Equation of a Straight Line

(i) A(-2, 1) and B(3, -4)

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow m = \frac{-4 - 1}{3 - (-2)} \\[1em] = \frac{-5}{3 + 2} \\[1em] = \frac{-5}{5} \\[1em] = -1.\end{aligned}$$

Hence, slope is -1.

(ii) A(0, –3) and B(2, 1)

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow m = \frac{1 - (-3)}{2 - 0} \\[1em] = \frac{1 + 3}{2} \\[1em] = \frac{4}{2} \\[1em] = 2.\end{aligned}$$

Hence, slope is 2.

(iii) A(4, –9) and B(–2, –1)

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow m = \frac{-1 - (-9)}{-2 - 4} \\[1em] = \frac{-1 + 9}{-6} \\[1em] = \frac{8}{-6} \\[1em] = -\frac{4}{3}.\end{aligned}$$

Hence, slope is $-\frac{4}{3}$.

(iv) A(2, 5) and B(–4, –4)

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow m = \frac{-4 - 5}{-4 - 2} \\[1em] = \frac{-9}{-6} \\[1em] = \frac{3}{2}.\end{aligned}$$

Hence, slope is $\frac{3}{2}$.

Given,

P(k, 3) and Q(8, –6)

Slope = $\frac{-3}{4}$

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow \text{Slope of PQ} = \frac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\frac{3}{4} = \frac{-6 - 3}{8 - k} \\[1em] \Rightarrow -\frac{3}{4} = \frac{-9}{8 - k} \\[1em] \Rightarrow 3(8 - k) = 4(9) \\[1em] \Rightarrow 24 -3k = 36 \\[1em] \Rightarrow -3k = 36 - 24 \\[1em] \Rightarrow -3k = 12 \\[1em] \Rightarrow k = \frac{12}{-3} \\[1em] \Rightarrow k = -4.\end{aligned}$$

Hence, k = -4.

To prove that the points A(1, 4), B(3, –2), and C(–3, 16) are collinear we must show that the slope between any pair of points is the same.

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow \text{Slope of AB} = \frac{-2 - 4}{3 - 1} \\[1em] = \frac{-6}{2} \\[1em] = -3. \\[1em] \Rightarrow \text{Slope of BC} = \frac{16 -(-2)}{-3 - 3} \\[1em] = \frac{16 + 2}{-6} \\[1em] = \frac{18}{-6} \\[1em] = -3.\end{aligned}$$

Slope of AB = Slope of BC

Hence, proved points A, B and C are collinear.

Given,

Points P, Q and R are collinear.

Thus, the slope of PQ equal to the slope of QR.

By formula,

Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

$$\begin{aligned}\Rightarrow \text{Slope of PQ} = \frac{-5 - 1}{2 - k} \\[1em] = \frac{-6}{2 - k} \\[1em] \Rightarrow \text{Slope of QR} = \frac{-3 -(-5)}{(k - 2) - 2} \\[1em] = \frac{-3 + 5}{k - 4} \\[1em] = \frac{2}{k - 4}.\end{aligned}$$

Slope of PQ = Slope of QR

$\Rightarrow \frac{-6}{2 - k} = \frac{2}{k - 4}$

⇒ -6(k - 4) = 2(2 - k)

⇒ -6k + 24 = 4 - 2k

⇒ 24 - 4 = -2k + 6k

⇒ 20 = 4k

⇒ k = $\frac{20}{4}$

⇒ k = 5

Hence, k = 5.

We know that the equation of straight line parallel to x-axis is

y = a

Since the line passes through the point (-3,2), we get

a = 2

∴ Equation of the line

⇒ y = 2 or y - 2 = 0.

Hence, equation of the line is y = 2.

We know that the equation of straight line parallel to y-axis is x = a.

Since the line passes through the point (-7, 5), we get

a = -7

∴ Equation of the line

⇒ x = -7 or x + 7 = 0.

Hence, equation of the line is x + 7 = 0.

Given,

θ = 30° and c = -2.

We know that,

m = tan θ = tan 30° = $\frac{1}{\sqrt3}$.

Substituting values of m and c in equation y = mx + c, we get :

$$\begin{aligned}\Rightarrow y = \frac{1}{\sqrt3} x + (-2) \\[1em] \Rightarrow \sqrt3y = x - 2\sqrt{3} \\[1em] \Rightarrow \sqrt3y - x + 2\sqrt{3} = 0.\end{aligned}$$

Hence, equation of the line is $\sqrt3y - x + 2\sqrt{3}$.

(i) The equation of the straight line is given by, y = mx + c, we get where m is the slope and c is the y-intercept.

Given slope = $\frac{3}{4}$ and y-intercept = -4. Substituting values in equation we get,

⇒ y = $\frac{3}{4}$x - 4.

⇒ y = $\frac{3x - 16}{4}$

⇒ 4y = 3x - 16

⇒ 3x - 4y = 16.

Hence, equation of the line is 3x - 4y = 16.

(ii) The equation of straight line is given by y = mx + c, where m is the slope and c is the y-intercept.

Given slope = $\sqrt{3}$ and y-intercept = $\frac{-2}{3}$.

Substituting values in equation we get,

$$\begin{aligned}\Rightarrow y = \sqrt{3}x + \Big(\frac{-2}{3}\Big) \\[1em] \Rightarrow y = \frac{3\sqrt{3}x -2}{3} \\[1em] \Rightarrow 3y = 3\sqrt{3}x - 2 \\[1em] \Rightarrow 3\sqrt{3}x -3y - 2 = 0.\end{aligned}$$

Hence, equation of the line is $3\sqrt{3}x -3y - 2 = 0$.

m = tan θ = tan 60° = $\sqrt3$

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

$$\begin{aligned}\Rightarrow y - (-3) = \sqrt3(x - 0) \\[1em] \Rightarrow y + 3 = \sqrt3x \\[1em] \Rightarrow \sqrt3x - y = 3.\end{aligned}$$

Hence, equation of the line is $\sqrt3x - y = 3$.

(i) By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 7 = 4(x - 3)

⇒ y - 7 = 4x - 12

⇒ y = 4x - 12 + 7

⇒ y = 4x - 5

⇒ 4x - y = 5.

Hence, equation of the line is 4x - y = 5.

(ii) By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 3 = -3(x - (-2))

⇒ y - 3 = -3(x + 2)

⇒ y - 3 = -3x - 6

⇒ y = -3x - 6 + 3

⇒ y = -3x - 3

⇒ 3x + y + 3 = 0

Hence, equation of the line is 3x + y + 3 = 0.

(i) Given,

⇒ 5x – 10y = 3

Converting 5x – 10y = 3 in the form y = mx + c we get,

⇒ -10y = 3 - 5x

⇒ y = $-\frac{3}{10} - \frac{5x}{-10}$

⇒ y = $-\frac{3}{10} + \frac{x}{2}$

⇒ y = $\frac{1}{2}x - \frac{3}{10}$

The equation of straight line is given by, y = mx + c, where m is the slope and c is the y-intercept.

Comparing, y = mx + c with y = $\frac{1}{2}x - \frac{3}{10}$, we get:

m = slope = $\frac{1}{2}$

c = y-intercept = $-\frac{3}{10}$

Hence, slope = $\frac{1}{2}, \text{y-intercept} = -\frac{3}{10}$.

(ii) Converting $\frac{x}{6} + \frac{y}{9} = 1$ in the form y = mx + c we get,

$$\begin{aligned}\Rightarrow \frac{x}{6} + \frac{y}{9} = 1 \\[1em] \Rightarrow \frac{y}{9} = 1 - \frac{x}{6} \\[1em] \Rightarrow y = 9 \Big(1 - \frac{x}{6}\Big) \\[1em] \Rightarrow y = \Big(9 - \frac{9x}{6}\Big) \\[1em] \Rightarrow y = -\frac{3x}{2} + 9.\end{aligned}$$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with $y = -\frac{3x}{2} + 9$, we get:

m = slope = $-\frac{3}{2}$

c = y-intercept = 9

Hence, slope = $-\frac{3}{2}$, y-intercept = 9.

(iii) Given,

x + 4 = 0

x = -4

This is a vertical line parallel to the y-axis, passing through the x-axis at x = -4.

We know that the inclination of a line parallel to y-axis is 90°.

∴ Slope of y-axis = tan 90° = infinity, which is not defined.

Since the line is parallel to the y-axis and passes through a negative x-value, it never crosses the y-axis. There is no y-intercept.

Hence, slope is not defined and line has no y-intercept.

(iv) Given,

y = 6

This is a horizontal line parallel to the x-axis, passing through the y-axis at y = 6.

We know that the inclination of a line parallel to x-axis is 0°.

∴ Slope of a line parallel to x-axis = tan 0° = 0.

y-intercept = 6

Hence, slope = 0 and y-intercept = 6.

(i) Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1}$

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

= $\frac{-5}{5}$

= -1.

Equation : y - y₁ = m(x - x₁)

⇒ y - 1 = -1(x - (-2))

⇒ y - 1 = -1(x + 2)

⇒ y - 1 = -x - 2

⇒ x + y - 1 + 2 = 0

⇒ x + y + 1 = 0.

Hence, slope = -1, equation of AB is x + y + 1 = 0.

(ii) Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1}$

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

= $\frac{-1}{-2}$

= $\frac{1}{2}$.

Equation : y - y₁ = m(x - x₁)

⇒ y - (-2) = $\frac{1}{2}$ (x - 4)

⇒ 2(y + 2) = (x - 4)

⇒ 2y + 4 = x - 4

⇒ x - 2y - 4 - 4 = 0

⇒ x - 2y - 8 = 0.

Hence, slope = $\frac{1}{2}$, equation of AB is x - 2y - 8 = 0.

(a) From graph,

Co-ordinates of A = (0, 6) and D = (-3, 0)

(b) By formula,

Co-ordinates of centroid = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

$$\begin{aligned}= \Big(\frac{0 + 4 + (-4)}{3}, \frac{6 + (-4) + (-2)}{3}\Big) \\[1em] = \Big(\frac{0}{3}, \frac{0}{3}\Big) \\[1em] = (0, 0).\end{aligned}$$

Hence, co-ordinates of centroid of ∆ABC = (0, 0).

(c) By section-formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Given,

D divides AC in the ratio k : 1.

$$\begin{aligned}\therefore (-3, 0) = \Big(\frac{k \times -4 + 1 \times 0}{k + 1}, \frac{k \times -2 + 1 \times 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\frac{-4k + 0}{k + 1}, \frac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow (-3, 0) = \Big(\frac{-4k}{k + 1}, \frac{-2k + 6}{k + 1}\Big) \\[1em] \Rightarrow -3 = -\frac{4k}{k + 1} \text{ and } 0 = \frac{-2k + 6}{k + 1} \\[1em] \Rightarrow -3(k + 1) = -4k \text{ and } 0 = -2k + 6 \\[1em] \Rightarrow 3(k + 1) = 4k \text{ and } 2k = 6 \\[1em] \Rightarrow 3k + 3 = 4k \text{ and } k = \frac{6}{2} \\[1em] \Rightarrow 4k - 3k = 3 \text{ and } k = 3 \\[1em] \Rightarrow k = 3.\end{aligned}$$

Hence, k = 3.

(d) By two point form,

Equation of line :

y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of BD :

⇒ y - (-4) = $\frac{0 - (-4)}{-3 - 4}(x - 4)$

⇒ y + 4 = $\frac{4}{-7}(x - 4)$

⇒ -7(y + 4) = 4(x - 4)

⇒ -7y - 28 = 4x - 16

⇒ 4x + 7y - 16 + 28 = 0

⇒ 4x + 7y + 12 = 0.

Hence, equation of BD is 4x + 7y + 12 = 0.

(i) Given points, P(-1, 4) and Q(5, -2)

$$\begin{aligned}\Rightarrow \text{ Slope of PQ } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-2 - 4}{5 - (-1)} \\[1em] = \frac{-6}{6} = -1.\end{aligned}$$

By point-slope form,

Equation of the line PQ, y - y₁ = m(x - x₁)

⇒ y - 4 = -1[x - (-1)]

⇒ y - 4 = -1[x + 1]

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y - 3 = 0.

Hence, equation of line is x + y - 3 = 0.

(ii) For point A (on x-axis), y = 0.

So, putting y = 0 in the equation of PQ, we have

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0).

For point B (on y-axis), x = 0.

So, putting x = 0 in the equation of PQ, we have

⇒ 0 + y = 3

⇒ y = 3

∴ B = (0, 3).

Hence, co-ordinates of A = (3, 0) and B = (0, 3).

(iii) M is the mid-point of AB.

∴ M = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

= $\Big(\frac{3 + 0}{2}, \frac{0 + 3}{2}\Big)$

= $\Big(\frac{3}{2}, \frac{3}{2}\Big)$

Hence, mid-point of AB = $\Big(\frac{3}{2}, \frac{3}{2}\Big)$.

