y\sin\alpha - \cos\dfrac{\phi - \theta}{2} = 0$$

The perpendicular distance from the origin $$(0,0)$$ is

$$d = \dfrac{\left|-\cos\frac{\phi - \theta}{2}\right|}{\sqrt{\cos^2\alpha + \sin^2\alpha}} = \left|\cos\dfrac{\phi - \theta}{2}\right|$$

Hence the perpendicular distance from the origin is $$\left|\cos\dfrac{\theta - \phi}{2}\right|$$.

First find the point of intersection of $$x - 7y + 5 = 0$$ and $$3x + y = 0$$.

From $$3x + y = 0$$: $$y = -3x$$. Substitute into $$x - 7y + 5 = 0$$:

$$x - 7(-3x) + 5 = 0 \;\Rightarrow\; x + 21x + 5 = 0 \;\Rightarrow\; 22x = -5 \;\Rightarrow\; x = -\dfrac{5}{22}$$

Then $$y = -3x = \dfrac{15}{22}$$. So the point of intersection is $$\left(-\dfrac{5}{22}, \dfrac{15}{22}\right)$$.

A line parallel to the $$y$$-axis has an equation of the form $$x = \text{constant}$$. Since it passes through the point of intersection, the constant is its $$x$$-coordinate:

$$x = -\dfrac{5}{22}$$

The given line is $$\dfrac{x}{4} + \dfrac{y}{6} = 1$$. It meets the $$y$$-axis where $$x = 0$$:

$$\dfrac{0}{4} + \dfrac{y}{6} = 1 \;\Rightarrow\; y = 6$$

So the point where it meets the $$y$$-axis is $$(0, 6)$$.

Find the slope of the given line. Multiplying $$\dfrac{x}{4} + \dfrac{y}{6} = 1$$ by $$12$$ gives $$3x + 2y = 12$$, i.e. $$y = -\dfrac{3}{2}x + 6$$, so its slope is $$-\dfrac{3}{2}$$.

The required line is perpendicular to it, so its slope is

$$m = -\dfrac{1}{-3/2} = \dfrac{2}{3}$$

Equation of the required line through $$(0, 6)$$:

$$y - 6 = \dfrac{2}{3}(x - 0)$$

$$3(y - 6) = 2x \;\Rightarrow\; 3y - 18 = 2x$$

$$2x - 3y + 18 = 0$$

The three lines are $$y = x$$, $$y = -x$$ and $$x = k$$. Find the vertices of the triangle as the pairwise intersections.

$$y = x$$ and $$y = -x$$: $$x = -x \Rightarrow x = 0$$, hence $$y = 0$$. Vertex $$O = (0, 0)$$.

$$y = x$$ and $$x = k$$: $$x = k$$, $$y = k$$. Vertex $$P = (k, k)$$.

$$y = -x$$ and $$x = k$$: $$x = k$$, $$y = -k$$. Vertex $$Q = (k, -k)$$.

Using the area formula for vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$:

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

With $$O(0, 0)$$, $$P(k, k)$$, $$Q(k, -k)$$:

$$\text{Area} = \dfrac{1}{2}\left|0(k - (-k)) + k(-k - 0) + k(0 - k)\right|$$

$$= \dfrac{1}{2}\left|0 - k^2 - k^2\right| = \dfrac{1}{2}\left(2k^2\right) = k^2$$

For the three lines to intersect at one point, that point must lie on all three lines.

Find the intersection of the first and third lines, $$3x + y - 2 = 0$$ and $$2x - y - 3 = 0$$. Adding them eliminates $$y$$:

$$(3x + y - 2) + (2x - y - 3) = 0 \;\Rightarrow\; 5x - 5 = 0 \;\Rightarrow\; x = 1$$

From $$3x + y - 2 = 0$$: $$y = 2 - 3x = 2 - 3 = -1$$. So the common point is $$(1, -1)$$.

The second line $$px + 2y - 3 = 0$$ must also pass through $$(1, -1)$$:

$$p(1) + 2(-1) - 3 = 0 \;\Rightarrow\; p - 2 - 3 = 0 \;\Rightarrow\; p = 5$$

The three lines are concurrent, so they pass through one common point.

