Equation of a Line

Given,

Point (m, -4) lies on the line x + y = 4.

Substituting value of point in equation, we get :

⇒ m + (-4) = 4

⇒ m - 4 = 4

⇒ m = 8.

Hence, Option 4 is the correct option.

Given,

Line kx - y = 9 passes through the point (6, 3).

Substituting value of point in equation, we get :

⇒ 6k - 3 = 9

⇒ 6k = 9 + 3

⇒ 6k = 12

⇒ k = $\frac{12}{6}$ = 2.

Hence, Option 1 is the correct option.

Substituting value of point in L.H.S. of the equation 3x - $\frac{y}{2}$ = 10, we get :

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

= 6 - (-2)

= 6 + 2

= 8.

Since,

8 ≠ 10.

∴ The statement, line 3x - $\frac{y}{2}$ = 10 passes through the point (2, -4) is false.

Hence, Option 1 is the correct option.

Given,

1st Equation :

⇒ x + y = 8

⇒ x = 8 - y ......(1)

2nd equation :

⇒ x - y = 0

⇒ x = y .........(2)

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

⇒ y = 8 - y

⇒ y + y = 8

⇒ 2y = 8

⇒ y = $\frac{8}{2}$ = 4.

From equation (2), we get :

⇒ x = y = 4.

Point of intersection of lines x + y = 8 and x - y = 0 is (4, 4).

Given, point of intersection lies on the line mx - 2y = 0.

Substituting values we get :

⇒ 4m - 2 × 4 = 0

⇒ 4m - 8 = 0

⇒ 4m = 8

⇒ m = $\frac{8}{4}$ = 2.

Hence, Option 3 is the correct option.

Given,

Points : (0, k) and (4, 0)

By formula,

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

Substituting values we get :

$$\begin{aligned}\text{Mid-point } = \Big(\frac{0 + 4}{2}, \frac{k + 0}{2}\Big) \\[1em] = \Big(\frac{4}{2}, \frac{k}{2}\Big) \\[1em] = \Big(2, \frac{k}{2}\Big).\end{aligned}$$

Given,

Line x + y = 4 bisects the line segment joining the points (0, k) and (4, 0).

∴ Line x + y = 4 passes through the point $\Big(2, \frac{k}{2}\Big)$.

Substituting value we get :

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

Hence, Option 2 is the correct option.

(i) Substituting x = 1 and y = 3 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 1 - 2 × 3 + 5

= 1 - 6 + 5

= -5 + 5

= 0.

Since, L.H.S. = R.H.S.

∴ Point (1, 3) satisfies the equation.

Hence, (1, 3) lies on the line represented by the equation x - 2y + 5 = 0.

(ii) Substituting x = 0 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 0 - 2 × 5 + 5

= 0 - 10 + 5

= -5

Since, L.H.S. ≠ R.H.S.

∴ Point (0, 5) does not satisfies the equation.

Hence, (0, 5) does not lies on the line represented by the equation x - 2y + 5 = 0.

(iii) Substituting x = -5 and y = 0 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = -5 - 2 × 0 + 5

= -5 - 0 + 5

= -5 + 5

= 0.

Since, L.H.S. = R.H.S.

∴ Point (-5, 0) satisfies the equation.

Hence, (-5, 0) lies on the line represented by the equation x - 2y + 5 = 0.

(iv) Substituting x = 5 and y = 5 in the L.H.S. of the equation x - 2y + 5 = 0, we get :

L.H.S. = 5 - 2 × 5 + 5

= 5 - 10 + 5

= 10 - 10

= 0.

Since, L.H.S. = R.H.S.

∴ Point (5, 5) satisfies the equation.

Hence, (5, 5) lies on the line represented by the equation x - 2y + 5 = 0.

Substituting x = 2 and y = 3 in the L.H.S. of the equation $\frac{x}{2} + \frac{y}{3} = 0$, we get :

L.H.S. = $\frac{2}{2} + \frac{3}{3}$

= 1 + 1

= 2

Since, L.H.S. ≠ R.H.S.

∴ The line $\frac{x}{2} + \frac{y}{3} = 0$ does not passes through the point (2, 3).

Hence, the statement is false.

Given,

(2, a) lies on the line 2x - y = 3.

∴ Substituting (2, a) in 2x - y = 3, satisfies the equation.

⇒ 2(2) - a = 3

⇒ 4 - a = 3

⇒ a = 4 - 3 = 1.

Hence, the statement is false.

In order for the point (3, -k) to lie on the line 9x + 4y = 3, it must satisfy the equation.

On putting x = 3 and y = -k, we have

⇒ 9(3) + 4(-k) = 3

⇒ 27 – 4k = 3

⇒ 4k = 27 – 3 = 24

⇒ k = $\frac{24}{4}$

⇒ k = 6.

Hence, k = 6.

Since, (m, 2m - 1) lies on the line $\frac{3x}{5} - \frac{2y}{3} + 1 = 0$, it will satisfy the equation.

Substituting x = m, y = 2m - 1 in the equation $\frac{3x}{5} - \frac{2y}{3} + 1 = 0$ we get,

$$\begin{aligned}\Rightarrow \frac{3m}{5} - \frac{2(2m - 1)}{3} + 1 = 0 \\[1em] \Rightarrow \frac{(3m \times 3) - [5 \times 2(2m - 1)] + 15}{15} = 0 \\[1em] \Rightarrow 9m - 10(2m - 1) + 15 = 0 \\[1em] \Rightarrow 9m - 20m + 10 + 15 = 0 \\[1em] \Rightarrow -11m = -25 \\[1em] \Rightarrow 11m = 25 \\[1em] \Rightarrow m = \frac{25}{11} = 2\frac{3}{11}.\end{aligned}$$

Hence, m = $2\frac{3}{11}.$

By mid-point formula,

Mid-point of (5, -2) and (-1, 2) = $\frac{5 + (-1)}{2}, \frac{-2 + 2}{2}$

= $\frac{4}{2}, \frac{0}{2}$

= (2, 0).

Substituting x = 2 and y = 0 in L.H.S. of the equation 3x - 5y = 6.

L.H.S. = 3 × 2 - 5 × 0

= 6.

Since, L.H.S. = R.H.S.,

∴ (2, 0) lies on the line 3x - 5y = 6.

Hence, the line 3x – 5y = 6 bisects the join of (5, -2) and (-1, 2).

By mid-point formula,

Mid-point of (a, 3) and (2, -5) = $\frac{a + 2}{2}, \frac{3 + (-5)}{2}$

= $\frac{a + 2}{2}, \frac{-2}{2}$

= $\Big(\frac{a + 2}{2}, -1\Big)$.