(i) Given points, A(2, 3) and B(–2, 5)

$$\begin{aligned}\Rightarrow \text{ Slope of AB } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{5 - 3}{-2 -2} \\[1em] = \frac{2}{-4} = -\frac{1}{2}.\end{aligned}$$

Hence, slope = $-\frac{1}{2}$.

(ii) By point-slope form,

Equation of the line AB, y - y₁ = m(x - x₁)

⇒ y - 3 = $-\frac{1}{2}$(x - 2)

⇒ 2(y - 3) = -1(x - 2)

⇒ 2y - 6 = -x + 2

⇒ x + 2y = 6 + 2

⇒ x + 2y = 8.

Hence, equation of line is x + 2y = 8.

(iii) The line intersects the x-axis when y = 0. Substituting y = 0 into the equation of the line, x + 2y = 8, we get :

⇒ x + 2(0) = 8

⇒ x = 8

Hence, coordinates of the point where AB intersects the x-axis are (8, 0).

(i) Given points, A(2, –4) and B(5, –2)

$$\begin{aligned}\Rightarrow \text{ Slope of AB } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-2 - (-4)}{5 - 2} \\[1em] = \frac{2}{3}.\end{aligned}$$

Hence, slope = $\frac{2}{3}$.

(ii) By point-slope form,

Equation of the line AB, y - y₁ = m(x - x₁)

⇒ y - (-4) = $\frac{2}{3}$(x - 2)

⇒ 3(y + 4) = 2(x - 2)

⇒ 3y + 12 = 2x - 4

⇒ 3y - 2x + 12 + 4 = 0

⇒ 3y - 2x + 16 = 0

Hence, equation of line is 3y - 2x + 16 = 0.

(iii) Since the line AB passes through the point P(k + 3, k - 4), the coordinates of P must satisfy the equation of the line, x - 2y - 10 = 0.

⇒ 3(k - 4) - 2(k + 3) + 16 = 0

⇒ 3k - 12 - 2k - 6 + 16 = 0

⇒ (3k - 2k) + (-12 - 6 + 16) = 0

⇒ k - 2 = 0

⇒ k = 2.

Hence, k = 2.

Let median through C be CX.

We know that, the median, CX through C will bisect the line AB.

By Mid-point formula,

Mid-point = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

The co-ordinates of point X are

$\Big(\frac{3 + 7}{2}, \frac{4 + (-2)}{2}\Big) = \Big(\frac{10}{2}, \frac{2}{2}\Big)$ = (5, 1).

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of CX = $\frac{1 - (-1)}{5 - (-2)} = \frac{2}{7}$.

Then, the required equation of the median CX is given by :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = $\frac{2}{7}$[x - (2)]

⇒ 7(y + 1) = 2(x + 2)

⇒ 7y + 7 = 2x + 4

⇒ 7y = 2x + 4 - 7

⇒ 2x - 7y - 3 = 0

Hence, equation of line is 2x - 7y - 3 = 0.

Given AM : MB = 1 : 2. By section-formula the coordinates of M are,

$\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \Big)$

Substituting values we get,

$$\begin{aligned}M = \Big(\frac{1(-1) + 2(2)}{1 + 2}, \frac{1(2) + 2(5)}{1 + 2} \Big) \\[1em] = \Big(\frac{-1 + 4}{3}, \frac{2 + 10}{3} \Big) \\[1em] = \Big(\frac{3}{3}, \frac{12}{3} \Big) \\[1em] = (1, 4).\end{aligned}$$

Equation of line CM can be given by two-point formula i.e.,

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Substituting values we get,

⇒ y − 8 = $\frac{4 - 8}{1 - 5}$(x − 5)

⇒ y − 8 = $\frac{-4}{-4}$(x − 5)

⇒ y − 8 = 1(x − 5)

⇒ y − 8 = x − 5

⇒ x − y − 5 + 8 = 0

⇒ x − y + 3 = 0.

Hence, the equation of CM is x - y + 3 = 0 and the coordinates of M are (1, 4).

Given,

Coordinates A(2, −11) and B(2, 13)

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of AB = $\frac{13 - (-11)}{2 - 2} = \frac{13 + 11}{0} =\frac{24}{0}$

Slope is not defined.

The line AB is a vertical line parallel to the y-axis.

Points have the same x-coordinate, x = 2.

Equation of line AB: x = 2

Given, Points: B(2,13), C(−12,1)

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of BC = $\frac{1 - (13)}{-12 - 2} = \frac{-12}{-14} =\frac{6}{7}$

By point-slope form,

Equation of the line BC, y - y₁ = m(x - x₁)

⇒ y - 13 = $\frac{6}{7}$ (x - 2)

⇒ 7(y - 13) = 6(x - 2)

⇒ 7y - 91 = 6x - 12

⇒ 7y - 6x = -12 + 91

⇒ 7y - 6x = 79

Equation of BC: 7y - 6x = 79

Given, Points: C(−12,1), A(2,−11)

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get,

Slope of CA = $\frac{-11 - 1}{2 - (-12)} = \frac{-12}{14} =-\frac{6}{7}$

By point-slope form,

Equation of the line CA, y - y₁ = m(x - x₁)

⇒ y - 1 = $-\frac{6}{7}$ (x + 12)

⇒ 7(y - 1) = -6(x + 12)

⇒ 7y - 7 = -6x - 72

⇒ 7y + 6x - 7 + 72 = 0

⇒ 7y + 6x + 65 = 0

Equation of the line CA: 7y + 6x + 65 = 0

Hence, the equation of AB, BC and CA are x = 2, 7y - 6x = 79, 7y + 6x + 65 = 0 respectively.

(i) By formula,

Mid-point (M) = = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Given,

D is the mid-point of BC.

∴ Co-ordinates of D

$$\begin{aligned}= \Big(\frac{0 + (-6)}{2}, \frac{4 + 4}{2}\Big) \\[1em] = \Big(\frac{-6}{2}, \frac{8}{2}\Big) \\[1em] = (-3, 4).\end{aligned}$$

Hence, coordinates of D = (-3, 4).

(ii) Slope = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{(-3) - 1} = -\frac{5}{4}$

Equation of a line :

y - y₁ = m(x - x₁)

Substituting values we get :

Equation of AD :

⇒ y - (-1) = $-\frac{5}{4}$ (x - 1)

⇒ -4(y + 1) = 5(x - 1)

⇒ -4y - 4 = 5x - 5

⇒ 5x + 4y = -4 + 5

⇒ 5x + 4y - 1 = 0.

Hence, equation of median AD is 5x + 4y - 1 = 0.

On the y-axis,

x = 0.

Substitute x = 0 into 2x − 3y = 6:

⇒ 2(0) − 3y = 6

⇒ −3y = 6

⇒ y = −2

So, the line 2x − 3y = 6 meets the y-axis at (0, -2).

Calculating slope for points (2, 3) and (0, -2).

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

m = $\frac{3 -(-2)}{2 - 0} = \frac{5}{2}$

By two-point form :

Equation of a line :

y - y₁ = m(x - x₁)

Substituting values we get :

⇒ y - (-2) = $\frac{5}{2}$ (x - 0)

⇒ 2(y + 2) = 5x

⇒ 2y + 4 = 5x

⇒ 5x - 2y - 4 = 0

Hence, equation of line 5x - 2y = 4.

When x-intercept = 5; corresponding point on the x-axis = (5, 0).

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of the line through (5, 0) and (4, -3) = $\frac{-3 - 0}{4 - 5} = \frac{-3}{-1} = 3$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 3(x − 5)

⇒ y = 3x - 15

⇒ 3x - y = 15.

Hence, equation of line is 3x - y = 15.

From figure,

The lines intersect at point I, J, K and L.

By two point formula,

Equation of line : $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

Equation of diagonal IK is

$$\begin{aligned}\Rightarrow y - 2 = \frac{-1 - 2}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y - 2 = -\frac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y - 2) = -3[x + 1] \\[1em] \Rightarrow 5y - 10 = -3x -3 \\[1em] \Rightarrow 5y + 3x -10 + 3 = 0 \\[1em] \Rightarrow 3x + 5y - 7 = 0 \\[1em] \Rightarrow 3x + 5y = 7.\end{aligned}$$

Equation of diagonal LJ is

⇒ y - (-1) = $\frac{2 - (-1)}{4 - (-1)}$ [x - (-1)]

⇒ y - (-1) = $\frac{3}{5}$ [x - (-1)]

⇒ 5(y + 1) = 3[x + 1]

⇒ 5y + 5 = 3x + 3

⇒ 5y − 3x + 5 − 3 = 0

⇒ 5y - 3x + 2 = 0

⇒ 3x - 5y = 2.

Hence, equation of diagonals are 3x - 5y = 2 and 3x + 5y = 7.

Let the line containing the point (3, 2) passes through x-axis at A(x, 0) and y-axis at B(0, y).

Given, the intercepts made on both the axes are equal.

∴ x = y

Slope of the line

$$\begin{aligned}m = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{0 - y}{x - 0} \\[1em] = \frac{-y}{x} \\[1em] = \frac{-x}{x} \\[1em] = -1.\end{aligned}$$

Hence, the equation of the line will be

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = -1(x - 3)

⇒ y - 2 = -x + 3

⇒ y + x - 2 - 3 = 0

⇒ y = -x + 5.

Comparing above equation with y = mx + c, we get :

c = 5.

Thus, y-intercept = 5.

∴ x-intercept = 5.

Hence, equation of line is x + y = 5 and length of x and y intercept is 5 units.

⇒ 5x + 7y = 3 ….(1)

⇒ 2x - 3y = 7 ….(2)

Multiplying equation (1) by 3, we get :

⇒ 3(5x + 7y) = 3.3

⇒ 15x + 21y = 9 ….(3)

Multiplying equation (2) by 7, we get :

⇒ 7(2x - 3y) = 7.7

⇒ 14x - 21y = 49 ….(4)

Adding equations (3) and (4) we get,

⇒ 15x + 21y + 14x - 21y = 9 + 49

⇒ 29x = 58

⇒ x = $\frac{58}{29}$

⇒ x = 2.

Substituting x = 2 in (1), we get :

⇒ 5(2) + 7y = 3

⇒ 10 + 7y = 3

⇒ 7y = 3 - 10

⇒ 7y = -7

⇒ y = $\frac{-7}{7}$

⇒ y = -1.

Hence, the point of intersection of lines is (2, -1).

The equation of the line joining (2, -1) and (0, 0) will be given by two-point form i.e.,

$y - y _1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$

Substituting values in above equation we get,

⇒ y - (-1) = $\frac{0 - (-1)}{0 - 2}$ (x - 2)

⇒ y + 1 = $\frac{1}{-2}$ (x - 2)

⇒ -2(y + 1) = (x - 2)

⇒ -2y - 2 = x - 2

⇒ x + 2y - 2 + 2 = 0

⇒ x + 2y = 0.

Hence, equation of line is x + 2y = 0.

(i) Let the coordinates of M and N be (x, 0) and (0, y).

By section formula the coordinates of P are,

$$\begin{aligned}\Rightarrow \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\frac{2(0) + 3x}{2 + 3}, \frac{2y + 3(0)}{2 + 3}\Big) \\[1em] = \Big(\frac{3x}{5}, \frac{2y}{5}\Big)\end{aligned}$$

Given, P(3, 2). Comparing two values of P we get,

⇒ 3 = $\frac{3x}{5}$ and 2 = $\frac{2y}{5}$

⇒ 3x = 15 and 2y = 10

⇒ x = 5 and y = 5.

Hence, the coordinates of M and N are (5, 0) and (0, 5) respectively.

(ii) Slope of line MN can be given by $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting value in above equation we get slope,

= $\frac{5 - 0}{0 - 5}$

= $\frac{5}{-5}$

= -1.

Hence, the slope of the line is -1.

Let P divides line AB in the ratio m : n.

By section formula,

$$\begin{aligned}\Rightarrow P(x, y) = \Big(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\Big) \\[1em] = \Big(\frac{m(-3) + n(2)}{m + n}, \frac{m(6) + n(1)}{m + n}\Big) \\[1em] = \Big(\frac{-3m + 2n}{m + n}, \frac{6m + n}{m + n}\Big).\end{aligned}$$

Since, point P lies on line x - 5y + 15 = 0, substituting values we get :

$$\begin{aligned}\Rightarrow \frac{-3m + 2n}{m + n} - 5 \Big(\frac{6m + n}{m + n}\Big) + 15 = 0 \\[1em] \Rightarrow \frac{-3m + 2n - 5(6m + n) + 15(m + n)}{(m + n)} = 0 \\[1em] \Rightarrow -3m + 2n - 30m - 5n + 15m + 15n = 0 \\[1em] \Rightarrow -18m + 12n = 0 \\[1em] \Rightarrow 6(-3m + 2n) = 0 \\[1em] \Rightarrow -3m + 2n = 0 \\[1em] \Rightarrow 2n = 3m \\[1em] \Rightarrow \frac{m}{n} = \frac{2}{3}.\end{aligned}$$

By section formula,

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\Big) \\[1em] \Rightarrow (x, y) = \Big(\frac{2(-3) + 3(2)}{2 + 3}, \frac{2(6) + 3(1)}{2 + 3}\Big) \\[1em] \Rightarrow (x, y) = \Big(\frac{-6 + 6}{5}, \frac{12 + 3}{5}\Big) \\[1em] \Rightarrow (x, y) = 0, \frac{15}{5} \\[1em] \Rightarrow (x, y) = (0, 3).\end{aligned}$$

Hence, the line x – 5y + 15 = 0 divides AB in the ratio 2 : 3 and the co-ordinates of their point of intersection are (0, 3).