Find the point of intersection of the first two lines. Setting $$m_1 x + c_1 = m_2 x + c_2$$:

$$x(m_1 - m_2) = c_2 - c_1 \;\Rightarrow\; x = \dfrac{c_2 - c_1}{m_1 - m_2}$$

and then

$$y = m_1 x + c_1 = \dfrac{m_1(c_2 - c_1)}{m_1 - m_2} + c_1 = \dfrac{m_1 c_2 - m_1 c_1 + c_1 m_1 - c_1 m_2}{m_1 - m_2} = \dfrac{m_1 c_2 - m_2 c_1}{m_1 - m_2}$$

For concurrency this point must lie on the third line $$y = m_3 x + c_3$$:

$$\dfrac{m_1 c_2 - m_2 c_1}{m_1 - m_2} = m_3 \cdot \dfrac{c_2 - c_1}{m_1 - m_2} + c_3$$

Multiply both sides by $$(m_1 - m_2)$$:

$$m_1 c_2 - m_2 c_1 = m_3(c_2 - c_1) + c_3(m_1 - m_2)$$

$$m_1 c_2 - m_2 c_1 = m_3 c_2 - m_3 c_1 + m_1 c_3 - m_2 c_3$$

Bring every term to the left-hand side:

$$m_1 c_2 - m_1 c_3 + m_2 c_3 - m_2 c_1 + m_3 c_1 - m_3 c_2 = 0$$

Group the terms containing $$m_1$$, $$m_2$$ and $$m_3$$:

$$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$$

which is the required result.

The slope of the line $$x - 2y = 3$$ is found by writing $$y = \dfrac{x - 3}{2}$$, which gives slope $$\dfrac{1}{2}$$.

Let the slope of a required line be $$m$$. The angle between it and the given line is $$45^\circ$$:

$$\tan 45^\circ = \left|\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m}\right| \;\Rightarrow\; 1 = \left|\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m}\right|$$

So $$\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = 1$$ or $$\dfrac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = -1$$.

Case 1: $$m - \dfrac{1}{2} = 1 + \dfrac{1}{2}m \;\Rightarrow\; \dfrac{1}{2}m = \dfrac{3}{2} \;\Rightarrow\; m = 3$$.

Case 2: $$m - \dfrac{1}{2} = -1 - \dfrac{1}{2}m \;\Rightarrow\; \dfrac{3}{2}m = -\dfrac{1}{2} \;\Rightarrow\; m = -\dfrac{1}{3}$$.

Both lines pass through $$(3, 2)$$. Using the point-slope form:

With $$m = 3$$: $$y - 2 = 3(x - 3) \;\Rightarrow\; y - 2 = 3x - 9 \;\Rightarrow\; 3x - y - 7 = 0$$.

With $$m = -\dfrac{1}{3}$$: $$y - 2 = -\dfrac{1}{3}(x - 3) \;\Rightarrow\; 3(y - 2) = -(x - 3) \;\Rightarrow\; x + 3y - 9 = 0$$.

First find the point of intersection of $$4x + 7y - 3 = 0$$ and $$2x - 3y + 1 = 0$$.

Write them as $$4x + 7y = 3$$ ...(A) and $$2x - 3y = -1$$ ...(B). Multiply (B) by $$2$$: $$4x - 6y = -2$$. Subtract this from (A):

$$(4x + 7y) - (4x - 6y) = 3 - (-2) \;\Rightarrow\; 13y = 5 \;\Rightarrow\; y = \dfrac{5}{13}$$

From (B): $$2x = -1 + 3y = -1 + \dfrac{15}{13} = \dfrac{2}{13} \;\Rightarrow\; x = \dfrac{1}{13}$$.

So the point of intersection is $$\left(\dfrac{1}{13}, \dfrac{5}{13}\right)$$.

A line with equal non-zero intercepts $$a$$ on both axes is $$\dfrac{x}{a} + \dfrac{y}{a} = 1$$, i.e. $$x + y = a$$. Since it passes through $$\left(\dfrac{1}{13}, \dfrac{5}{13}\right)$$:

$$\dfrac{1}{13} + \dfrac{5}{13} = a \;\Rightarrow\; a = \dfrac{6}{13}$$

So the line is $$x + y = \dfrac{6}{13}$$, i.e. $$13x + 13y - 6 = 0$$.

(If the intercepts are allowed to be equal and zero, the line passes through the origin; since $$\dfrac{y}{x} = \dfrac{5/13}{1/13} = 5$$, that line is $$y = 5x$$, i.e. $$5x - y = 0$$.)