Given, line y = 3x - 2 bisects the join of (a, 3) and (2, -5).

∴ $\Big(\frac{a + 2}{2}, -1\Big)$ satisfies the equation y = 3x - 2.

$$\begin{aligned}\therefore -1 = 3 \times \frac{a + 2}{2} - 2 \\[1em] \Rightarrow -1 = \frac{3a + 6}{2} - 2 \\[1em] \Rightarrow -1 = \frac{3a + 6 - 4}{2} \\[1em] \Rightarrow -2 = 3a + 2 \\[1em] \Rightarrow 3a = -4 \\[1em] \Rightarrow a = -\frac{4}{3}.\end{aligned}$$

Hence, a = $-\frac{4}{3}$.

By mid-point formula,

Mid-point of (8, -1) and (0, k) = $\frac{8 + 0}{2}, \frac{-1 + k}{2}$

= $\frac{8}{2}, \frac{k - 1}{2}$

= $\Big(4, \frac{k - 1}{2}\Big)$.

Given, line x - 6y + 11 bisects the join of (8, -1) and (0, k).

∴ $\Big(4, \frac{k - 1}{2}\Big)$ satisfies the equation x - 6y + 11 = 0.

$$\begin{aligned}\therefore 4 - 6 \times \frac{k - 1}{2} + 11 = 0 \\[1em] \Rightarrow 4 - 3(k - 1) + 11 = 0 \\[1em] \Rightarrow 4 - 3k + 3 + 11 = 0 \\[1em] \Rightarrow 18 - 3k = 0 \\[1em] \Rightarrow 3k = 18 \\[1em] \Rightarrow k = \frac{18}{3} \\[1em] \Rightarrow k = 6.\end{aligned}$$

Hence, k = 6.

Since, (-3, 2) lies on the line ax + 3y + 6 = 0, so it satisfies the equation.

Substituting x = -3 and y = 2 in the equation ax + 3y + 6 = 0 we have,

⇒ -3a + 3(2) + 6 = 0

⇒ -3a + 6 + 6 = 0

⇒ -3a + 12 = 0

⇒ 3a = 12

⇒ a = $\frac{12}{3}$

⇒ a = 4.

Hence, a = 4.

Since, (-4, 4) lies on the line y = mx + 8, so it satisfies the equation.

Substituting x = -4 and y = 4 in the equation y = mx + 8 we have,

⇒ 4 = -4m + 8

⇒ 4m = 8 - 4

⇒ 4m = 4

⇒ m = $\frac{4}{4}$

⇒ m = 1.

Hence, m = 1.

By section-formula co-ordinates of,

P = $\frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$$\begin{aligned}P = \Big(\frac{2 \times -3 + 3 \times 2}{2 + 3}, \frac{2 \times 6 + 3 \times 1}{2 + 3}\Big) \\[1em] = \Big(\frac{-6 + 6}{5}, \frac{12 + 3}{5}\Big) \\[1em] = \Big(\frac{0}{5}, \frac{15}{5}\Big) \\[1em] = (0, 3).\end{aligned}$$

If P will lie on the line x - 5y + 15 = 0, it will satisfy the equation.

Substituting x = 0 and y = 3 in L.H.S. of the equation x - 5y + 15 = 0.

L.H.S. = 0 - 5 × 3 + 15

= 0 - 15 + 15

= 0.

Since, L.H.S. = R.H.S.

∴ P lies on the line x - 5y + 15 = 0.

Hence, P = (0, 3) and it lies on the line x - 5y + 15 = 0.

By section-formula co-ordinates of,

Q = $\frac{m_1x_2 + m_2x_1}{m_1 + m_2}$

Substituting values we get,

$$\begin{aligned}Q = \Big(\frac{1 \times 2 + 2 \times 5}{1 + 2}, \frac{1 \times 2 + 2 \times -4}{1 + 2}\Big) \\[1em] = \Big(\frac{2 + 10}{3}, \frac{2 + (-8)}{3}\Big) \\[1em] = \Big(\frac{12}{3}, \frac{-6}{3}\Big) \\[1em] = (4, -2).\end{aligned}$$

If Q will lie on the line x - 2y = 0, it will satisfy the equation.

Substituting x = 4 and y = -2 in L.H.S. of the equation x - 2y = 0.

L.H.S. = 4 - 2 × (-2)

= 4 + 4

= 8.

Since, L.H.S. ≠ R.H.S.

∴ Q does not lies on the line x - 2y = 0.

Hence, Q = (4, -2) and it does not lie on the line x - 2y = 0.

Solving,

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

3x - y + 9 = 0 ..........(2)

⇒ y = 3x + 9

Substituting value of y in equation 1 we get,

⇒ 4x + 3(3x + 9) = 1

⇒ 4x + 9x + 27 = 1

⇒ 13x = -26

⇒ x = $\frac{-26}{13}$

⇒ x = -2.

Substituting x = -2, in y = 3x + 9 we get,

⇒ y = 3(-2) + 9 = -6 + 9 = 3.

Point of intersection = (-2, 3).

Given, (-2, 3) lies on the line (2k - 1)x - 2y = 4

∴ (2k - 1)(-2) - 2 × 3 = 4

⇒ -4k + 2 - 6 = 4

⇒ -4k - 4 = 4

⇒ 4k = -8

⇒ k = $\frac{-8}{4}$

⇒ k = -2.

Hence, point of intersection = (-2, 3) and k = -2.

When lines are concurrent they intersect at same point.

Solving, 2x + 5y = 1 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 6 + 3y

Substituting x = 3y + 6 in 2x + 5y = 1 we get,

⇒ 2(3y + 6) + 5y = 1

⇒ 6y + 12 + 5y = 1

⇒ 11y = -11

⇒ y = -1.

Substituting y = -1 in x = 6 + 3y we get,

x = 6 + 3(-1) = 6 - 3 = 3.

∴ 2x + 5y = 1 and x - 3y = 6 intersect in the point (3, -1).

Solving, x + 5y + 2 = 0 and x - 3y = 6 simultaneously.

⇒ x - 3y = 6

⇒ x = 3y + 6

Substituting x = 3y + 6 in x + 5y + 2 = 0 we get,

⇒ 3y + 6 + 5y + 2 = 0

⇒ 8y + 8 = 0

⇒ 8y = -8

⇒ y = -1.

Substituting y = -1 in x = 3y + 6 we get,

x = 3(-1) + 6 = -3 + 6 = 3.

∴ x + 5y + 2 = 0 and x - 3y = 6 intersect in the point (3, -1).

Hence, proved that 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Given,

⇒ Slope = $\sqrt{3}$

⇒ tan θ = $\sqrt{3}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, Option 3 is the correct option.