By using Midpoint formula,

(x, y) = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

D (Midpoint of BC): B(2, 1)and C(0, 4)

D = $\Big(\frac{2 + 0}{2}, \frac{1 + 4}{2}\Big) = \Big(1, \frac{5}{2})$

E (Midpoint of AC): A(-1, 2) and C(0, 4)

E = $\Big(\frac{-1 + 0}{2}, \frac{2 + 4}{2}\Big) = \Big(-\frac{1}{2}, 3\Big)$

F (Midpoint of AB): A(-1, 2) and B(2, 1)

F = $\Big(\frac{-1 + 2}{2}, \frac{2 + 1}{2}\Big) = \Big(\frac{1}{2}, \frac{3}{2}\Big)$

Median AD through A(-1, 2) and $D\Big(1, \frac{5}{2}\Big)$

By slope formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substitute values we get:

$m_{AD} = \frac{\frac{5}{2} - 2}{1 - (-1)} = \frac{\frac{1}{2}}{2} = \frac{1}{4}$

The equation of the line will be given by two-point form i.e.,

y - y₁ = m(x - x₁)

Substituting values in above equation we get,

⇒ y - 2 = $\frac{1}{4}$ (x + 4)

⇒ 4(y - 2) = (x + 1)

⇒ 4y - 8 = x + 1

⇒ x - 4y + 9 = 0

Median BE through B(2, 1) and $E\Big(-\frac{1}{2}, 3\Big)$

$m_{BE} = \frac{3 - 1}{-\frac{1}{2} - 2} = \frac{2}{-\frac{5}{2}} = -\frac{4}{5}$

The equation of the line will be given by two-point form i.e.,

⇒ y - 1 = $-\frac{4}{5}$ (x - 2)

⇒ 5(y - 1) = 4(x - 2)

⇒ 5y - 5 = 4x - 8

⇒ 4x + 5y - 13 = 0

Median CF through C(0, 4) and $F\Big(\frac{1}{2}, \frac{3}{2}\Big)$

$m_{CF} = \frac{\frac{3}{2} - 4}{\frac{1}{2} - 0} = \frac{-\frac{5}{2}}{\frac{1}{2}} = -5$

Since C(0, 4) is the y-intercept, we use y = mx + c:

⇒ y = -5x + 4

⇒ 5x + y - 4 = 0

By using centroid formula,

$G = \Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Using A(-1, 2), B(2, 1), and C(0, 4):

$$\begin{aligned}\Rightarrow G = \Big(\frac{-1 + 2 + 0}{3}, \frac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\frac{-1 + 2 + 0}{3}, \frac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\frac{1}{3}, \frac{7}{3}\Big).\end{aligned}$$

Hence, equations of medians are x - 4y + 9 = 0, 4x + 5y - 13 = 0 and 5x + y - 4 = 0, coordinates of centroid $\Big(\frac{1}{3}, \frac{7}{3}\Big)$.

By formula,

Centroid of triangle = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Centroid of △ ABC} = \Big(\frac{-1 + 3 + 0}{3}, \frac{3 + (-1) + 0}{3}\Big) \\[1em] \Rightarrow P = \Big(\frac{2}{3}, \frac{2}{3}\Big).\end{aligned}$$

By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

$$\begin{aligned}\text{Slope of AB} = \frac{-1 - 3}{3 - (-1)} \\[1em] = \frac{-4}{4} \\[1em] = -1.\end{aligned}$$

We know that,

Slope of parallel lines are equal.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get :

Equation of line passing through P and parallel to AB :

$$\begin{aligned}\Rightarrow y - \frac{2}{3} = -1\Big(x - \frac{2}{3}\Big) \\[1em] \Rightarrow \frac{3y - 2}{3} = -1 \times \frac{3x - 2}{3} \\[1em] \Rightarrow 3y - 2 = -1(3x - 2) \\[1em] \Rightarrow 3y - 2 = -3x + 2 \\[1em] \Rightarrow 3y + 3x = 2 + 2 \\[1em] \Rightarrow 3y + 3x = 4.\end{aligned}$$

Hence, required equation is 3x + 3y = 4.

The parallelogram ABCD is shown in the figure below:

(i) We know that the diagonals of a parallelogram bisect each other. Let (x, y) be the coordinates of D.

Mid-point of diagonal AC = $\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} = \frac{3 + 3}{2}, \frac{6 + 2}{2} = (3, 4)$

Mid-point of diagonal BD = $\frac{5 + x}{2}, \frac{10 + y}{2}$

These two should be same. On equating we get,

$\frac{5 + x}{2} = 3, \frac{10 + y}{2} = 4$

⇒ 5 + x = 6 and 10 + y = 8

⇒ x = 6 − 5 and y = 8 − 10

⇒ x = 1 and y = −2.

Hence, coordinates of D are (1, -2).

(ii) By distance formula the distance between B(5, 10) and D(1, -2) is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values we get BD,

$$\begin{aligned}= \sqrt{(1 - 5)^2 + (-2 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-12)^2} \\[1em] = \sqrt{16 + 144} \\[1em] = \sqrt{160} \\[1em] = 4\sqrt{10}\end{aligned}$$

Hence, the length of diagonal BD is $4\sqrt{10}$ units.

(iii) Equation of side AB can be given by two point formula i.e.,

$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$

Substituting values we get,

⇒ y − 6 = $\frac{10 - 6}{5 - 3}$ (x - 3)

⇒ y − 6 = $\frac{4}{2}$ (x - 3)

⇒ (y − 6) = 2(x - 3)

⇒ (y − 6) = 2x - 6

⇒ 2x - y -6 + 6 = 0

⇒ 2x - y = 0

Hence, equation of the line is 2x - y = 0.

(i) From figure,

The co-ordinates of A = (4, 0).

(ii) By distance formula,

D = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting values we get,

$$\begin{aligned}AB = \sqrt{(-2 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (3)^2} \\[1em] = \sqrt{36 + 9} \\[1em] = \sqrt{45} \\[1em] = 3\sqrt{5}. \\[1em] AC = \sqrt{(-2 - 4)^2 + (-4 - 0)^2} \\[1em] = \sqrt{(-6)^2 + (-4)^2} \\[1em] = \sqrt{36 + 16} \\[1em] = \sqrt{52} \\[1em] = 2\sqrt{13}.\end{aligned}$$

Hence, length of $AB = 3\sqrt{5}\text{ and } AC = 2\sqrt{13}$.

(iii) From figure,

Q lies on y-axis.

∴ x co-ordinate of Q = 0.

Let co-ordinate of Q are (0, a).

Let ratio in which Q divides AC be k : 1.

By section-formula,

$Q = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Comparing x-coordinate we get :

$$\begin{aligned}\Rightarrow 0 = \frac{k \times -2 + 1 \times 4}{k + 1} \\[1em] \Rightarrow 0 = -2k + 4 \\[1em] \Rightarrow 2k = 4 \\[1em] \Rightarrow k = 2.\end{aligned}$$

k : 1 = 2 : 1.

Hence, Q divides AC in the ratio 2 : 1.

(iv) By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of AC = $\frac{-4 - 0}{-2 - 4} = \frac{-4}{-6} = \frac{2}{3}$

By point-slope form,

Equation of AC is :

⇒ y - y₁ = m (x - x₁)

⇒ y - 0 = $\frac{2}{3}$ (x - 4)

⇒ 3y = 2x - 8

⇒ 2x - 3y = 8

Hence, equation of AC is 2x - 3y = 8.

AB is parallel to line segment CD means they must have the same gradient

By using slope formula,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Given, points A(2, –3), B(–5, 1)

Substituting values we get,

$m_{AB} = \frac{1 - (-3)}{-5 - 2} = \frac{1 + 3}{-7} = -\frac{4}{7}$

Given, points C(7, –1) and D(0, k)

Substituting values we get,

$m_{CD} = \frac{k - (-1)}{0 - 7} = \frac{k + 1}{-7}$

Equate the Gradients:

$$\begin{aligned}\Rightarrow -\frac{4}{7} = \frac{k + 1}{-7} \\[1em] \Rightarrow -\frac{4}{7} \times -7 = k + 1 \\[1em] \Rightarrow k + 1 = 4 \\[1em] \Rightarrow k = 4 - 1 \\[1em] \Rightarrow k = 3.\end{aligned}$$

Hence, value of k = 3.

Given,

⇒ x + 2y – 5 = 0

Converting x + 2y - 5 = 0 in the form y = mx + c we get,

⇒ 2y = -x + 5

⇒ y = $\frac{-x}{2} + \frac{5}{2}$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with y = $\frac{-x}{2} + \frac{5}{2}$, we get:

⇒ m₁ = $-\frac{1}{2}$

Given,

⇒ 2x + 4y + 9 = 0

Converting 2x + 4y + 9 = 0 in the form y = mx + c we get,

⇒ 4y = -2x - 9

⇒ y = $\frac{-2x}{4} - \frac{9}{4}$

⇒ y = $\frac{-1x}{2} - \frac{9}{4}$

Comparing y = mx + c with y = $\frac{-1x}{2} - \frac{9}{4}$, we get:

⇒ m₂ = $-\frac{1}{2}$

Since the gradient of the first line is equal to the gradient of the second line.

The lines are parallel to each other.

Hence, proved that lines are parallel.

Since, the lines are parallel they have same gradient.

Given, kx + 2y + 3 = 0

Converting kx + 2y + 3 = 0 in the form y = mx + c, we get :

⇒ 2y = -kx - 3

⇒ y = $-\frac{k}{2}x - \frac{3}{2}$

The equation of straight line is given by,

y = mx + c, where m is the slope and c is the y-intercept.

Comparing y = mx + c with y = $-\frac{k}{2}x - \frac{3}{2}$, we get :

⇒ m₁ = $-\frac{k}{2}$

Given,

8x + ky - 1 = 0

Converting 8x + ky - 1 = 0 in the form y = mx + c we get,

⇒ ky = -8x + 1

⇒ y = $\frac{-8x}{k} + \frac{1}{k}$

Comparing y = mx + c with y = $\frac{-8x}{k} + \frac{1}{k}$, we get :

⇒ m₂ = $-\frac{8}{k}$

Since, lines are parallel, equating the gradients :

$$\begin{aligned}\Rightarrow -\frac{k}{2} = -\frac{8}{k} \\[1em] \Rightarrow k(k) = 8(2) \\[1em] \Rightarrow k^2 = 16 \\[1em] \Rightarrow k = \sqrt{16} \\[1em] \Rightarrow k = \pm 4.\end{aligned}$$

Hence, value of k = ± 4.

Converting 2x - by + 5 = 0 in the form y = mx + c we get,

⇒ 2x - by + 5 = 0

⇒ by = 2x + 5

⇒ y = $\frac{2}{b}x + \frac{5}{b}$

m₁ = $\frac{2}{b}$

Converting ax + 3y = 2 in the form y = mx + c we get,

⇒ ax + 3y = 2

⇒ 3y = -ax + 2

⇒ y = $-\frac{a}{3}x + \frac{2}{3}$

m₂ = $-\frac{a}{3}$

Given, two lines are parallel so their slopes will be equal,

m₁ = m₂

⇒ $\frac{2}{b} = -\frac{a}{3}$

⇒ 3 × 2 = −a × b

⇒ 6 = −ab

⇒ ab + 6 = 0

⇒ ab = −6.

Hence, the relation connecting a and b is ab = -6.

The slope of the line passing through two points (x₁, y₁) and (x₂, y₂) is given by

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope (m₁) of line joining (-2, 6) and (4, 8) is,

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

Slope (m₂) of line joining (8, 12) and (4, 24) is,

=$\frac{24 - 12}{4 - 8} = \frac{12}{-4} = -3$

m₁ × m₂ = $\frac{1}{3} \times -3$ = -1.

Product of slopes = -1.

Hence, the lines are perpendicular to each other.