Let the required line through the origin have slope $$M$$. Since it passes through the origin, its equation is $$y = Mx$$, so $$M = \dfrac{y}{x}$$.

The given line $$y = mx + c$$ has slope $$m$$. The angle between the two lines is $$\theta$$, so

$$\tan\theta = \left|\dfrac{M - m}{1 + mM}\right|$$

Removing the modulus,

$$\dfrac{M - m}{1 + mM} = \pm\tan\theta$$

$$M - m = \pm\tan\theta\,(1 + mM) = \pm\tan\theta \pm mM\tan\theta$$

Collect the terms containing $$M$$ on the left:

$$M \mp mM\tan\theta = m \pm \tan\theta$$

$$M(1 \mp m\tan\theta) = m \pm \tan\theta$$

$$M = \dfrac{m \pm \tan\theta}{1 \mp m\tan\theta}$$

Since $$M = \dfrac{y}{x}$$, the equation of the line through the origin is

$$\dfrac{y}{x} = \dfrac{m \pm \tan\theta}{1 \mp m\tan\theta}$$

which is the required result.

Let the line $$x + y = 4$$ divide the segment joining $$A(-1, 1)$$ and $$B(5, 7)$$ in the ratio $$k : 1$$.

By the section formula, the dividing point is

$$\left(\dfrac{k(5) + 1(-1)}{k + 1}, \dfrac{k(7) + 1(1)}{k + 1}\right) = \left(\dfrac{5k - 1}{k + 1}, \dfrac{7k + 1}{k + 1}\right)$$

This point lies on the line $$x + y = 4$$:

$$\dfrac{5k - 1}{k + 1} + \dfrac{7k + 1}{k + 1} = 4$$

$$\dfrac{(5k - 1) + (7k + 1)}{k + 1} = 4 \;\Rightarrow\; \dfrac{12k}{k + 1} = 4$$

$$12k = 4(k + 1) \;\Rightarrow\; 12k = 4k + 4 \;\Rightarrow\; 8k = 4 \;\Rightarrow\; k = \dfrac{1}{2}$$

So the ratio is $$k : 1 = \dfrac{1}{2} : 1 = 1 : 2$$. Since $$k > 0$$, the division is internal.

First note that the point $$(1, 2)$$ lies on the line $$2x - y = 0$$, since $$2(1) - 2 = 0$$. So the required distance is measured from $$(1, 2)$$ along the line $$2x - y = 0$$ up to the point where it meets $$4x + 7y + 5 = 0$$.

Find the intersection of $$2x - y = 0$$ and $$4x + 7y + 5 = 0$$. From $$2x - y = 0$$: $$y = 2x$$. Substitute:

$$4x + 7(2x) + 5 = 0 \;\Rightarrow\; 4x + 14x + 5 = 0 \;\Rightarrow\; 18x = -5 \;\Rightarrow\; x = -\dfrac{5}{18}$$

Then $$y = 2x = -\dfrac{5}{9}$$. So the lines meet at $$\left(-\dfrac{5}{18}, -\dfrac{5}{9}\right)$$.

The required distance is the distance between $$(1, 2)$$ and this point:

$$1 - \left(-\dfrac{5}{18}\right) = \dfrac{23}{18}, \qquad 2 - \left(-\dfrac{5}{9}\right) = \dfrac{23}{9} = \dfrac{46}{18}$$

$$d = \sqrt{\left(\dfrac{23}{18}\right)^2 + \left(\dfrac{46}{18}\right)^2} = \dfrac{1}{18}\sqrt{23^2 + 46^2} = \dfrac{1}{18}\sqrt{23^2(1 + 4)} = \dfrac{23\sqrt{5}}{18}$$

Let the line through $$(-1, 2)$$ make an inclination $$\theta$$ with the positive $$x$$-axis. A point on this line at a distance $$r$$ from $$(-1, 2)$$ has coordinates

$$x = -1 + r\cos\theta, \qquad y = 2 + r\sin\theta$$

This point should lie on $$x + y = 4$$ when $$r = 3$$:

$$(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$$

$$1 + 3\cos\theta + 3\sin\theta = 4 \;\Rightarrow\; 3(\cos\theta + \sin\theta) = 3 \;\Rightarrow\; \cos\theta + \sin\theta = 1$$

Write $$\cos\theta + \sin\theta = \sqrt{2}\sin(\theta + 45^\circ)$$:

$$\sqrt{2}\sin(\theta + 45^\circ) = 1 \;\Rightarrow\; \sin(\theta + 45^\circ) = \dfrac{1}{\sqrt{2}}$$

Hence $$\theta + 45^\circ = 45^\circ$$ or $$\theta + 45^\circ = 135^\circ$$, which gives

$$\theta = 0^\circ \qquad \text{or} \qquad \theta = 90^\circ$$

So the line must be drawn either parallel to the $$x$$-axis ($$\theta = 0^\circ$$) or parallel to the $$y$$-axis ($$\theta = 90^\circ$$).