We know that,

Product of slope of two perpendicular lines = -1.

Let slope of perpendicular line be m.

∴ 5 × m = -1

⇒ m = $-\frac{1}{5}$.

Hence, Option 4 is the correct option.

By formula,

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

Substituting values we get :

Slope of line passing through (0, 0) and (-3, 4)

= $\frac{4 - 0}{-3 - 0} = \frac{4}{-3} = -\frac{4}{3}$.

Hence, Option 2 is the correct option.

By formula,

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

Substituting values we get :

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

By formula,

⇒ Slope = tan θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, Option 3 is the correct option.

By formula,

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

Substituting values we get :

Slope of line passing through (8, -5) and (-4, 7)

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

Let slope of line perpendicular to line segment joining points (8, -5) and (-4, 7) be m.

We know that,

Product of slopes of two perpendicular lines = -1.

∴ m × -1 = -1

⇒ m = $\frac{-1}{-1}$ = 1.

Hence, Option 2 is the correct option.

By formula,

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

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

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

Let inclination be θ,

∴ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, slope = 1 and inclination = 45°.

(ii) A = (0, -$\sqrt{3}$) and B = (3, 0)

$$\begin{aligned}\text{Slope of AB} = \frac{0 - (-\sqrt{3})}{3 - 0} \\[1em] = \frac{\sqrt{3}}{3} \\[1em] = \frac{1}{\sqrt{3}}. \\[1em]\end{aligned}$$

Let inclination be θ,

∴ tan θ = $\frac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, slope = $\frac{\sqrt{3}}{3}$ and inclination = 30°.

(iii) A = (-1, $2\sqrt{3}$) and B = (-2, $\sqrt{3}$)

$$\begin{aligned}\text{Slope of AB} = \frac{\sqrt{3} - 2\sqrt{3}}{-2 - (-1)} \\[1em] = \frac{-\sqrt{3}}{-2 + 1} \\[1em] = \frac{-\sqrt{3}}{-1} = \sqrt{3}. \\[1em]\end{aligned}$$

Let inclination be θ,

∴ tan θ = $\sqrt3{}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, slope = $\sqrt{3}$ and inclination = 60°.

By formula,

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

Let m₁ be the slope of line passing through (0, 2) and (-3, -1), and m₂ be the slope of line passing through (-1, 5) and (4, a).

Since, lines are parallel.

∴ m₁ = m₂

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

Hence, a = 10.

By formula,

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

Let m₁ be the slope of line passing through (-4, -2) and (2, -3), and m₂ be the slope of line passing through (a, 5) and (2, -1).

Since, lines are perpendicular.

∴ m₁.m₂ = -1

$$\begin{aligned}\Rightarrow \frac{-3 - (-2)}{2 - (-4)} \times \frac{-1 - 5}{2 - a} = -1 \\[1em] \Rightarrow \frac{-3 + 2}{2 + 4} \times \frac{-6}{2 - a} = -1 \\[1em] \Rightarrow \frac{-1}{6} \times \frac{-6}{2 - a} = -1 \\[1em] \Rightarrow \frac{1}{2 - a} = -1 \\[1em] \Rightarrow -(2 - a) = 1 \\[1em] \Rightarrow a - 2 = 1 \\[1em] \Rightarrow a = 1 + 2 = 3.\end{aligned}$$

Hence, a = 3.

By formula,

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

$$\begin{aligned}\text{Slope of AB }(m_1) = \frac{4 - (-2)}{-4 - 4} \\[1em] = \frac{4 + 2}{-8} \\[1em] = -\frac{6}{8}. \\[1em] \text{Slope of AC }(m_2) = \frac{6 - (-2)}{10 - 4} \\[1em] = \frac{6 + 2}{6} \\[1em] = \frac{8}{6}. \\[1em] m_1 \times m_2 = -\frac{6}{8} \times \frac{8}{6} = -1.\end{aligned}$$

Since, m₁.m₂ = -1.

∴ AB ⊥ AC.

Hence, proved that ABC is a right-angled triangle at A.

By formula,

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

$$\begin{aligned}\Rightarrow \text{Slope of AB} = \frac{2 - 5}{1 - 4} \\[1em] = \frac{-3}{-3} \\[1em] = 1. \\[1em] \Rightarrow \text{Slope of CD} = \frac{6 - 3}{7 - 4} \\[1em] = \frac{3}{3} \\[1em] = 1. \\[1em] \Rightarrow \text{Slope of BC} = \frac{3 - 2}{4 - 1} \\[1em] = \frac{1}{3}. \\[1em] \Rightarrow \text{Slope of AD} = \frac{6 - 5}{7 - 4} \\[1em] = \frac{1}{3}.\end{aligned}$$

From above,

Slope of AB = Slope of CD and Slope of BC = Slope of AD.

∴ AB || CD and BC || AD.

Hence, proved ABCD is a parallelogram.

Let the given points be A(-2, 4), B(4, 8), C(10, 7) and D(11, -5).

And, let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

By mid-point formula,

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

So,

$$\begin{aligned}\Rightarrow P = \Big(\frac{-2 + 4}{2}, \frac{4 + 8}{2}\Big) \\[1em] = \Big(\frac{2}{2}, \frac{12}{2}\Big) = (1, 6). \\[1em] \Rightarrow Q = \Big(\frac{4 + 10}{2}, \frac{8 + 7}{2}\Big) \\[1em] = \Big(\frac{14}{2}, \frac{15}{2}\Big) = (7, 7.5). \\[1em] \Rightarrow R = \Big(\frac{10 + 11}{2}, \frac{7 + (-5)}{2}\Big) \\[1em] = \Big(\frac{21}{2}, \frac{2}{2}\Big) = (10.5, 1). \\[1em] \Rightarrow S = \Big(\frac{11 + (-2)}{2}, \frac{-5 + 4}{2}\Big) \\[1em] = (4.5, -0.5).\end{aligned}$$

By formula,

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

$$\begin{aligned}\Rightarrow \text{Slope of PQ} = \frac{7.5 - 6}{7 - 1} \\[1em] = \frac{1.5}{6} \\[1em] = \frac{1}{4}. \\[1em] \Rightarrow \text{Slope of QR} = \frac{1 - 7.5}{10.5 - 7} \\[1em] = \frac{-6.5}{3.5} \\[1em] = -\frac{65}{35} = -\frac{13}{7}. \\[1em] \Rightarrow \text{Slope of RS} = \frac{-0.5 - 1}{4.5 - 10.5} \\[1em] = \frac{-1.5}{-6} \\[1em] = \frac{1}{4}. \\[1em] \Rightarrow \text{Slope of PS} = \frac{-0.5 - 6}{4.5 - 1} \\[1em] = \frac{-6.5}{3.5} = -\frac{65}{35} \\[1em] = -\frac{13}{7}.\end{aligned}$$

From above calculation we get,

Slope of PQ = Slope of RS and Slope of QR = Slope of PS

∴ PQ || RS and QR || PS.