By using slope formula,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Given, points A(2, –5), B(–2, 5)

Substituting values we get,

$m_{AB} = \frac{5 - (-5)}{-2 - 2} = \frac{5 + 5}{-4} = -\frac{10}{4} = -\frac{5}{2}$

Given, points C(k, 3) and D(1, 1)

Substituting values we get,

$m_{CD} = \frac{1 - 3}{1 - k} = \frac{-2}{1 - k}$

Since the lines are perpendicular the product of the gradients is equal to -1:

$$\begin{aligned}\Rightarrow \Big(-\frac{5}{2}\Big) \times \Big(\frac{-2}{1 - k}\Big) = -1 \\[1em] \Rightarrow \Big(\frac{10}{2(1 - k)}\Big) = -1 \\[1em] \Rightarrow \Big(\frac{5}{1 - k}\Big) = -1 \\[1em] \Rightarrow 5 = -1(1 - k) \\[1em] \Rightarrow 5 = -1 + k \\[1em] \Rightarrow k = 5 + 1 \\[1em] \Rightarrow k = 6.\end{aligned}$$

Hence, value of k = 6.

Converting 2x + 3y + 8 = 0 in the form y = mx + c we get,

⇒ 3y = -2x - 8

⇒ y = $-\frac{2}{3}x - \frac{8}{3}$

Comparing, we get slope of this line : m₁ = $-\frac{2}{3}$

Converting 27x – 18y + 10 = 0 in the form y = mx + c we get,

⇒ -18y = -27x - 10

⇒ y = $\frac{-27x}{-18} - \frac{10}{-18}$

⇒ y = $\frac{3x}{2} + \frac{5}{9}$

Comparing, we get slope of this line : m₂ = $\frac{3}{2}$

Product of Gradients

m₁ × m₂ = $-\frac{2}{3} \times \frac{3}{2}$

= $-\frac{6}{6}$

= -1

Since the product of the gradients is -1. The lines are perpendicular to each other.

Hence, proved that lines 2x + 3y + 8 = 0 and 27x – 18y + 10 = 0 are perpendicular to each other.

Given lines,

⇒ y = 3x + 7 and 2y + px = 3

⇒ y = 3x + 7 and 2y = -px + 3

⇒ y = 3x + 7 and y = $-\frac{p}{2}x + \frac{3}{2}$

Comparing above equations with y = mx + c we get,

Slope of 1st line = 3

Slope of 2nd line = $-\frac{p}{2}$

Since,

Product of slopes of perpendicular lines = -1.

$$\begin{aligned}\therefore 3 \times -\frac{p}{2} = -1 \\[1em] \Rightarrow -p = \frac{-1 \times 2}{3} \\[1em] \Rightarrow -p = \frac{-2}{3} \\[1em] \Rightarrow p = \frac{2}{3}.\end{aligned}$$

Hence, p = $\frac{2}{3}$.

Converting 3x - 5y + 7 = 0 in the form y = mx + c we get,

⇒ 3x - 5y + 7 = 0

⇒ 5y = 3x + 7

⇒ y = $\frac{3}{5}x + \frac{7}{5}$

Comparing, we get slope of first line = m₁ = $\frac{3}{5}$

Converting 4x + ay + 9 = 0 in the form y = mx + c we get,

⇒ 4x + ay + 9 = 0

⇒ ay = -4x - 9

⇒ y = $-\frac{4}{a}x - \frac{9}{a}$

Comparing, we get slope of second line = m₂ = $-\frac{4}{a}$

Given, two lines are perpendicular so product of their slopes will be equal to -1,

$$\begin{aligned}\Rightarrow m_1 \cdot m_2 = -1 \\[1em] \Rightarrow \frac{3}{5} \times -\frac{4}{a} = -1 \\[1em] \Rightarrow -\frac{12}{5a} = -1 \\[1em] \Rightarrow a = \frac{12}{5}.\end{aligned}$$

Hence,the value of a = $\frac{12}{5}$.

By using slope formula,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Given, points A(1, 3), B(3, –1)

Substituting values we get,

$m_{AB} = \frac{-1 - 3}{3 - 1} = -\frac{4}{2} = -2$

Given, points B(3, –1) and C(–5, –5)

Substituting values we get,

$m_{BC} = \frac{-5 - (-1)}{-5 - 3} = \frac{-5 + 1}{-8} = \frac{-4}{-8} = \frac{1}{2}$

Check for perpendicularity,

$\Rightarrow m_{AB} \times m_{BC} = -2 \times \frac{1}{2} = -1.$

Since the product of the gradients of AB and BC is -1, the side AB is perpendicular to the side BC.

∠ABC = 90°.

Hence, proved the points A(1, 3), B(3, –1) and C(–5, –5) are the vertices of a right-angled triangle.

By using slope formula,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Given, points A(1, –2), B(3, 6)

Substituting values we get,

$m_{AB} = \frac{6 - (-2)}{3 - 1} = \frac{8}{2} = 4$

Given, points C(5, 10) and D(3, 2)

Substituting values we get,

$m_{DC} = \frac{10 - 2}{5 - 3} = \frac{8}{2} = 4$

Given, points B(3, 6), C(5, 10)

Substituting values we get,

$m_{BC} = \frac{10 - 6}{5 - 3} = \frac{4}{2} = 2$

Given, points A(1, –2), D(3, 2)

Substituting values we get,

$m_{AD} = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2$

m_AB = m_CD and m_BC = m_AD

∴ AB is parallel to CD and BC is parallel to AD

Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

Hence, proved that ABCD is a parallelogram.

(i) Since, AD is median. So, D is the mid-point of BC.

By using formula,

(x, y) = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big)$

Substitute values we get,

D = $\Big(\frac{-3 + 1}{2}, \frac{-2 + (-8)}{2}\Big) = \Big(\frac{-2}{2}, \frac{-10}{2}\Big) = (-1, -5)$

By using slope formula,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Slope of AD = $\frac{-5 - 4}{-1 - 5} = \frac{-9}{-6} = \frac{3}{2}$

Hence, slope of the median AD = $\frac{3}{2}$.

(ii) The altitude BM is perpendicular to the side AC. Therefore, the product of their slopes is -1.

Slope of AC = $\frac{-8 - 4}{1 - 5} = \frac{-12}{-4} = 3$

m_BM × 3 = -1

m_BM = $\frac{-1}{3}$

Hence, slope of the BM = $-\frac{1}{3}$.

Given,

⇒ 3x + 2y = 8

Converting 3x + 2y = 8 in the form y = mx + c we get,

⇒ 2y = -3x + 8

⇒ y = $-\frac{3}{2}x + \frac{8}{2}$

Comparing above equations with y = mx + c we get,

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

Since, parallel lines have equal slope.

∴ Slope of line parallel to 3x + 2y = 8 is $-\frac{3}{2}$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

Thus, equation of line with slope $-\frac{3}{2}$ and passing through (0, 1) is :

⇒ y - 1 = $-\frac{3}{2}$(x - 0)

⇒ 2(y - 1) = -3x

⇒ 2y - 2 = -3x

⇒ 3x + 2y = 2 .

Hence, equation of the line passing through (0, 1) and parallel to 3x + 2y = 8 is 3x + 2y = 2.

Simultaneously solving equations :

⇒ x + y - 4 = 0 …….(1)

⇒ 2x - y = 8 ……..(2)

Solving equation (1), we get :

⇒ x = 4 - y ………..(3)

Substituting value of x from (3) in (2), we get :

⇒ 2(4 - y) - y = 8

⇒ 8 - 2y - y = 8

⇒ 8 - 3y = 8

⇒ 3y = 0

⇒ y = 0.

Substituting value of y in (3), we get :

⇒ x = 4 - 0 = 4.

Point of intersection = (4, 0).

Given,

Equation :

⇒ 2x + y - 7 = 0

⇒ y = -2x + 7

Comparing above equation with y = mx + c, we get :

⇒ m = -2.

We know that,

Slope of parallel lines are equal.

∴ Slope of line parallel to 2x + y - 7 = -2.

By point-slope formula,

Equation of line :

⇒ y - y₁ = m(x - x₁)

Thus, equation of line with slope = -2 and passing through (4, 0).

⇒ y - 0 = -2(x - 4)

⇒ y = -2x + 8

⇒ 2x + y - 8 = 0.

Hence, the equation of required line is 2x + y - 8 = 0.

(i) Centroid of the triangle is given by,

$$\begin{aligned}G = \Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big) \\[1em] = \Big(\frac{-1 + 4 + 3}{3}, \frac{3 + 2 - 2}{3}\Big) \\[1em] = \Big(\frac{6}{3}, \frac{3}{3}\Big) \\[1em] = (2, 1).\end{aligned}$$

Hence, the coordinates of the centroid G of the triangle is (2, 1).

(ii) Slope of AC = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{3 - (-1)} = -\frac{5}{4}$

So, the slope of the line parallel to AC is also $-\frac{5}{4}$. and it passes through (2, 1). Hence, its equation can be given by point-slope form i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $-\frac{5}{4}$(x - 2)

⇒ 4(y − 1) = −5(x − 2)

⇒ 4y − 4 = −5x + 10

⇒ 4y + 5x = 14

⇒ 5x + 4y − 14 = 0.

Hence, the equation of the required line is 5x + 4y - 14 = 0.

Since the required line is parallel to the line segment AB, they must have the same gradient.

Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-3)}{-1 - 4} = -\frac{8}{5}$

Using the point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $-\frac{8}{5}$[x - (-2)]

⇒ 5(y - 1) = -8(x + 2)

⇒ 5y - 5 = -8x - 16

⇒ 8x + 5y - 5 + 16 = 0

⇒ 8x + 5y + 11 = 0

Hence, the equation of the required line is 8x + 5y + 11 = 0.

(i) 1st equation :

⇒ kx - y + 4 = 0

⇒ y = kx + 4

Slope (s₁) : k

2nd equation :

⇒ 2y = 6x + 7

⇒ y = $\frac{6}{2}x + \frac{7}{2}$

⇒ y = 3x + $\frac{7}{2}$

Slope (s₂) : 3

We know that,

Product of slope of perpendicular lines = -1

⇒ k × 3 = -1

⇒ k = $-\frac{1}{3}$

Hence, k = $-\frac{1}{3}$.

(ii) We know that,

Slope of parallel lines are equal.

Slope of line parallel to line 2y = 6x + 7 is 3.

By point-slope form :

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = 3[x - (-1)]

⇒ y - 1 = 3[x + 1]

⇒ y - 1 = 3x + 3

⇒ y = 3x + 3 + 1

⇒ y = 3x + 4.

Hence, equation of line parallel to 2y = 6x + 7 and passing through (–1, 1) is y = 3x + 4.

Converting y + 5x = 3 in the form y = mx + c we get,

⇒ y = -5x + 3

Comparing above equation with y = mx + c we get, m = -5

For two lines to be perpendicular, the product of their gradients must be -1.

Let slope of required line be m₂, then :

⇒ -5 × m₂ = -1

⇒ m₂ = $\frac{-1}{-5}$

⇒ m₂ = $\frac{1}{5}$

Using the slope-intercept form y = mx + c. Since the line passes through the origin, the y-intercept is 0.

⇒ y = $\frac{1}{5}$x + 0

⇒ 5y = x

⇒ x - 5y = 0.

Hence, the equation of the required line is x - 5y = 0.

Converting 3x - 2y + 4 = 0 in the form y = mx + c we get,

⇒ -2y = -3x - 4

⇒ y = $\frac{-3x}{-2} - \frac{4}{-2}$

⇒ y = $\frac{3x}{2}$ + 2

Comparing above equation with y = mx + c we get, m = $\frac{3}{2}$

Since the required line is parallel to the given line, they must have the same gradient:

⇒ Slope of parallel line = $\frac{3}{2}$

Using the slope-intercept form y = mx + c. Since the line passes through the origin, the y-intercept is 0.

⇒ y = $\frac{3}{2}$x + 0

⇒ 2y = 3x

⇒ 3x - 2y = 0.

Hence, the equation of the required line is 3x - 2y = 0.

Let point where line touches x-axis be A. So, A = (-3, 0)

Given equation of line,

⇒ 3x + 5y = 1

⇒ 5y = -3x + 1

⇒ y = $-\frac{3}{5}x + \frac{1}{5}$

Comparing above equation with y = mx + c we get, m₁ = $-\frac{3}{5}$

Let slope of line perpendicular to 3x + 5y = 1 be m₂

⇒ m₁ × m₂ = -1

⇒ $-\frac{3}{5} \times m_2 = -1$

⇒ $m_2 = \frac{5}{3}$

By point-slope form,

Equation of line with slope = $\frac{5}{3}$and passing through (-3, 0) is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\frac{5}{3}$[x - (-3)]

⇒ 3y = 5[x + 3]

⇒ 3y = 5x + 15

⇒ 5x - 3y + 15 = 0.

Hence, equation of required line is 5x - 3y + 15 = 0.

A line perpendicular to the x-axis, has equation x = k.

Since, line passes through (2, 4).

∴ Equation : x = 2.

Hence, equation of required line is x = 2.

Let P = (-1, 2)

Let A and B be the points (1, 4) and (2, 3).

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

We know that,

Product of slope of perpendicular lines is -1.

Let slope of line through P and perpendicular to AB be m.

∴ m × Slope of AB = -1

⇒ m × -1 = -1

⇒ -m = -1

⇒ m = 1.