The two legs (perpendicular sides) are parallel to the coordinate axes, so they are a vertical line and a horizontal line meeting at the right-angle vertex.

The right-angle vertex shares its $$x$$-coordinate with one end of the hypotenuse and its $$y$$-coordinate with the other end. The ends of the hypotenuse are $$(1, 3)$$ and $$(-4, 1)$$, so the right-angle vertex is either $$(1, 1)$$ or $$(-4, 3)$$.

Case 1 — right angle at $$(1, 1)$$: the vertical leg passes through $$(1, 3)$$ and $$(1, 1)$$, so it is $$x = 1$$; the horizontal leg passes through $$(-4, 1)$$ and $$(1, 1)$$, so it is $$y = 1$$.

Case 2 — right angle at $$(-4, 3)$$: the vertical leg passes through $$(-4, 1)$$ and $$(-4, 3)$$, so it is $$x = -4$$; the horizontal leg passes through $$(1, 3)$$ and $$(-4, 3)$$, so it is $$y = 3$$.

Let the image of $$(3, 8)$$ in the line $$x + 3y - 7 = 0$$ be $$(h, k)$$. Since the line acts as a mirror, it is the perpendicular bisector of the segment joining $$(3, 8)$$ and $$(h, k)$$.

Condition 1 — mid-point lies on the line. The mid-point is $$\left(\dfrac{3 + h}{2}, \dfrac{8 + k}{2}\right)$$. Putting it in $$x + 3y - 7 = 0$$:

$$\dfrac{3 + h}{2} + 3 \cdot \dfrac{8 + k}{2} - 7 = 0$$

Multiply by $$2$$: $$(3 + h) + 3(8 + k) - 14 = 0 \;\Rightarrow\; h + 3k + 13 = 0 \qquad \text{...(1)}$$

Condition 2 — the segment is perpendicular to the line. The slope of $$x + 3y - 7 = 0$$ is $$-\dfrac{1}{3}$$, so the slope of the segment joining $$(3, 8)$$ and $$(h, k)$$ is $$3$$:

$$\dfrac{k - 8}{h - 3} = 3 \;\Rightarrow\; k - 8 = 3(h - 3) \;\Rightarrow\; 3h - k - 1 = 0 \qquad \text{...(2)}$$

From (2), $$k = 3h - 1$$. Substitute into (1):

$$h + 3(3h - 1) + 13 = 0 \;\Rightarrow\; h + 9h - 3 + 13 = 0 \;\Rightarrow\; 10h + 10 = 0 \;\Rightarrow\; h = -1$$

Then $$k = 3(-1) - 1 = -4$$.

Hence the image of $$(3, 8)$$ is $$(-1, -4)$$.

The slopes are: line 1, $$y = 3x + 1$$, slope $$m_1 = 3$$; line 2, $$2y = x + 3$$ i.e. $$y = \dfrac{x}{2} + \dfrac{3}{2}$$, slope $$m_2 = \dfrac{1}{2}$$; and the line $$y = mx + 4$$ has slope $$m$$.

'Equally inclined' means the angle between line 1 and $$y = mx + 4$$ equals the angle between line 2 and $$y = mx + 4$$:

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

$$\left|\dfrac{3 - m}{1 + 3m}\right| = \left|\dfrac{\frac{1}{2} - m}{1 + \frac{1}{2}m}\right| = \left|\dfrac{1 - 2m}{2 + m}\right|$$

So $$\dfrac{3 - m}{1 + 3m} = \pm\dfrac{1 - 2m}{2 + m}$$.

Case 1 (with $$+$$): $$(3 - m)(2 + m) = (1 - 2m)(1 + 3m)$$

$$6 + m - m^2 = 1 + m - 6m^2 \;\Rightarrow\; 6 - m^2 = 1 - 6m^2 \;\Rightarrow\; 5m^2 = -5$$

This gives $$m^2 = -1$$, which has no real solution.