Hence, proved that the quadrilateral, obtained on joining the mid-points of sides of quadrilateral with vertices (-2, 4), (4, 8), (10, 7) and (11, -5), is a parallelogram.

By formula,

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

$$\begin{aligned}\text{Slope of PQ} = \frac{c + a - (b + c)}{b - a} \\[1em] = \frac{c + a - b - c}{b - a} \\[1em] = \frac{a - b}{b - a} \\[1em] = \frac{-(b - a)}{b - a} = -1. \\[1em] \text{Slope of QR} = \frac{a + b - (c + a)}{c - b} \\[1em] = \frac{a + b - c - a}{c - b} \\[1em] = \frac{b - c}{c - b} \\[1em] = \frac{-(c - b)}{c - b} = -1.\end{aligned}$$

Since, Slope of PQ = QR.

Hence, proved that P, Q and R are collinear.

By formula,

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

Given, slope of line joining (x, 2) and (8, -11) is $-\frac{3}{4}$.

$$\begin{aligned}\therefore -\frac{3}{4} = \frac{-11 - 2}{8 - x} \\[1em] \Rightarrow -3(8 - x) = 4 \times -13 \\[1em] \Rightarrow -24 + 3x = -52 \\[1em] \Rightarrow 3x = -52 + 24 \\[1em] \Rightarrow 3x = -28 \\[1em] \Rightarrow x = -\frac{28}{3}.\end{aligned}$$

Hence, x = $-\frac{28}{3}$.

(i) By formula,

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

$$\begin{aligned}\text{Slope of AB }= \frac{-2 - 4}{-3 - 5} \\[1em] = \frac{-6}{-8} \\[1em] = \frac{3}{4}.\end{aligned}$$

We know that,

Product of slope of perpendicular lines = -1.

∴ Slope of AB × Slope of altitude = -1

⇒ $\frac{3}{4}$ x Slope of altitude = -1

⇒ Slope of altitude = $-\frac{4}{3}$

Hence, slope of the altitude of AB = $-\frac{4}{3}$.

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

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

$$\begin{aligned}\text{Slope of AD }= \frac{-5 - 4}{-1 - 5} \\[1em] = \frac{-9}{-6} \\[1em] = \frac{3}{2}.\end{aligned}$$

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

(iii) $$\begin{aligned}\text{Slope of AC }= \frac{-8 - 4}{1 - 5} \\[1em] = \frac{-12}{-4} \\[1em] = 3.\end{aligned}$$

Since, slope of parallel lines are equal.

Hence, slope of line parallel to AC = 3.

Rectangle ABCD is shown in the figure below:

Since, product of slope of perpendicular line = -1.

Slope of AB x Slope of BC = -1

$$\begin{aligned}\Rightarrow \text{Slope of AB} \times \frac{2}{3} = -1 \\[1em] \Rightarrow \text{Slope of AB} = -\frac{3}{2}.\end{aligned}$$

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

(ii) AD is parallel to BC and slope of parallel lines are equal.

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

Let points be A(K, 3), B(2, -4) and C(-K + 1, -2).

Since, points are collinear.

∴ Slope of AB = Slope of BC

$$\begin{aligned}\Rightarrow \frac{-4 - 3}{2 - K} = \frac{-2 - (-4)}{-K + 1 - 2} \\[1em] \Rightarrow \frac{-7}{2 - K} = \frac{2}{-K - 1} \\[1em] \Rightarrow -7(-K - 1) = 2(2 - K) \\[1em] \Rightarrow 7K + 7 = 4 - 2K \\[1em] \Rightarrow 7K + 2K = 4 - 7 \\[1em] \Rightarrow 9K = -3 \\[1em] \Rightarrow K = -\frac{3}{9} = -\frac{1}{3}.\end{aligned}$$

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

From graph,

AC appears to be steeper.

$$\begin{aligned}\text{Slope of AB } = \frac{7 - 1}{4 - 1} \\[1em] = \frac{6}{3} = 2. \\[1em] \text{Slope of AC } = \frac{10 - 1}{4 - 1} \\[1em] = \frac{9}{3} = 3.\end{aligned}$$

Since, AC has greater slope.

Hence, proved AC has steeper slope.

By formula,

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

Substituting values we get :

Slope of line passing through (-7, 4) and (5, 4)

= $\frac{4 - 4}{5 - (-7)} = \frac{0}{5 + 7} = \frac{0}{12}$ = 0.

We know that,

Slope of parallel lines are equal.

We know that,

Slope of x-axis = 0.

∴ Slope of line passing through (-7, 4) and (5, 4) = Slope of x-axis.

Hence, Option 1 is the correct option.

Given,

The line intersects the y-axis at point (0, 2) and makes an angle of 45° with x-axis.

Slope (m) = tan θ = tan 45° = 1.

Equation of line passing through a point is given by :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = 1(x - 0)

⇒ y - 2 = x

⇒ y - x = 2.

Hence, Option 2 is the correct option.

We know that,

x co-ordinate at y-axis = 0.

Let co-ordinates of A are (0, b).

Since,

Line 5x - 2y + 8 = 0, intersects the y-axis at point A(0, b).

∴ Point A satisfies the equation 5x - 2y + 8 = 0.

Substituting values we get :

⇒ 5(0) - 2b + 8 = 0

⇒ 0 - 2b + 8 = 0

⇒ 2b = 8

⇒ b = $\frac{8}{2}$ = 4.

∴ A = (0, b) = (0, 4).

Hence, Option 4 is the correct option.

By intercept form,

Equation of line : $\frac{x}{a} + \frac{y}{b} = 1$

Given,

x-intercept (a) : 7

y-intercept (b) : -7

Substituting values in equation, we get :

$$\begin{aligned}\Rightarrow \frac{x}{7} + \frac{y}{-7} = 1 \\[1em] \Rightarrow \frac{x}{7} - \frac{y}{7} = 1 \\[1em] \Rightarrow x - y = 7.\end{aligned}$$

Hence, Option 1 is the correct option.