By point-slope form,

Equation of line through P,

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = 1[x - (-1)]

⇒ y - 2 = 1(x + 1)

⇒ y - 2 = x + 1

⇒ y - x = 1 + 2

⇒ y - x = 3

⇒ x - y + 3 = 0.

Hence, equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3) is x - y + 3 = 0.

Given line equations are,

⇒ 7x + 6y - 71 = 0

⇒ 7x + 6y = 71 .......(1)

and

⇒ 5x - 8y + 23 = 0

⇒ 5x - 8y = -23 .......(2)

Multiplying (1) by 4, we get :

⇒ 28x + 24y = 284 .......(3)

Multiplying (2) by 3, we get :

⇒ 15x – 24y = -69 .......(4)

On adding (3) and (4), we get :

⇒ 28x + 24y + 15x - 24y = 284 + (-69)

⇒ 28x + 15x + 24y - 24y = 284 - 69

⇒ 43x = 215

⇒ x = $\frac{215}{43}$

⇒ x = 5.

Substituting value of x in equation (2), we get :

⇒ 5.5 - 8y = -23

⇒ 25 - 8y = -23

⇒ 8y = 25 + 23

⇒ 8y = 48

⇒ y = 6.

Hence, the required line passes through the point (5, 6).

Given,

⇒ 4x – 2y = 3

⇒ 2y = 4x – 3

⇒ y = 2x – $\frac{3}{2}$.

Comparing above equation with y = mx + c we get,

Slope (m) = 2

Let slope of required line be m₁.

As, the lines are perpendicular to each other so product of their slopes = -1.

⇒ m × m₁ = -1

⇒ 2 × m₁ = -1

⇒ m₁ = $-\frac{1}{2}$.

Thus, equation of the line with slope $-\frac{1}{2}$ and passing through (5, 6) is :

⇒ y – y₁ = m(x – x₁)

⇒ y – 6 = $-\frac{1}{2}$(x – 5)

⇒ 2(y – 6) = -1(x - 5)

⇒ 2y - 12 = -x + 5

⇒ 2y + x = 5 + 12

⇒ x + 2y = 17.

Hence, equation of required line is x + 2y = 17.

Let the line passing through the origin O(0, 0) be L1 and slope m₁, and the given line be L2 : 2x = 3y + 13 and slope be m₂.

Given equation of line,

⇒ 2x = 3y + 13

⇒ 3y = 2x - 13

⇒ y = $\frac{2}{3}x - \frac{13}{3}$

Comparing above equation with y = mx + c we get, Slope (m₂) = $\frac{2}{3}$.

Since L1 is perpendicular to L2, the product of their slopes is -1.

⇒ m₁ × m₂ = -1

⇒ m₁ × $\frac{2}{3}$ = -1

⇒ m₁ = $-\frac{3}{2}$.

The equation of the line L1 having slope m₁ and passing through the origin can be given by point-slope form i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{3}{2}$(x - 0)

⇒ y = $-\frac{3x}{2}$

⇒ 2y = -3x

⇒ 2y + 3x = 0.

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines.

-3y + 2x - 13 = 0 .......(i)

2y + 3x = 0 ..........(ii)

On multiplying equation (i) by 2, we get :

-6y + 4x - 26 = 0 ..........(iii)

On multiplying equation (ii) by 3 we get,

6y + 9x = 0 .........(iv)

Adding (iii) and (iv) we get,

⇒ -6y + 4x - 26 + 6y + 9x = 0

⇒ 13x - 26 = 0

⇒ 13x = 26

⇒ x = $\frac{26}{13}$

⇒ x = 2.

Substituting value of x in (ii), we get :

⇒ 2y + 3(2) = 0

⇒ 2y + 6 = 0

⇒ 2y = -6

⇒ y = $\frac{-6}{2}$

⇒ y = -3.

∴ Coordinates = (2, -3)

Hence, coordinates of Q = (2, -3).

Solving,

⇒ 4x - 3y - 5 = 0

⇒ 3y = 4x - 5

⇒ y = $\frac{4}{3}x - \frac{5}{3}$

Comparing above equation with y = mx + c we get, Slope of the line (m₁) = $\frac{4}{3}$

Let the slope of the line perpendicular to 4x - 3y - 5 = 0 be m₂.

Then,

$$\begin{aligned}\Rightarrow m_1 × m_2 = -1 \\[1em] \Rightarrow \frac{4}{3} × m_2 = -1 \\[1em] \Rightarrow m_2 = -\frac{3}{4}.\end{aligned}$$

The equation of the line having slope m₂ and passing through the point (1, -2) can be given by point-slope form i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $-\frac{3}{4}(x - 1)$

⇒ 4(y + 2) = −3(x − 1)

⇒ 4y + 8 = −3x + 3

⇒ 3x + 4y + 5 = 0.

For finding the coordinates of the foot of the perpendicular which is the point of intersection of the lines. Let point of intersection of lines be Q.

4x - 3y - 5 = 0 .......(1)

3x + 4y + 5 = 0 ........(2)

On multiplying equation (1) by 4, we get :

16x - 12y - 20 = 0 ...........(3)

On multiplying equation (2) by 3, we get :

9x + 12y + 15 = 0 ..........(4)

Adding equations (3) and (4) we get,

⇒ 16x - 12y - 20 + 9x + 12y + 15 = 0

⇒ 25x - 5 = 0

⇒ x = $\frac{5}{25}$

⇒ x = $\frac{1}{5}$.

Substituting value of x in (1), we get :

$$\begin{aligned}\Rightarrow 4\times \frac{1}{5} - 3y - 5 = 0 \\[1em] \Rightarrow \frac{4}{5} - 3y - 5 = 0 \\[1em] \Rightarrow 3y = \frac{4}{5} - \frac{25}{5} \\[1em] \Rightarrow 3y = -\frac{21}{5} \\[1em] \Rightarrow y = -\frac{7}{5}.\end{aligned}$$

∴ Q = $\Big(\frac{1}{5}, -\frac{7}{5}\Big)$

Hence, the equation of the new line is 3x + 4y + 5 = 0 and coordinates of the foot of perpendicular are $\Big(\frac{1}{5}, -\frac{7}{5}\Big)$.

Let A be the point where the line $\frac{x}{a} - \frac{y}{b} = 1$ meets y-axis.

So, x-co-ordinate of point A will be zero.

Substituting x = 0 in equation we get,

$$\begin{aligned}\Rightarrow \frac{0}{a} - \frac{y}{b} = 1 \\[1em] \Rightarrow -\frac{y}{b} = 1 \\[1em] \Rightarrow y = -b\end{aligned}$$

A = (0, -b).

The given line equation is,

$$\begin{aligned}\Rightarrow \frac{x}{a} - \frac{y}{b} = 1 \\[1em] \Rightarrow \frac{y}{b} = \frac{x}{a} - 1 \\[1em] \Rightarrow y= \frac{bx}{a} - b\end{aligned}$$

Comparing above equation with y = mx + c we get,

Slope (m) = $\frac{b}{a}$

Let slope of perpendicular line be m₁.

As product of slope of perpendicular lines is -1,

∴ m × m₁ = -1

⇒ $\frac{b}{a}$ × m₁ = -1

⇒ m₁ = $-\frac{a}{b}$

Equation of line through A (0, -b) and slope = $-\frac{a}{b}$ is :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-b) = $-\frac{a}{b}$(x - 0)

⇒ b(y + b) = -ax

⇒ by + b² = -ax

⇒ ax + by + b² = 0

Hence, equation of required line is ax + by + b² = 0.

(i) Given,

⇒ x + 2y + 6 = 0

⇒ 2y = -x - 6

⇒ y = $\frac{-x - 6}{2}$

⇒ y = -$\frac{1}{2}x$ - 3

Comparing y = $-\frac{1}{2}x$ - 3 with y = mx + c, we get :

Slope (m_AB) = $-\frac{1}{2}$

Given,

PQ is perpendicular to AB.

∴ Product of their slopes = -1

⇒ m_PQ × m_AB = -1

⇒ m_PQ × $\Big(-\frac{1}{2}\Big)$ = -1

⇒ m_PQ = -1 × -2

⇒ m_PQ = 2.

By point-slope formula,

Equation of PQ : y - y₁ = m(x − x₁)

⇒ y - (-2) = 2(x - 3)

⇒ y + 2 = 2x - 6

⇒ y = 2x - 8.

Hence, the equation of PQ is y = 2x - 8.

(ii) The point Q is the intersection of line AB and line PQ.

Equation of AB

⇒ x + 2y + 6 = 0 .....(1)

Equation of PQ

⇒ y = 2x − 8 ....(2)

Substituting the value of y from (2) in (1), we get :

⇒ x + 2(2x - 8) + 6 = 0

⇒ x + 4x - 16 + 6 = 0

⇒ 5x - 10 = 0

⇒ 5x = 10

⇒ x = $\frac{10}{5}$

⇒ x = 2.

Substituting the value of x in equation (2), we get :

⇒ y = 2x − 8

⇒ y = 2(2) − 8

⇒ y = 4 − 8

⇒ y = -4.

Q = (x, y) = (2, -4).

Hence, the coordinates of the point Q are (2, -4).

Given,

A(1, 3) and C(6, 8)

$m_{AC}= \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 3}{6 - 1} = \frac{5}{5} = 1.$

We know that diagonal AC is a perpendicular bisector of diagonal BD.

⇒ m_AC × m_BD = -1

⇒ 1 × m_BD = -1

⇒ m_BD = -1

Let O be the point of intersection of diagonals, which is the mid-point of both the diagonals.

$$\begin{aligned}O = \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{1 + 6}{2}, \frac{3 + 8}{2}\Big) \\[1em] = \Big(\frac{7}{2}, \frac{11}{2}\Big).\end{aligned}$$

By point-slope formula equation of AC is

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - \Big(\frac{11}{2}\Big) = -1 \Big[x - \Big(\frac{7}{2}\Big)\Big] \\[1em] \Rightarrow y - \frac{11}{2} = -x + \frac{7}{2} \\[1em] \Rightarrow y + x - \frac{11}{2} - \frac{7}{2} = 0 \\[1em] \Rightarrow y + x - \frac{18}{2} = 0 \\[1em] \Rightarrow x + y - 9 = 0.\end{aligned}$$

Hence, the equation of the required line is x + y - 9 = 0.

(i) By formula,

Centroid of triangle = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Substituting values we get,

Centroid = $\Big(\frac{1 + 3 + 7}{3}, \frac{4 + 2 + 5}{3}\Big) = \Big(\frac{11}{3}, \frac{11}{3}\Big)$

Hence, centroid of triangle = $\Big(\frac{11}{3}, \frac{11}{3}\Big)$.

(ii) Calculating,

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

Slope of line parallel to AB will also be equal to -1, as slope of parallel lies are equal.

By point-slope form,

Equation of a line, through the centroid and parallel to AB,

⇒ y - y₁ = m(x - x₁)

⇒ y - $\Big(\frac{11}{3}\Big) = -1 \Big[x - \Big(\frac{11}{3}\Big)\Big]$

⇒ $\Big(\frac{3y - 11}{3}\Big) = -1 \Big(\frac{3x - 11}{3}\Big)$

⇒ 3y − 11 = −1(3x − 11)

⇒ 3y − 11 = −3x + 11

⇒ 3y + 3x = 11 + 11

⇒ 3x + 3y = 22.

Hence, the equation of a line, through the centroid and parallel to AB is 3x + 3y = 22.

(i) Slope of AD = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{4 - (-4)} = \frac{-1}{8}$

A median joins a vertex to the midpoint of the opposite side. D is the midpoint of BC.

$$\begin{aligned}= \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{6 + 2}{2}, \frac{4 + (-2)}{2}\Big) \\[1em] = \Big(\frac{8}{2}, \frac{2}{2}\Big) = (4, 1).\end{aligned}$$

By point-slope form,

Equation of a median AB, given by:

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\frac{-1}{8}$ [x - (-4)]

⇒ 8(y - 2) = -1(x + 4)

⇒ 8y - 16 = -x - 4

⇒ x + 8y - 12 = 0

Hence, the equation of a line AD x + 8y - 12 = 0.

(ii) Slope of AC = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 2}{2 - (-4)} = \frac{-4}{6} = -\frac{2}{3}$

We know that altitude BM is a perpendicular AC.

Let the slope of BM be m₁,

⇒ m_AC × m₂ = -1

⇒ $-\frac{2}{3}$ × m₁ = -1

⇒ m₁ = $\frac{3}{2}$

Equation of a line BM,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = $\frac{3}{2}$ (x - 6)

⇒ 2(y - 4) = 3(x - 6)

⇒ 2y - 8 = 3x - 18

⇒ 3x - 2y - 10 = 0

Hence, the equation of a line BM 3x - 2y - 10 = 0.