Case 2 (with $$-$$): $$(3 - m)(2 + m) = -(1 - 2m)(1 + 3m)$$

$$6 + m - m^2 = -(1 + m - 6m^2) = -1 - m + 6m^2$$

$$6 + m - m^2 + 1 + m - 6m^2 = 0 \;\Rightarrow\; 7 + 2m - 7m^2 = 0 \;\Rightarrow\; 7m^2 - 2m - 7 = 0$$

By the quadratic formula:

$$m = \dfrac{2 \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)} = \dfrac{2 \pm \sqrt{4 + 196}}{14} = \dfrac{2 \pm \sqrt{200}}{14} = \dfrac{2 \pm 10\sqrt{2}}{14}$$

$$m = \dfrac{1 \pm 5\sqrt{2}}{7}$$

Let $$P(x, y)$$ be the variable point. Its perpendicular distances from the lines $$x + y - 5 = 0$$ and $$3x - 2y + 7 = 0$$ are

$$d_1 = \dfrac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \dfrac{|x + y - 5|}{\sqrt{2}}, \qquad d_2 = \dfrac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \dfrac{|3x - 2y + 7|}{\sqrt{13}}$$

Their sum is always $$10$$:

$$\dfrac{|x + y - 5|}{\sqrt{2}} + \dfrac{|3x - 2y + 7|}{\sqrt{13}} = 10$$

Consider any region of the plane in which the signs of $$x + y - 5$$ and $$3x - 2y + 7$$ stay fixed (for instance, both non-negative). In that region the modulus signs can be dropped, and the equation becomes

$$\dfrac{x + y - 5}{\sqrt{2}} + \dfrac{3x - 2y + 7}{\sqrt{13}} = 10$$

Multiplying through by $$\sqrt{2}\cdot\sqrt{13} = \sqrt{26}$$:

$$\sqrt{13}\,(x + y - 5) + \sqrt{2}\,(3x - 2y + 7) = 10\sqrt{26}$$

$$\left(\sqrt{13} + 3\sqrt{2}\right)x + \left(\sqrt{13} - 2\sqrt{2}\right)y + \left(7\sqrt{2} - 5\sqrt{13} - 10\sqrt{26}\right) = 0$$

This is an equation of the first degree, of the form $$Ax + By + C = 0$$, in which the constants $$A$$ and $$B$$ are not both zero. An equation of the first degree in $$x$$ and $$y$$ always represents a straight line. (For other sign-combinations of the two expressions we similarly obtain first-degree equations.)

Hence the point $$P$$ must move along a straight line.

First make the coefficients of $$x$$ and $$y$$ the same in both equations. Multiply $$3x + 2y + 6 = 0$$ by $$3$$:

$$9x + 6y + 18 = 0$$

So the two parallel lines are $$9x + 6y - 7 = 0$$ and $$9x + 6y + 18 = 0$$.

The line equidistant from them is parallel to both and lies midway between them, so it has the form $$9x + 6y + c = 0$$, where $$c$$ is the average of the two constant terms:

$$c = \dfrac{-7 + 18}{2} = \dfrac{11}{2}$$

So the required line is

$$9x + 6y + \dfrac{11}{2} = 0$$

Multiplying both sides by $$2$$ to clear the fraction:

$$18x + 12y + 11 = 0$$

When a ray reflects off the $$x$$-axis, the reflected ray travels as if it came in a straight line from the image of the source in the $$x$$-axis.

The image of the source $$(1, 2)$$ in the $$x$$-axis is $$(1, -2)$$.

Therefore the points $$(1, -2)$$, $$A$$ and $$(5, 3)$$ are collinear, with $$A$$ lying on the $$x$$-axis. Find the line through $$(1, -2)$$ and $$(5, 3)$$.

Slope $$= \dfrac{3 - (-2)}{5 - 1} = \dfrac{5}{4}$$. Equation through $$(5, 3)$$:

$$y - 3 = \dfrac{5}{4}(x - 5)$$

The point $$A$$ is where this line meets the $$x$$-axis, i.e. where $$y = 0$$:

$$0 - 3 = \dfrac{5}{4}(x - 5) \;\Rightarrow\; -3 = \dfrac{5}{4}(x - 5)$$

$$x - 5 = -\dfrac{12}{5} \;\Rightarrow\; x = 5 - \dfrac{12}{5} = \dfrac{13}{5}$$

Hence $$A = \left(\dfrac{13}{5}, 0\right)$$.