Given, equation :

⇒ 5x - 5y = 12

⇒ -5y = -5x + 12

⇒ 5y = 5x - 12

⇒ y = $\frac{5x}{5} - \frac{12}{5}$

⇒ y = x - $\frac{12}{5}$

Comparing above equation with y = mx + c (where m is slope and c is the y-intercept), we get :

⇒ m = 1 and c = $-\frac{12}{5}$

To find x-intercept of the line, we put y = 0 in above equation:

0 = x - $\frac{12}{5}$

⇒ x = $\frac{12}{5}$

∴ Slope = 1 and x-intercept = $\frac{12}{5}$

Hence, Option 3 is the correct option.

m = tan θ = tan 45° = 1.

Equation of a line : y = mx + c

Substituting values we get,

y = x - 1.

Hence, equation of line is y = x - 1.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 4 = $-\frac{4}{3}$[x - (-3)]

⇒ 3(y - 4) = -4(x + 3)

⇒ 3y - 12 = -4x - 12

⇒ 4x + 3y = -12 + 12

⇒ 4x + 3y = 0.

Hence, equation of line is 4x + 3y = 0.

m = tan θ = tan 60° = $\sqrt{3}$

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 4 = $\sqrt{3}(x - 5)$

⇒ y - 4 = $\sqrt{3}x - 5\sqrt{3}$

⇒ y = $\sqrt{3}x + 4 - 5\sqrt{3}$.

Hence, equation of line is y = $\sqrt{3}x + 4 - 5\sqrt{3}$.

(i) (0, 1) and (1, 2)

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

Substituting values we get,

$$\begin{aligned}m = \frac{2 - 1}{1 - 0} = \frac{1}{1} = 1. \\[1em] \text{Equation } : y - y_1= m(x - x_1) \\[1em] \Rightarrow y - 1 = 1(x - 0) \\[1em] \Rightarrow y - 1 = x \\[1em] \Rightarrow y = x + 1.\end{aligned}$$

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

(ii) (-1, -4) and (3, 0)

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

Substituting values we get,

$$\begin{aligned}m = \frac{0 - (-4)}{3 - (-1)} = \frac{0 + 4}{3 + 1} = \frac{4}{4} = 1. \\[1em] \text{Equation } : y - y_1= m(x - x_1) \\[1em] \Rightarrow y - (-4) = 1(x - (-1)) \\[1em] \Rightarrow y + 4 = x + 1 \\[1em] \Rightarrow y = x + 1 - 4 \\[1em] \Rightarrow y = x - 3.\end{aligned}$$

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

(i) By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of PQ } = \frac{5 - 6}{-3 - 2} \\[1em] = \frac{-1}{-5} \\[1em] = \frac{1}{5}.\end{aligned}$$

Hence, gradient of slope PQ = $\frac{1}{5}$.

(ii) By point slope form,

Equation : y - y₁ = m(x - x₁)

⇒ y - 6 = $\frac{1}{5}$(x - 2)

⇒ 5(y - 6) = x - 2

⇒ 5y - 30 = x - 2

⇒ 5y = x - 2 + 30

⇒ 5y = x + 28

Hence, equation of PQ is 5y = x + 28.

(iii) The point where PQ intersects x-axis, there y co-ordinate = 0.

Substituting y = 0 in equation of PQ we get,

⇒ 5 × 0 = x + 28

⇒ 0 = x + 28

⇒ x = -28.

Point = (x, y) = (-28, 0).

Hence, co-ordinates of the point where PQ intersects the x-axis = (-28, 0).

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

Equation : y - y₁ = m(x - x₁)

⇒ y - 4 = (-1)[x - (-3)]

⇒ y - 4 = -1(x + 3)

⇒ y - 4 = -x - 3

⇒ y + x = -3 + 4

⇒ x + y = 1.

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

(ii) The point where AB intersects y-axis, there x co-ordinate = 0.

Substituting x = 0 in equation of AB we get,

⇒ 0 + y = 1

⇒ y = 1.

Point = (x, y) = (0, 1).

Hence, co-ordinates of the point where AB intersects the y-axis = (0, 1).

From figure,

Slope of AB = tan 45° = 1.

Slope of CD = tan 60° = $\sqrt{3}$.

By point-slope form,

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

Since, line AB passes through point P(3, 4) and slope = 1. Substituting values in point-slope form,

⇒ y - 4 = 1(x - 3)

⇒ y - 4 = x - 3

⇒ y - x = -3 + 4

⇒ y - x = 1

⇒ y = x + 1.

Since, line CD passes through point P(3, 4) and slope = $\sqrt{3}$. Substituting values in point-slope form,

⇒ y - 4 = $\sqrt{3}$(x - 3)

⇒ y - 4 = $\sqrt{3}x - 3\sqrt{3}$

⇒ y = $\sqrt{3}x - 3\sqrt{3} + 4$.

Hence, equation of AB is y = x + 1 and equation of CD is y = $\sqrt{3}x - 3\sqrt{3} + 4$.

Let AD be the median.

Since, AD is the median so D will be the mid-point of BC.

Co-ordinates of D = $\Big(\frac{7 + 1}{2}, \frac{8 + (-10)}{2}\Big) = \Big(\frac{8}{2}, \frac{-2}{2}\Big)$ = (4, -1).

$$\begin{aligned}\text{Slope of AD } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-1 - 5}{4 - 3} \\[1em] = \frac{-6}{1} \\[1em] = -6.\end{aligned}$$

By point-slope form,

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

Substituting values we get,

⇒ y - 5 = -6(x - 3)

⇒ y - 5 = -6x + 18

⇒ y + 6x = 18 + 5

⇒ 6x + y = 23.

Hence, equation of median through A is 6x + y = 23.

Given, ∠A = 60° and vertex C = (7, 5)

As, ABCD is a parallelogram, we have

∠A + ∠B = 180° [Sum of adjacent angles in a || gm = 180°]

∠B = 180° – 60° = 120°

So, the anticlockwise angle of BC from x-axis is (180° - 120°) = 60°.

Slope of BC = tan 60° = $\sqrt{3}$

By point-slope form,

Equation of line BC is :

⇒ y – y₁ = m(x – x₁)

⇒ y – 5 = $\sqrt{3}$(x – 7)

⇒ y – 5 = $\sqrt{3}x - 7\sqrt{3}$

⇒ y = $\sqrt{3}x - 7\sqrt{3} + 5$

As, CD || AB and AB || x-axis

Slope of CD = Slope of AB = 0 [As slope of x-axis is zero]

By point-slope form,

Equation of line CD is :

⇒ y – y₁ = m(x – x₁)

⇒ y – 5 = 0(x – 7)

⇒ y = 5.

Hence, equation of BC is y = $\sqrt{3}x - 7\sqrt{3} + 5$ and CD is y = 5.