(iii) Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 2}{6 - (-4)} = \frac{2}{10} = \frac{1}{5}$

Right bisector of AB is perpendicular to AB

Let the slope of Right bisector of AB be m₂,

⇒ m_AB × m₂ = -1

⇒ $\frac{1}{5}$ × m₂ = -1

⇒ m₂ = -5

Coordinates of Midpoint of AB

$$\begin{aligned}= \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{-4 + 6}{2}, \frac{2 + 4}{2}\Big) \\[1em] = \Big(\frac{2}{2}, \frac{6}{2}\Big) = (1, 3).\end{aligned}$$

Equation of a line BM,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -5 (x - 1)

⇒ (y - 3) = -5x + 5

⇒ 5x + y - 8 = 0

Hence, the equation of right bisector of AB 5x + y - 8 = 0.

(iv) Centroid of triangle ABC = $\Big(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\Big)$

Substitute values we get,

$= \Big(\frac{-4 + 6 + 2}{3}, \frac{2 + 4 + (-2)}{3}\Big) = \Big(\frac{4}{3}, \frac{4}{3}\Big)$

Hence, coordinates of centroid are $\Big(\frac{4}{3}, \frac{4}{3}\Big)$.

Given line y = 3x + 4...(1)

Comparing above equation with y = mx + c we get m₁ = 3

Since line is perpendicular to y = 3x + 4, the product of their gradients must be -1.

Let the slope of required line be m ₂

⇒ m₁ × m₂ = -1

⇒ 3 × m₂ = -1

⇒ m₂ = $-\frac{1}{3}$

By point slope formula,

Equation of a line,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $-\frac{1}{3}$ (x - 2)

⇒ 3(y - 3) = -1(x - 2)

⇒ 3y - 9 = -x + 2

⇒ x + 3y - 11 = 0...(2)

The equation of the perpendicular drawn from P(2, 3) is x + 3y - 11 = 0.

Substitute y into x + 3y - 11 = 0:

⇒ x + 3(3x + 4) - 11 = 0

⇒ x + 9x + 12 - 11 = 0

⇒ 10x + 1 = 0

⇒ 10x = -1

⇒ x = $-\frac{1}{10}$

Substitute value of x in y = 3x + 4:

⇒ y = 3 $\Big(-\frac{1}{10}\Big)$ + 4

⇒ y = $\Big(-\frac{3}{10} + \frac{40}{10}\Big)$

⇒ y = $\frac{37}{10}$

Hence, equation of required line is x + 3y - 11 = 0 and coordinates of foot of perpendicular $\Big(-\frac{1}{10}, \frac{37}{10}\Big)$.

(i) Slope of AC = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{4 - 1} = \frac{1}{3}$

The altitude through B is perpendicular to the side AC.

Let the slope of altitude be m₁,

⇒ m_AC × m₂ = -1

⇒ $\frac{1}{3}$ × m₁ = -1

⇒ m₁ = -3

By point slope formula,

Equation of altitude B,

⇒ y - 3 = -3(x - 2)

⇒ y - 3 = -3x + 6

⇒ 3x + y - 9 = 0 ...(1)

Hence, the equation of the altitude through B is 3x + y - 9 = 0.

(ii) Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 2}{2 - 1} = \frac{1}{1} = 1$

The altitude through C is perpendicular to the side AB.

Let the slope of altitude be m₂,

⇒ m_AB × m₂ = -1

⇒ 1 × m₂ = -1

⇒ m₂ = -1

By point slope formula,

Equation of altitude B,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -1(x - 4)

⇒ y - 3 = -x + 4

⇒ x + y - 7 = 0 ...(2)

Hence, the equation of the altitude through C is x + y - 7 = 0.

(iii) Subtract Equation (2) from Equation (1):

⇒ (3x + y - 9) - (x + y - 7) = 0 - 0

⇒ 3x + y - 9 - x - y + 7 = 0 - 0

⇒ 3x - x + y - y - 9 + 7 = 0

⇒ 2x - 2 = 0

⇒ 2x = 2

⇒ x = 1

Substitute x = 1 into Equation (2):

⇒ 1 + y - 7 = 0

⇒ y - 6 = 0

⇒ y = 6.

Hence, the co-ordinates of the orthocentre of ΔABC are (1, 6).

(i) Slope of BC = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - (-4)}{5 - 3} = \frac{-2}{2} = -1$

The right bisector is perpendicular to BC.

Let the slope of right bisector be m₁,

⇒ m_AB × m₁ = -1

⇒ -1 × m₁ = -1

⇒ m₁ = 1

The right bisector (perpendicular bisector) passes through the midpoint of BC

$$\begin{aligned}x, y = \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{3 + 5}{2}, \frac{-4 + (-6)}{2}\Big) \\[1em] = \Big(\frac{8}{2}, \frac{-10}{2}\Big) = (4, -5).\end{aligned}$$

By point slope formula,

Equation of right bisector of BC,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = 1(x - 4)

⇒ y + 5 = x - 4

⇒ x - y - 9 = 0....(1)

Hence, equation of right bisector of BC is x - y - 9 = 0.

(ii) Slope of CA = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-6)}{1 - 5} = \frac{8}{-4} = -2$

The right bisector is perpendicular to CA.

Let the slope of right bisector of CA be m₂,

⇒ m_CA × m₂ = -1

⇒ -2 × m₂ = -1

⇒ m₂ = $\frac{1}{2}$

Midpoint of CA

$$\begin{aligned}x, y = \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{5 + 1}{2}, \frac{-6 + 2 }{2}\Big) \\[1em] = \Big(\frac{6}{2}, \frac{-4}{2}\Big) = (3, -2).\end{aligned}$$

By point slope formula,

Equation of right bisector of CA,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $\frac{1}{2}$ (x - 3)

⇒ 2(y + 2) = (x - 3)

⇒ 2y + 4 = (x - 3)

⇒ x - 2y - 7 = 0 ...(2)

Hence, equation of right bisector of AC is x - 2y - 7 = 0.

(iii) Subtract Equation (2) from Equation (1):

⇒ (x - y) - (x - 2y) = 9 - 7

⇒ x - x - y + 2y = 2

⇒ y = 2.

Substitute y = 2 into Equation (1):

⇒ x - 2 = 9

⇒ x = 11.

Hence, coordinates of the circumcenter are (11, 2).

(i) Let the point P(x, 0) divide the line segment joining A(-6, 2) and B(9, -4) in the ratio m : n.

By section formula,

$y = \frac{my_2 + ny_1}{m + n}$

Substituting the values we get :

$$\begin{aligned}\Rightarrow 0 = \frac{m(-4) + n(2)}{m+n} \\[1em] \Rightarrow 0 = -4m + 2n \\[1em] \Rightarrow 4m = 2n \\[1em] \Rightarrow \frac{m}{n} = \frac{2}{4} \\[1em] \Rightarrow \frac{m}{n} = \frac{1}{2}.\end{aligned}$$

Hence, the ratio in which P divides the line segment AB is 1 : 2.

(ii) From part (a),

A (-6, 2) and B (9, - 4)

m : n = 1 : 2

By section-formula,

$x = \frac{mx_2 + nx_1}{m + n}$

$$\begin{aligned}\Rightarrow x = \frac{1(9) + 2(-6)}{1+2} \\[1em] \Rightarrow x = \frac{9 - 12}{3} \\[1em] \Rightarrow x = \frac{-3}{3} \\[1em] \Rightarrow x = -1.\end{aligned}$$

P = (x, 0) = (-1, 0).

Hence, the coordinates of the point P are (-1, 0).

(iii) By formula,

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

$$\begin{aligned}\Rightarrow m_{AB} = \frac{-4 - 2}{9 - (-6)} \\[1em] \Rightarrow m_{AB} = \frac{-6}{15} \\[1em] \Rightarrow m_{AB} = \frac{-2}{5}\end{aligned}$$

Line parallel to AB will have the same slope, so the slope of the new line is also m = $\frac{-2}{5}$.

By point-slope formula,

y - y₁ = m(x - x₁)

Line passing through (-3, -2) and parallel to AB is :

$$\begin{aligned}\Rightarrow y - (-2) = \frac{-2}{5}(x - (-3)) \\[1em] \Rightarrow y + 2 = \frac{-2}{5}(x + 3) \\[1em] \Rightarrow 5(y + 2) = -2(x + 3) \\[1em] \Rightarrow 5y + 10 = -2x - 6 \\[1em] \Rightarrow 5y + 10 + 2x + 6 = 0 \\[1em] \Rightarrow 2x + 5y + 16 = 0 \\[1em]\end{aligned}$$

Hence, the equation of the line is 2x + 5y + 16 = 0.

(i) As A lies on x-axis let its co-ordinates be (a, 0) and B lies on y-axis so, co-ordinates = (0, b).

By section-formula,

$$\begin{aligned}\Rightarrow P = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \Big) \\[1em] \Rightarrow (4, -1) = \Big(\frac{1(0) + 2a}{1 + 2}, \frac{1(b) + 2(0)}{1 + 2} \Big) \\[1em] \Rightarrow (4, -1) = \Big(\frac{2a}{3}, \frac{b}{3} \Big) \\[1em] \Rightarrow \frac{2a}{3} = 4 \text{ and } \frac{b}{3} = -1 \\[1em] \Rightarrow a = \frac{12}{2} = 6 \text{ and } b = -3.\end{aligned}$$

Hence, A = (6, 0) and B = (0, -3).

(ii) By formula,

Slope of AB = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 0}{0 - 6} = \frac{-3}{-6} = \frac{1}{2}.$

Let slope of perpendicular line be m.

Since the product of the slopes of perpendicular lines = -1.

⇒ m × Slope of AB = -1

⇒ m × $\frac{1}{2}$ = -1

⇒ m = -2.

By point-slope from,

Equation of line passing through P and slope = -2 is :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = -2(x - 4)

⇒ y + 1 = -2x + 8

⇒ 2x + y = 7.

Hence, equation of required line is 2x + y = 7.

(i) Let the slope of BC be m₁. Slope of BC is given by,

$$\begin{aligned}m_1 = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-6 - 2}{6 - (-1)} \\[1em] = \frac{-8}{7}.\end{aligned}$$

Hence, the slope of BC is $-\frac{8}{7}$.

(ii) Let slope of line perpendicular to BC be m₂

So,m₁ × m₂ = -1

⇒ $-\frac{8}{7}$ × m₂ = -1

⇒ m₂ = $\frac{7}{8}$

Equation of the line having the slope = $\frac{7}{8}$ and passing through A(3, 8) can be given bu point slope formula i.e.,

⇒ y - y₁ = m(x - x₁)

⇒ y - 8 = $\frac{7}{8}$ (x - 3)

⇒ 8(y − 8) = 7(x − 3)

⇒ 8y − 64 = 7x − 21

⇒ 7x − 8y − 21 + 64 = 0

⇒ 7x − 8y + 43 = 0.

Hence, the equation of the required line is 7x - 8y + 43 = 0.

(i) By formula,

Slope of a line = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of CD = $\frac{3 - (-1)}{4 - 0} = \frac{4}{4} = 1$

Hence, slope of CD = 1.

(ii) We know that,

The product of slope of two perpendicular lines equals to -1.

∴ Slope of AB × Slope of CD = -1

⇒ Slope of AB × 1 = -1

⇒ Slope of AB = -1.

By point-slope formula,

Equation of line

⇒ y - y₁ = m(x - x₁)

Equation of AB :

⇒ y - 0 = -1(x - 4)

⇒ y = -x + 4

⇒ x + y = 4.

Hence, equation of AB is x + y = 4.

(i) From figure,

A = (4, 0) and B = (0, 4).

(ii) Let coordinates of P be (x, y).

By section formula,

(x, y) = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{3(0) + 1(4)}{3 + 1}, \frac{3(4) + 1(0)}{3 + 1}\Big) \\[1em] = \Big(\frac{0 + 4}{4}, \frac{12 + 0}{4}\Big) \\[1em] = \Big(\frac{4}{4}, \frac{12}{4}\Big) \\[1em] = (1,3).\end{aligned}$$

Hence, coordinates of P = (1, 3).

(iii) By formula,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substituting values we get :

Slope of AB = $\frac{4 - 0}{0 - 4} = \frac{4}{-4} = -1$

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of line perpendicular to AB = -1

⇒ -1 × Slope of line perpendicular to AB = -1

⇒ Slope of line perpendicular to AB = $\frac{-1}{-1} = 1$

Line passing through P and perpendicular to AB :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 1)

⇒ y - 3 = x - 1

⇒ y = x - 1 + 3

⇒ x - y + 2 = 0

Hence, required equation is x - y + 2 = 0.

(i) By formula,

Slope of line = $\Big(\frac{y_2 - y_1}{x_2 - x_1}\Big)$

Substituting values we get :

$$\begin{aligned}\text{Slope of line PQ} = \Big(\frac{-1 - (-3)}{0 - 2}\Big) \\[1em] = \frac{-1 + 3}{-2} \\[1em] = \frac{2}{-2} \\[1em] = -1.\end{aligned}$$

Equation of line :

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = -1(x - 2)

⇒ y + 3 = -x + 2

⇒ x + y + 3 - 2 = 0

⇒ x + y + 1 = 0.

Hence, equation of line PQ is x + y + 1 = 0.