Write the line $$\dfrac{x}{a}\cos\theta + \dfrac{y}{b}\sin\theta = 1$$ in the general form:

$$\dfrac{\cos\theta}{a}x + \dfrac{\sin\theta}{b}y - 1 = 0$$

Let $$D = \sqrt{\dfrac{\cos^2\theta}{a^2} + \dfrac{\sin^2\theta}{b^2}}$$ denote the denominator that appears in the distance formula. The perpendicular distances from $$\left(\sqrt{a^2 - b^2}, 0\right)$$ and $$\left(-\sqrt{a^2 - b^2}, 0\right)$$ are

$$p_1 = \dfrac{\left|\dfrac{\cos\theta}{a}\sqrt{a^2 - b^2} - 1\right|}{D}, \qquad p_2 = \dfrac{\left|-\dfrac{\cos\theta}{a}\sqrt{a^2 - b^2} - 1\right|}{D}$$

Writing $$u = \dfrac{\cos\theta}{a}\sqrt{a^2 - b^2}$$, the numerators are $$|u - 1|$$ and $$|u + 1|$$, whose product is $$|u^2 - 1|$$. So

$$p_1 p_2 = \dfrac{|u^2 - 1|}{D^2} = \dfrac{\left|\dfrac{\cos^2\theta}{a^2}(a^2 - b^2) - 1\right|}{D^2}$$

Numerator.

$$\dfrac{\cos^2\theta(a^2 - b^2)}{a^2} - 1 = \dfrac{a^2\cos^2\theta - b^2\cos^2\theta - a^2}{a^2} = \dfrac{-a^2(1 - \cos^2\theta) - b^2\cos^2\theta}{a^2} = \dfrac{-(a^2\sin^2\theta + b^2\cos^2\theta)}{a^2}$$

So the numerator equals $$\dfrac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}$$.

Denominator.

$$D^2 = \dfrac{\cos^2\theta}{a^2} + \dfrac{\sin^2\theta}{b^2} = \dfrac{b^2\cos^2\theta + a^2\sin^2\theta}{a^2 b^2}$$

Product.

$$p_1 p_2 = \dfrac{\dfrac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2}}{\dfrac{a^2\sin^2\theta + b^2\cos^2\theta}{a^2 b^2}} = \dfrac{a^2 b^2}{a^2} = b^2$$

Hence the product of the lengths of the two perpendiculars is $$b^2$$.

To reach the path $$6x - 7y + 8 = 0$$ in the least time, the person must travel along the shortest route, which is the perpendicular from the junction point to that line.

Step 1 — find the junction point (intersection of $$2x - 3y + 4 = 0$$ and $$3x + 4y - 5 = 0$$).

Multiply the first equation by $$4$$ and the second by $$3$$:

$$8x - 12y + 16 = 0, \qquad 9x + 12y - 15 = 0$$

Adding them: $$17x + 1 = 0 \;\Rightarrow\; x = -\dfrac{1}{17}$$.

From $$2x - 3y + 4 = 0$$: $$3y = 2x + 4 = -\dfrac{2}{17} + 4 = \dfrac{66}{17} \;\Rightarrow\; y = \dfrac{22}{17}$$.

So the junction is $$\left(-\dfrac{1}{17}, \dfrac{22}{17}\right)$$.

Step 2 — find the path. The slope of the line $$6x - 7y + 8 = 0$$ is $$\dfrac{6}{7}$$. The required path is perpendicular to it, so its slope is

$$m = -\dfrac{7}{6}$$

Equation of the path through the junction $$\left(-\dfrac{1}{17}, \dfrac{22}{17}\right)$$:

$$y - \dfrac{22}{17} = -\dfrac{7}{6}\left(x + \dfrac{1}{17}\right)$$

Multiply both sides by $$6$$:

$$6y - \dfrac{132}{17} = -7x - \dfrac{7}{17}$$

$$7x + 6y - \dfrac{132}{17} + \dfrac{7}{17} = 0 \;\Rightarrow\; 7x + 6y - \dfrac{125}{17} = 0$$

Multiplying by $$17$$:

$$119x + 102y - 125 = 0$$