Solving x + 2y = 7 and x - y = 4 simultaneously,

⇒ x + 2y = 7

⇒ x = 7 - 2y .......(1)

Substituting above value of x in x - y = 4 we get,

⇒ 7 - 2y - y = 4

⇒ -3y = 4 - 7

⇒ -3y = -3

⇒ y = 1.

Substituting y = 1 in equation 1 we get,

⇒ x = 7 - 2(1) = 5.

Point of intersection = (5, 1).

Slope of line passing through (0, 0) and (5, 1) = $\frac{1 - 0}{5 - 0} = \frac{1}{5}.$

By point-slope form,

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

Substituting values we get,

⇒ y - 0 = $\frac{1}{5}(x - 0)$

⇒ 5(y - 0) = 1(x - 0)

⇒ 5y = x

⇒ x - 5y = 0.

Hence, equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4 is x - 5y = 0.

Let AD be the median through A. So, D will be the mid-point of BC.

Co-ordinates of D = $\Big(\frac{-2 + 0}{2}, \frac{3 + 1}{2}\Big) = \Big(\frac{-2}{2}, \frac{4}{2}\Big) = (-1, 2)$.

By formula,

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

$$\begin{aligned}\text{Slope of AD} = \frac{2 - 7}{-1 - 4} \\[1em] = \frac{-5}{-5} = 1.\end{aligned}$$

By point-slope form,

Equation : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 7 = 1(x - 4)

⇒ y - 7 = x - 4

⇒ x - y - 4 + 7 = 0

⇒ x - y + 3 = 0.

$$\begin{aligned}\text{Slope of AC} = \frac{1 - 7}{0 - 4} \\[1em] = \frac{-6}{-4} = \frac{3}{2}.\end{aligned}$$

Since, parallel lines have equal slope, equation of line passing through B and parallel to AC is

⇒ y - 3 = $\frac{3}{2}$[x - (-2)]

⇒ 2(y - 3) = 3(x + 2)

⇒ 2y - 6 = 3x + 6

⇒ 3x - 2y + 12 = 0.

Hence, equation of median through A is x - y + 3 = 0 and equation of line passing through B and parallel to AC is 3x - 2y + 12 = 0.

Let P = (-1, 2)

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

$$\begin{aligned}\text{Slope of AB }= \frac{3 - 4}{2 - 1} \\[1em] = \frac{-1}{1} = -1.\end{aligned}$$

We know that,

Product of slope of perpendicular lines is -1.

Let slope of line through P 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

⇒ y = x + 3

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

(i) x-intercept = 5 and y-intercept = 3

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

When y-intercept = 3; corresponding point on the y-axis = (0, 3).

$\text{Slope }= \frac{3 - 0}{0 - 5} = -\frac{3}{5}.$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{3}{5}$(x - 5)

⇒ 5y = -3(x - 5)

⇒ 5y = -3x + 15

⇒ 3x + 5y = 15.

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

(ii) x-intercept = -4 and y-intercept = 6

When x-intercept = -4; corresponding point on the x-axis = (-4, 0)

When y-intercept = 6; corresponding point on the y-axis = (0, 6).

$\text{Slope }= \frac{6 - 0}{0 - (-4)} = \frac{6}{4}.$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\frac{6}{4}$[x - (-4)]

⇒ y = $\frac{3}{2}$(x + 4)

⇒ 2y = 3x + 12

⇒ 2y - 3x = 12

⇒ 2y = 3x + 12

Hence, equation of line is 2y = 3x + 12.

When x-intercept = 6; corresponding point on the x-axis = (6, 0).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{5}{6}(x - 6)$

⇒ 6y = -5(x - 6)

⇒ 6y = -5x + 30

⇒ 5x + 6y = 30.

Hence, equation of line is 5x + 6y = 30.

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

Slope of the line through (5, 0) and (-3, 2) = $\frac{2 - 0}{-3 - 5} = \frac{2}{-8} = -\frac{1}{4}$.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{1}{4}(x - 5)$

⇒ 4y = -1(x - 5)

⇒ 4y = -x + 5

⇒ x + 4y = 5.

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

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

Slope of the line through (0, 5) and (1, 3) = $\frac{3 - 5}{1 - 0} = \frac{-2}{1} = -2$.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = -2(x - 0)

⇒ y - 5 = -2x

⇒ 2x + y = 5.

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

Let there be two lines AB and CD equally inclined to co-ordinate axes and passing through E(-2, 0).

From figure,

AB is inclined at an angle of 45°.

Slope = tan 45° = 1.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1[x - (-2)]

⇒ y = x + 2

⇒ x - y + 2 = 0.

From figure,

CD is inclined at an angle of -45° (As measured clockwise).

Slope = tan (-45°) = -1.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = -1[x - (-2)]

⇒ y = -[x + 2]

⇒ y = -x - 2

⇒ x + y + 2 = 0.

Hence, equation of lines are x - y + 2 = 0 and x + y + 2 = 0.

(i) From figure,

Inclination of the line = 45°

Slope = tan 45° = 1.

Hence, slope of the line = 1.

(ii) By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = 1(x - 5)

⇒ y - 3 = x - 5

⇒ x - y - 5 + 3 = 0

⇒ x - y - 2 = 0

⇒ x - y = 2

⇒ y = x - 2.

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

(iii) Let co-ordinates of point Q be (0, a).

Substituting values in equation we get,

a = 0 - 2

⇒ a = -2.

Hence, co-ordinates of point Q = (0, -2).

By section formula,

Co-ordinates of P = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$$\begin{aligned}P = \Big(\frac{3 \times 12 + 1 \times 4}{3 + 1}, \frac{3 \times 0 + 1 \times -8}{3 + 1}\Big) \\[1em] = \Big(\frac{36 + 4}{4}, \frac{0 - 8}{4}\Big) \\[1em] = \Big(\frac{40}{4}, \frac{-8}{4}\Big) \\[1em] = (10, -2).\end{aligned}$$

By point-slope form,

Equation of line having slope = $-\frac{2}{5}$ and passing through P,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $-\frac{2}{5}$[x - 10]

⇒ 5(y + 2) = -2(x - 10)

⇒ 5y + 10 = -2x + 20

⇒ 5y + 2x = 20 - 10

⇒ 2x + 5y = 10.

Hence, 2x + 5y = 10.

(i) 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,

$$\begin{aligned}\text{Centroid } = \Big(\frac{1 + 3 + 7}{3}, \frac{4 + 2 + 5}{3}\Big) \\[1em] = \Big(\frac{11}{3}, \frac{11}{3}\Big).\end{aligned}$$

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

(ii) $$\begin{aligned}\text{Slope of AB} = \frac{2 - 4}{3 - 1} \\[1em] = \frac{-2}{2} = -1.\end{aligned}$$

Slope of line parallel to AB will also be equal to -1.