(ii) We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of PQ × Slope of AB = -1

⇒ -1 × Slope of AB = -1

⇒ Slope of AB = $\frac{-1}{-1}$ = 1.

Equation of line :

⇒ y - y₁ = m(x - x₁)

Equation of line AB :

⇒ y - 6 = 1(x - 0)

⇒ y - 6 = x

⇒ x - y + 6 = 0

Hence, equation of line AB is x - y + 6 = 0.

(iii) Given,

Line PQ cuts x-axis at point R.

Let R = (a, 0)

Equation of line PQ = x + y + 1 = 0

Since, point R lies on line PQ,

⇒ a + 0 + 1 = 0

⇒ a + 1 = 0

⇒ a = -1.

R = (a, 0) = (-1, 0)

Given,

AB is perpendicular to PQ at point S.

∴ Point S is the intersection point of AB and PQ.

PQ : x + y + 1 = 0

AB : y - x = 6 or y = x + 6

Substituting value of y from equation AB in equation PQ, we get :

⇒ x + (x + 6) + 1 = 0

⇒ 2x + 7 = 0

⇒ 2x = -7

⇒ x = $-\frac{7}{2}$

Substituting value of x in equation AB, we get :

y = $-\frac{7}{2} + 6 = \frac{-7 + 12}{2} = \frac{5}{2}$.

S = $\Big(-\frac{7}{2}, \frac{5}{2}\Big)$.

Hence, coordinates of R = (-1, 0) and S = $\Big(-\frac{7}{2}, \frac{5}{2}\Big)$.

(i) From graph,

A(3, 3), B(0, −2), C(−4, −2), D(−1, 3).

Hence, A(3, 3), B(0, −2), C(−4, −2), D(−1, 3).

(ii) Given,

The diagonals of a parallelogram bisect each other.

Hence, the intersection point P is the midpoint of both AC and BD.

By mid-point formula,

$\Rightarrow M = \Big(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\Big)$

Mid-point of AC :

$$\begin{aligned}\Rightarrow P = \Big(\frac{3 + (-4)}{2}\ , \frac{3 + (-2)}{2}\Big) \\[1em] \Rightarrow P = \Big(\frac{-1}{2}\ , \frac{1}{2}\Big) \\[1em] \Rightarrow P = (-0.5, 0.5).\end{aligned}$$

Hence, P = (-0.5, 0.5).

(iii) By formula,

$\text{Slope (m)} = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting values for line CB,

$$\begin{aligned}\Rightarrow m_{CB} = \frac{-2-(-2)}{0-(-4)} \\[1em] = \frac{0}{4} \\[1em] = 0.\end{aligned}$$

Substituting values for line DA,

$$\begin{aligned}\Rightarrow m_{DA} = \frac{3-3}{3-(-1)} \\[1em] = \frac{0}{4} \\[1em] = 0 .\end{aligned}$$

Since, slope of parallel lines are equal, thus CB || DA.

Hence, proved that CB || DA.

(iv) By formula,

Slope of line = $\frac{y_2-y_1}{x_2-x_1}$

Slope of AC = $\frac{-2 - 3}{-4 - 3} = \frac{-5}{-7} = \frac{5}{7}$.

Using point-slope formula,

y - y₁ = m(x - x₁)

Equation of AC :

$$\begin{aligned}\Rightarrow y - 3 = \frac{5}{7}(x - 3) \\[1em] \Rightarrow 7(y - 3) = 5(x - 3) \\[1em] \Rightarrow 7y - 21 = 5x - 15 \\[1em] \Rightarrow 5x - 7y - 15 + 21 = 0 \\[1em] \Rightarrow 5x - 7y + 6 = 0.\end{aligned}$$

Hence, equation of the diagonal AC is 5x - 7y + 6 = 0.

The angle that a horizontal line makes with the x-axis is 0°.

The angle between the line and the x-axis is 0°.

Hence, option 1 is the correct option.

A vertical line makes a right angle with the horizontal x-axis.

Therefore, the angle of inclination of the y-axis is 90°.

Hence, option 3 is the correct option.

The slope (m) of a line is related to its inclination (θ) by the formula:

m = tan θ

m = tan 30° = $\frac{1}{\sqrt3}$.

Hence, option 4 is the correct option.

A horizontal line is a line parallel to the x-axis.

Thus θ = 0°

We know that,

m = tan θ

m = tan 0° = 0

Hence, option 1 is the correct option.

A vertical line is perpendicular to x-axis.

Thus, θ = 90°.

We know that,

m = tan θ

m = tan 90° = not defined

Hence, option 4 is the correct option.

The correct formula for slope is: $m = \frac{y_2 - y_1}{x_2 - x_1}$

Hence, option 4 is the correct option.

A horizontal line has equation : y = c.

It does not crosses x-axis at any point, thus x-intercept = 0.

Hence, option 1 is the correct option.

A y-intercept is the point where a line crosses the y-axis.

A vertical line: A vertical line has the equation x = a.

If a = 0 , it has an infinite number of y-intercepts.

If a ≠ 0 , the line is parallel to the y-axis and never crosses it. Therefore, it has no y-intercept.

Hence, option 2 is the correct option.

The equation of x-axis is :

y = 0

Every point on the x-axis has a y-coordinate of 0, regardless of its x-coordinate.

Hence, option 2 is the correct option.

The equation of y-axis is:

x = 0

Every point on the y-axis has an x-coordinate of 0, regardless of its y-coordinate.

Hence, option 1 is the correct option.

Any line parallel to the x-axis is a horizontal line and has the equation of form y = k.

The distance is 5 units below x-axis, thus the y-coordinate (k) of every point on the line is -5.

y = -5

y + 5 = 0.

Hence, option 4 is the correct option.

Any line parallel to the y-axis is a vertical line and has the equation form x = k.

The distance is 8 units to the right, thus the x-coordinate (k) of every point on the line is +8.

x = 8

x - 8 = 0

Hence, option 1 is the correct option.

y = mx + c, is the standard form of a linear equation known as the slope-intercept form.

Hence, option 3 is the correct option.

The slope (m) of a line is defined as the tangent of its angle of inclination.

m = tan α

Hence, option 3 is the correct option.

By point-slope form,

Equation of line :

y - y₁ = m (x - x₁)

Thus, equation of line passing through (a, b) and slope m is :

y - b = m(x - a)

Hence, option 4 is the correct option.

By two-point formula,

Equation of line :

(y – y₁) = $\frac{y_2 - y_1}{x_2 - x_1}$ (x – x₁)

Hence, option 3 is the correct option.

Given,

⇒ 3x + 4y + 7 = 0

⇒ 4y = -3x - 7

⇒ y = $-\frac{3}{4}x - \frac{7}{4}$

Comparing above equation with y = mx + c, we get :

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

By point-slope formula,

Equation of line : y - y₁ = m(x - x₁)

Thus, equation of line parallel to 3x + 4y + 7 = 0 and passing through origin (0, 0) is :

⇒ y - 0 = $-\frac{3}{4}$ (x - 0)

⇒ y = $-\frac{3x}{4}$

⇒ 4y = -3x

⇒ 4y + 3x = 0.

Hence, option 4 is the correct option.

We know that,

$$\begin{aligned}m = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] m_{AB} = \frac{-6 - 4}{2-(-3)} \\[1em] = \frac{-10}{5} = -2.\end{aligned}$$

Hence, option 2 is the correct option.

The slope m is determined by the inclination θ = 30°.

m = tan θ

m = tan 30°

m = $\frac{1}{\sqrt3}$

Slope-intercept form:

y = mx + c

$$\begin{aligned}\Rightarrow y = \frac{1}{\sqrt3}x - 4 \\[1em] \Rightarrow \sqrt3y = x - 4\sqrt3 \\[1em] \Rightarrow x - \sqrt3y - 4\sqrt3 = 0.\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

$$\begin{aligned}m_{AB} = \frac{-4 - (-2)}{3 - 3} \\[1em] = \frac{-4 + 2}{0} \\[1em] = \frac{-2}{0}.\end{aligned}$$

Slope is not defined.

Hence, option 4 is the correct option.

Comparing equation, y = $\sqrt{3}$ x – 5 with y = mx + c, we get :

⇒ m = $\sqrt3$

⇒ tan θ = $\sqrt{3}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, option 4 is the correct option.

We know that,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Substitute values we get:

$$\begin{aligned}\Rightarrow 2 = \frac{3 - 5}{k - 2} \\[1em] \Rightarrow 2 = \frac{-2}{k - 2} \\[1em] \Rightarrow 2(k - 2) = -2 \\[1em] \Rightarrow (k - 2) = \frac{-2}{2} \\[1em] \Rightarrow (k - 2) = -1 \\[1em] \Rightarrow k = -1 + 2 \\[1em] \Rightarrow k = 1.\end{aligned}$$

Hence, option 3 is the correct option.

If the slope is negative, then tan θ must be negative.

tan θ < 0

This happens only when the angle θ made with the positive x-axis is between 90° and 180°.

And angles between 90° and 180° are called obtuse angles.

Hence, option 2 is the correct option.

The equation of horizontal line is of the form y = c.

Since the line must pass through the point (-6, 8), its y-coordinate must be equal to 8.

y = 8

y - 8 = 0.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3x - 2\sqrt{2}y - 4 = 0 \\[1em] \Rightarrow -2\sqrt{2}y = -3x + 4 \\[1em] \Rightarrow y = \frac{-3}{-2\sqrt{2}}x + \frac{4}{-2\sqrt{2}} \\[1em] \Rightarrow y = \frac{3}{2\sqrt{2}}x - \frac{2}{\sqrt{2}}.\end{aligned}$$

Comparing above equation with y = mx + c, we get :

m = $\frac{3}{2\sqrt2}$

Rationalizing the Denominator,

$$\begin{aligned}m = \frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\[1em] = \frac{3\sqrt{2}}{2 \times 2} \\[1em] = \frac{3\sqrt{2}}{4}.\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

$$\begin{aligned}m = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] m_{AB} = \frac{6 - 3}{-2 - 4} \\[1em] = \frac{3}{-6} \\[1em] = -\frac{1}{2}.\end{aligned}$$

By point-slope form :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $-\frac{1}{2}$ (x - 4)

⇒ 2(y - 3) = -1(x - 4)

⇒ 2y - 6 = -x + 4

⇒ x + 2y - 6 - 4 = 0

⇒ x + 2y - 10 = 0.

Hence, option 1 is the correct option.

The line intersects the y-axis 8 units below the x-axis. This means the line intersect y-axis at (0, -8).

Thus, y-intercept of lie (c) = -8

Thus, line passes through (0, -8) and (5, -3).

We know that,

$$\begin{aligned}m = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-3 - (-8)}{5 - 0} \\[1em] = \frac{5}{5} \\[1em] = 1.\end{aligned}$$

Substitute m = 1 and c = -8 into y = mx + c, we get :

⇒ y = 1.x + (-8)

⇒ y = x - 8

⇒ x - y - 8 = 0.

Hence, option 2 is the correct option.

For three points A, B, and C to be collinear.

m_AB = m_BC

We know that,

m = $\frac{y_2 - y_1}{x_2 - x_1}$

Using A(5, -2) and B(8, -3) :

$$\begin{aligned}m_{AB} = \frac{-3 - (-2)}{8 - 5} \\[1em] = \frac{-1}{3}.\end{aligned}$$

Using B(8, -3) and C(m, -12):

$$\begin{aligned}m_{BC} = \frac{-12 - (-3)}{m - 8} \\[1em] = \frac{-9}{m - 8}\end{aligned}$$

m_AB = m_BC

⇒ $\frac{-1}{3} = \frac{-9}{m - 8}$

⇒ -1(m - 8) = -9 × 3

⇒ -m + 8 = -27

⇒ -m = -27 - 8

⇒ -m = -35

⇒ m = 35.

Hence, option 3 is the correct option.

The equation of line parallel to y-axis is :

x = c

Since the line must pass through the point (9, -9), its x-coordinate must be x = 9.

x - 9 = 0

Hence, option 4 is the correct option.

Two distinct non-vertical lines are parallel if and only if their slopes are equal.

m₁ = m₂

Hence, option 4 is the correct option.

Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

m₁.m₂ = –1

Hence, option 2 is the correct option.

Given, points A(3, –7) and B(5, –7)

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$

Substitute values we get,

$$\begin{aligned}m_{AB} = \frac{-7 - (-7)}{5 - 3} \\[1em] = \frac{-7 + 7}{2} \\[1em] = \frac{0}{2} = 0.\end{aligned}$$

As, slope of parallel lines are equal.

Thus, slope of line parallel to AB = 0.

Hence, option 4 is the correct option.

Slope of the line passing through P(3, –4) and Q(1, –8).

$$\begin{aligned}m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-8 - (-4)}{1 - 3} \\[1em] = \frac{-8 + 4}{-2} \\[1em] = \frac{-4}{-2} = 2.\end{aligned}$$

Since the required line is perpendicular to line PQ, the product of their slopes must be -1.

Let slope of required line be m.