By point-slope form,

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

⇒ y - y₁ = m(x - x₁)

$$\begin{aligned}\Rightarrow y - \frac{11}{3} = -1\Big(x - \frac{11}{3}\Big) \\[1em] \Rightarrow \frac{3y - 11}{3} = -1 \times \frac{3x - 11}{3} \\[1em] \Rightarrow 3y - 11 = -1(3x - 11) \\[1em] \Rightarrow 3y - 11 = -3x + 11 \\[1em] \Rightarrow 3y + 3x = 11 + 11 \\[1em] \Rightarrow 3x + 3y = 22.\end{aligned}$$

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

By section formula,

Co-ordinates of P = $\Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get,

$$\begin{aligned}P = \Big(\frac{2 \times -3 + 3 \times 7}{2 + 3}, \frac{2 \times 4 + 3 \times -1}{2 + 3}\Big) \\[1em] = \Big(\frac{-6 + 21}{5}, \frac{8 - 3}{5}\Big) \\[1em] = \Big(\frac{15}{5}, \frac{5}{5}\Big) \\[1em] = (3, 1).\end{aligned}$$

Slope of BP = $\frac{1 - 1}{3 - 4} = 0.$

By point-slope form,

Equation of BP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = 0(x - 3)

⇒ y - 1 = 0

⇒ y = 1.

Hence, equation of BP is y = 1.

By intercept form,

Equation of line : $\frac{x}{a} + \frac{y}{b} = 1$

Given,

x-intercept (a) : 5

y-intercept (b) : 5

Substituting values in equation, we get :

$$\begin{aligned}\Rightarrow \frac{x}{5} + \frac{y}{5} = 1 \\[1em] \Rightarrow x + y = 5.\end{aligned}$$

Hence, Option 2 is the correct option.

Let AB be the line not passing through the origin and intersecting x-axis at D and y-axis at E.

We know that,

y-coordinate = 0 at x axis and x-coordinate = 0 at y-axis.

Let D = (a, 0) and E = (0, b).

By formula,

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

From figure,

Slope of DE = Slope of AB = 1

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Slope of DE} = \frac{b - 0}{0 - a} \\[1em] \Rightarrow 1 = \frac{b}{-a} \\[1em] \Rightarrow b = -a.\end{aligned}$$

From figure,

a and b are intercepts on x-axis and y-axis respectively.

Hence, Option 3 is the correct option.

By slope-intercept form :

Equation of line : y = mx + c ......(1)

Given,

Line passes through point (1, 1) and makes a y-intercept of 3.

Substituting value in equation (1), we get :

⇒ 1 = m(1) + 3

⇒ 1 = m + 3

⇒ m = 1 - 3 = -2.

∴ Slope of line (m) = -2 and y-intercept (c) = 3

Substituting value of m and c in equation (1), we get :

⇒ y = -2x + 3

⇒ 2x + y = 3.

Hence, Option 3 is the correct option.

Given,

Equations :

⇒ x - y = 3 ........(1)

⇒ x + y = 0 ........(2)

Adding equation (1) and (2), we get :

⇒ (x - y) + (x + y) = 3 + 0

⇒ x + x - y + y = 3

⇒ 2x = 3

⇒ x = $\frac{3}{2}$

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

⇒ $\frac{3}{2}$ + y = 0

⇒ y = $-\frac{3}{2}$.

∴ Point of intersection of lines x - y = 3 and x + y = 0 is $\Big(\frac{3}{2}, -\frac{3}{2}\Big)$.

Given,

Inclination (θ) = 45°

Slope = tan 45° = 1.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Equation of line passing through $\Big(\frac{3}{2}, -\frac{3}{2}\Big)$ and slope = 1 is :

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

Hence, Option 2 is the correct option.

By intercept form :

Equation of line : $\frac{x}{a} + \frac{y}{b} = 1$ .........(1)

Given,

Line cuts equal intercepts with both the axes.

∴ x-intercept (a) = y-intercept (b) = p (let)

Given,

Line passes through point (6, 6). So it will satisfy the equation (1).

Substituting values in equation we get :

$$\begin{aligned}\Rightarrow \frac{6}{p} + \frac{6}{p} = 1 \\[1em] \Rightarrow \frac{12}{p} = 1 \\[1em] \Rightarrow p = 12.\end{aligned}$$

∴ a = b = 12.

Substituting value of a and b in equation (1), we get :

$$\begin{aligned}\Rightarrow \frac{x}{12} + \frac{y}{12} = 1 \\[1em] \Rightarrow \frac{x + y}{12} = 1 \\[1em] \Rightarrow x + y = 12.\end{aligned}$$

Hence, Option 4 is the correct option.

(i) Given,

y = 4

Comparing it with y = mx + c, we get,

m = 0 and c = 4.

Hence, slope = 0 and y-intercept = 4.

(ii) Given,

⇒ ax - by = 0

⇒ by = ax

⇒ y = $\frac{a}{b}$x

Comparing it with y = mx + c, we get,

m = $\frac{a}{b}$ and c = 0.

Hence, slope = $\frac{a}{b}$ and y-intercept = 0.

(iii) Given,

⇒ 3x - 4y = 5

⇒ 4y = 3x - 5

⇒ y = $\frac{3}{4}x - \frac{5}{4}$

Comparing it with y = mx + c, we get,

m = $\frac{3}{4}$ and c = $-\frac{5}{4}$.

Hence, slope = $\frac{3}{4}$ and y-intercept = $-\frac{5}{4}$.

Given,

⇒ x - y = 4

⇒ -y = -x + 4

⇒ y = x - 4

Comparing it with y = mx + c, we get,

m = 1 and c = -4.

⇒ m = tan θ

⇒ 1 = tan θ

⇒ tan 45° = tan θ

⇒ θ = 45°.

Hence, slope = 1, y-intercept = -4 and inclination = 45°.

Given lines,

⇒ 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0

⇒ 4y = -3x - 7 and 21y = 28x + 50

⇒ y = $-\frac{3}{4}x - \frac{7}{4}$ and y = $\frac{28}{21}x + \frac{50}{21}$

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

Slope of 1st line = $-\frac{3}{4}$

Slope of 2nd line = $\frac{28}{21} = \frac{4}{3}$

Since,

Slope of 1st line × Slope of 2nd line = $-\frac{3}{4} \times \frac{4}{3} = -1.$

Hence, the lines 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0 are perpendicular to each other.