⇒ m_PQ × m = -1

⇒ 2 × m = -1

⇒ m = $\frac{-1}{2}$

Hence, option 1 is the correct option.

Solving,

⇒ 3x - 5y + 8 = 0

⇒ 3x + 8 = 5y

⇒ $y = \frac{3}{5}x + \frac{8}{5}$.

The slope of the given line is $m = \frac{3}{5}$.

Since the required line is parallel, its slope must be the same as m.

Let slope of required line be m₁,

m₁ = m = $\frac{3}{5}$.

Hence, option 4 is the correct option.

Solving,

⇒ 3x = 4y - 10

⇒ 4y = 3x + 10

$\Rightarrow y = \frac{3}{4}x + \frac{10}{4}$.

The slope of the given line is $m = \frac{3}{4}$.

Since the required line is perpendicular, the product of the slopes must be -1.

Let slope of required line be a, then :

⇒ m × a = -1

⇒ $\frac{3}{4}$ × a = -1

⇒ a = $\frac{-4}{3}$.

Hence, option 2 is the correct option.

For two lines to be parallel, their slopes must be equal.

Line 1 : 2x + 3y = 5

First, convert the equation 2x + 3y = 5 into the slope-intercept form, y = mx + c, to find its slope, m.

$$\begin{aligned}3y = -2x + 5 \\[1em] y = -\frac{2}{3}x + \frac{5}{3} \\[1em] m_1 = -\frac{2}{3}\end{aligned}$$

Line 2: kx - 6y = 7

First, convert the equation kx - 6y = 7 into the slope-intercept form, y = mx + c, to find its slope, m.

$$\begin{aligned}-6y = -kx + 7 \\[1em] y = \frac{-k}{-6}x + \frac{7}{-6} = \frac{k}{6}x - \frac{7}{6} \\[1em] m_2 = \frac{k}{6}\end{aligned}$$

The slopes of parallel lines are equal:

$$\begin{aligned}\Rightarrow m_1 = m_2 \\[1em] \Rightarrow -\frac{2}{3} = \frac{k}{6} \\[1em] \Rightarrow k = -\frac{2 \times 6}{3} \\[1em] \Rightarrow k = -\frac{12}{3} \\[1em] \Rightarrow k = -4.\end{aligned}$$

Hence, option 1 is the correct option.

For two lines to be perpendicular, the product of their slopes must be -1.

Line 1: x - my + 3 = 0

First, convert the equation x - my + 3 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$$\begin{aligned}-my = -x - 3 \\[1em] y = \frac{-1}{-m}x + \frac{-3}{-m} = \frac{1}{m}x + \frac{3}{m} \\[1em] m_1 = \frac{1}{m}.\end{aligned}$$

Line 2: 2x + 3y - 7 = 0

First, convert the equation 2x + 3y - 7 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$$\begin{aligned}3y = -2x + 7 \\[1em] y = -\frac{2}{3}x + \frac{7}{3} \\[1em] m_2 = -\frac{2}{3}\end{aligned}$$

The product of slopes of perpendicular lines is equal to -1:

$$\begin{aligned}\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow \Big(\frac{1}{m}\Big) \times \Big(-\frac{2}{3}\Big) = -1 \\[1em] \Rightarrow \frac{-2}{3m} = -1 \\[1em] \Rightarrow -2 = -3m \\[1em] \Rightarrow m = \frac{2}{3}.\end{aligned}$$

Hence, option 2 is the correct option.

The given line is $y = 3x + 1$, so its slope is m = 3.

Since the required line is parallel, its slope is also m = 3.

Using the point-slope form y - y₁ = m(x - x₁) with the point (x₁, y₁) = (1, 2) and m=3:

y - 2 = 3(x - 1)

y - 2 = 3x - 3

Rearranging the equation to the standard form (Ax + By + C = 0):

0 = 3x - y - 3 + 2

3x - y - 1 = 0

Hence, option 3 is the correct option.

Convert the general linear equation ax + by + c = 0 into the slope-intercept form, y = mx + c, to find the slope, m.

by = -ax - c

$y = \Big(\frac{-a}{b}\Big)x - \frac{c}{b}$

The slope, m, is the coefficient of x.

m = $-\frac{a}{b}$

Hence, option 1 is the correct option.

Equation:

⇒ x - y = 7

⇒ y = x - 7

Comparing above equation with y = mx + c, we get :

m = 1.

A line is equally inclined to the axes if slope = 1.

Hence, Option 3 is the correct option.

Slope of line l₁ = tan 45° = 1.

We know that,

Slope of parallel lines are equal.

Slope of line l₂ = Slope of line l₁ = 1.

We know that,

Product of slope of perpendicular lines is equal to -1.

⇒ Slope of line l₂ × Slope of line l₃ = -1

⇒ 1 × Slope of line l₃ = -1

⇒ Slope of line l₃ = -1.

Hence, Option 3 is the correct option.

Substituting x = 0 in first equation,

⇒ 3(0) + 3y = 6

⇒ 3y = 6

⇒ y = $\frac{6}{3}$

⇒ y = 2.

The line touches y-axis at point (0, 2).

Substituting y = 0 in first equation,

⇒ 3x + 3(0) = 6

⇒ 3x = 6

⇒ x = $\frac{6}{3}$

⇒ x = 2.

The line touches y-axis at point (2, 0).

∴ Line 3x + 3y = 6 cuts positive x-axis and positive y-axis at equal distance i.e. 2 units form the origin.

Hence, Option 1 is the correct option.

From graph,

A = (4, 0) and B = (0, 2).

Given,

P is the mid-point of AB.

P = $\Big(\frac{4 + 0}{2}, \frac{0 + 2}{2}\Big) = \Big(\frac{4}{2}, \frac{2}{2}\Big)$ = (2, 1).

By two-point form,

⇒ y - y₁ = $\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

⇒ y - 1 = $\frac{1 - 0}{2 - 0}(x - 2)$

⇒ y - 1 = $\frac{1}{2}(x - 2)$

⇒ 2(y - 1) = x - 2

⇒ 2y - 2 = x - 2

⇒ 2y = x - 2 + 2

⇒ 2y = x.

Hence, Option 2 is the correct option.

Slope = $\frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-2)}{-7 - 3} = \frac{0}{-10} = 0$.

So, Assertion (A) is true.

The correct slope formula is $\frac{y_2 - y_1}{x_2 - x_1}$, whereas the Reason gives $\frac{x_2 - x_1}{y_2 - y_1}$, which is incorrect.

So, Reason (R) is false.

Hence, Option 3 is the correct option.

Comparing $y = \frac{1}{\sqrt{3}}x - 5$ with y = mx + c, we get $m = \frac{1}{\sqrt{3}}$.

$$\begin{aligned}\Rightarrow \tan \theta = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \tan \theta = \tan 30° \\[1em] \Rightarrow \theta = 30°.\end{aligned}$$

The inclination is 30°, not 60°, so Assertion (A) is false.

The gradient m of a line is given by m = tan θ, where θ is the angle of inclination, so Reason (R) is true.

Hence, Option 4 is the correct option.

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

$\Rightarrow m_1 = \frac{6 - 5}{-3 - 2} = -\frac{1}{5}.$

The product of the slopes of two perpendicular lines is -1. Let the slope of the perpendicular line be m₂.

$$\begin{aligned}\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\frac{1}{5} \times m_2 = -1 \\[1em] \Rightarrow m_2 = 5.\end{aligned}$$

So, Assertion (A) is true and Reason (R) is true, and the Reason is the principle used to obtain the slope in the Assertion.

Hence, Option 1 is the correct option.

Slope m = tan 45° = 1, and the y-intercept is c = -4.

By slope-intercept form,

$$\begin{aligned}\Rightarrow y = mx + c \\[1em] \Rightarrow y = (1)x + (-4) \\[1em] \Rightarrow y = x - 4 \\[1em] \Rightarrow x - y = 4.\end{aligned}$$

So, Assertion (A) is true.

The correct slope-intercept form is y = mx + c, whereas the Reason gives y = cx + m, which is incorrect.

So, Reason (R) is false.

Hence, Option 3 is the correct option.

The line 2x - 3y = 7 can be written as $y = \frac{2}{3}x - \frac{7}{3}$, so its slope is $\frac{2}{3}$.

The required line is parallel to it, so its slope is $\frac{2}{3}$, and it passes through (0, 4), so c = 4.

By slope-intercept form,

$$\begin{aligned}\Rightarrow y = \frac{2}{3}x + 4 \\[1em] \Rightarrow 3y = 2x + 12 \\[1em] \Rightarrow 2x - 3y + 12 = 0.\end{aligned}$$

The correct equation is 2x - 3y + 12 = 0, not 2x - 7y + 28 = 0, so Assertion (A) is false.

The slope of the line y = mx + c is indeed m, so Reason (R) is true.

Hence, Option 4 is the correct option.

(a) Given,

BD is equally inclined to both axes.

∴ Slope of BD = tan 45° = 1.

Hence, slope of BD = 1.

(b) Since, diagonals of a square are perpendicular and product of slope of perpendicular lines equals to -1.

∴ Slope of BD × Slope of AC = -1

⇒ 1 × Slope of AC = -1

⇒ Slope of AC = -1.

Since, diagonals meet at point (-2, -2).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = -1[x - (-2)]

⇒ y + 2 = -1[x + 2]

⇒ y + 2 = -x - 2

⇒ x + y + 2 + 2 = 0

⇒ x + y + 4 = 0.

Hence, equation of AC is x + y + 4 = 0.

Given,

Equation : x + 2y = 4

⇒ 2y = -x + 4

⇒ y = $-\frac{1}{2}x + 4$

Comparing above equation with y = mx + c we get :

⇒ m = $-\frac{1}{2}$

Let slope of line perpendicular to line x + 2y = 4 be m₁.

We know that,

Product of slope of perpendicular lines = -1.

$$\begin{aligned}\Rightarrow m_1 \times -\frac{1}{2} = -1 \\[1em] \Rightarrow m_1 = \frac{-1 \times 2}{-1} = 2.\end{aligned}$$

Substituting values we get :

⇒ y = mx + c

⇒ y = 2x + 2.

Simultaneously solving equation :

⇒ x + 2y = 4 .........(1)

⇒ y = 2x + 2 .......(2)

Substituting value of y from equation (2) in (1), we get :

⇒ x + 2(2x + 2) = 4

⇒ x + 4x + 4 = 4

⇒ 5x = 4 - 4

⇒ 5x = 0

⇒ x = $\frac{0}{5}$ = 0.

Substituting value of x in equation (2), we get :

⇒ y = 2(0) + 2 = 2.

Hence, equation of required line is y = 2x + 2 and point of intersection = (0, 2).

(a) (a) L₁ : x - y = 1

⇒ y = x - 1

Comparing above equation with y = mx + c, we get :

⇒ m = 1.

We know that,

Slope of parallel lines are equal.

∴ Slope of L₃ = 1.

Given,

L₃ has y-intercept = 3.

∴ y = mx + c

⇒ y = 1.x + 3

⇒ y = x + 3.

Hence, equation of line L₃ : y = x + 3.

(b) L₂ : x + y = 5 and L₃ : y = x + 3

⇒ x + y = 5 .........(1)

⇒ y = x + 3 .........(2)

Substituting value of y from equation (2) in (1), we get :

⇒ x + (x + 3) = 5

⇒ 2x + 3 = 5

⇒ 2x = 5 - 3

⇒ 2x = 2

⇒ x = $\frac{2}{2}$

⇒ x = 1.

Substituting value of x in equation (2), we get :

⇒ y = 1 + 3 = 4.

⇒ (x, y) = (1, 4)

∴ P(k, 4) = (1, 4)

Hence, value of k = 1.

(c) Given,

L₂ meets x-axis at point R.

At point on x-axis, y-coordinate = 0.

L₂ : x + y = 5

⇒ x + 0 = 5

⇒ x = 5

⇒ R = (x, y) = (5, 0).

P = (1, 4), Q = (3, 2) and R = (5, 0)

Let PQ : QR = k : 1

By section formula,

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\frac{k \times 5 + 1 \times 1}{k + 1}, \frac{k \times 0 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\frac{5k + 1}{k + 1}, \frac{4}{k + 1}\Big) \\[1em] \Rightarrow \frac{5k + 1}{k + 1} = 3 \text{ and } 2 = \frac{4}{k + 1} \\[1em] \Rightarrow 5k + 1 = 3(k + 1) \text{ and } 4 = 2(k + 1) \\[1em] \Rightarrow 5k + 1 = 3k + 3 \text{ and } 4 = 2k + 2 \\[1em] \Rightarrow 5k - 3k = 3 - 1 \text{ and } 2k = 4 - 2 \\[1em] \Rightarrow 2k = 2 \text{ and } 2k = 2 \\[1em] \Rightarrow k = \frac{2}{2} = 1.\end{aligned}$$

∴ PQ : QR = 1 : 1.

Hence, R = (5, 0) and PQ : QR = 1 : 1.