Given lines,

⇒ x - 3y = 4 and 3x - y = 7

⇒ 3y = x - 4 and y = 3x - 7

⇒ y = $\frac{1}{3}x - \frac{4}{3}$ and y = 3x - 7

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

Slope of 1st line = $\frac{1}{3}$

Slope of 2nd line = 3

Since,

Slope of 1st line × Slope of 2nd line = $\frac{1}{3} \times 3 = 1$ which is not equal to -1.

Hence, the lines x - 3y = 4 and 3x - y = 7 are not perpendicular to each other.

Given lines,

⇒ 3x + 2y = 5 and x + 2y = 1

⇒ 2y = -3x + 5 and 2y = -x + 1

⇒ y = $-\frac{3}{2}x + \frac{5}{2}$ and y = $-\frac{1}{2}x + \frac{1}{2}$

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

Slope of 1st line = $-\frac{3}{2}$

Slope of 2nd line = $-\frac{1}{2}$

Since,

Slope of 1st line ≠ Slope of 2nd line

Hence, the lines 3x + 2y = 5 and x + 2y = 1 are not parallel to each other.

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

Substituting values we get,

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

Hence, x = 0.

Given,

⇒ x + 2y + 3 = 0

⇒ 2y = -x - 3

⇒ y = $-\frac{1}{2}x - \frac{3}{2}$

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

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

Since, slope of parallel lines are equal.

Hence, slope of line parallel to x + 2y + 3 = 0 is $-\frac{1}{2}$.

Given,

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

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

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

Since, slope of parallel lines are equal.

Hence, slope of line parallel to $\frac{x}{2} - \frac{y}{3} - 1 = 0$ is $\frac{3}{2}$.

Given,

$$\begin{aligned}\Rightarrow x - \frac{y}{2} + 3 = 0 \\[1em] \Rightarrow \frac{y}{2} = x + 3 \\[1em] \Rightarrow y = 2x + 6.\end{aligned}$$

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

Slope = 2.

Let slope of perpendicular line be m.

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

∴ m × 2 = -1

⇒ m = $-\frac{1}{2}$.

Hence, slope of the line perpendicular to the line $x - \frac{y}{2} + 3 = 0 \text{ is } -\frac{1}{2}.$

Given,

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

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

Slope = $\frac{1}{6}$.

Let slope of perpendicular line be m.

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

∴ m × $\frac{1}{6}$ = -1

⇒ m = -6.

Hence, slope of the line perpendicular to the line $\frac{x}{3} - 2y = 4$ is -6.

Given lines,

⇒ 2x – by + 5 = 0 and ax + 3y = 2

⇒ by = 2x + 5 and 3y = -ax + 2

⇒ y = $\frac{2}{b}x + \frac{5}{2}$ and $y = -\frac{a}{3}x + \frac{2}{3}$

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

Slope of 1st line = $\frac{2}{b}$

Slope of 2nd line = $-\frac{a}{3}$

Since,

Slopes of parallel lines are equal.

$$\begin{aligned}\therefore \frac{2}{b} = -\frac{a}{3} \\[1em] \Rightarrow 6 = -ab \\[1em] \Rightarrow ab = -6.\end{aligned}$$

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

Given lines,

⇒ mx + 3y = -7 and 5x - ny = 3

⇒ 3y = mx + 7 and ny = 5x - 3

⇒ y = $\frac{m}{3}x + \frac{7}{3}$ and $y = \frac{5}{n}x - \frac{3}{n}$

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

Slope of 1st line = $\frac{m}{3}$

Slope of 2nd line = $\frac{5}{n}$

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

$$\begin{aligned}\therefore \frac{m}{3} \times \frac{5}{n} = -1 \\[1em] \Rightarrow \frac{5m}{3n} = -1 \\[1em] \Rightarrow 5m = -3n \\[1em] \Rightarrow 5m + 3n = 0.\end{aligned}$$

Hence, relation connecting m and n is 5m + 3n = 0.

Given lines,

⇒ 2x - y + 5 = 0 and px + 3y = 4

⇒ y = 2x + 5 and 3y = -px + 4

⇒ y = 2x + 5 and y = $-\frac{p}{3}x + \frac{4}{3}$.

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

Slope of 1st line = 2

Slope of 2nd line = $-\frac{p}{3}$

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

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

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

(i) Given,

⇒ 2x - 2y + 3 = 0

⇒ 2y = 2x + 3

⇒ y = $x + \frac{3}{2}$.

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

Slope = 1.

Hence, slope of line AB = 1.

(ii) Let angle be θ.

By formula,

⇒ Slope = tan θ

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, angle that line makes with positive direction of x-axis = 45°.

Given lines,

⇒ 4x + 3y = 9 and px - 6y + 3 = 0

⇒ 3y = -4x + 9 and 6y = px + 3

⇒ y = $-\frac{4}{3}x + 3$ and $y = \frac{p}{6}x + \frac{1}{2}$

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

Slope of 1st line = $-\frac{4}{3}$

Slope of 2nd line = $\frac{p}{6}$

Since,

Slopes of parallel lines are equal.

$$\begin{aligned}\therefore -\frac{4}{3} = \frac{p}{6} \\[1em] \Rightarrow p = \frac{-4 \times 6}{3} \\[1em] \Rightarrow p = -8.\end{aligned}$$

Hence, p = -8.

Given,

⇒ 2x - 4y = 5

⇒ 4y = 2x - 5

⇒ y = $\frac{2}{4}x - \frac{5}{4}$

⇒ y = $\frac{1}{2}x - \frac{5}{4}$

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

Slope (m₁) = $\frac{1}{2}$

$$\begin{aligned}\text{Slope of AB }(m_2) = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{b - 3}{4 - (-2)} \\[1em] = \frac{b - 3}{6}.\end{aligned}$$

Since,

Product of slopes of perpendicular lines = -1.

$$\begin{aligned}\therefore m_1 \times m_2 = -1 \\[1em] \Rightarrow \frac{1}{2} \times \frac{b - 3}{6} = -1 \\[1em] \Rightarrow \frac{b - 3}{12} = -1 \\[1em] \Rightarrow b - 3 = -12 \\[1em] \Rightarrow b = -12 + 3 = -9.\end{aligned}$$

Hence, b = -9.

(i) We know,

Slope of line parallel to x-axis = 0.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = 0[x - (-5)]

⇒ y - 7 = 0

⇒ y = 7.

Hence, equation of the line parallel to x-axis is y = 7.

(ii) We know that,

Eqiuation of line parallel to y-axis is given by,

⇒ x - x₁ = 0

⇒ x - (-5) = 0

⇒ x + 5 = 0.

Hence, equation of the line parallel to y-axis is x + 5 = 0.

Given,

⇒ x - 3y = 4

⇒ 3y = x - 4

⇒ y = $\frac{1}{3}x - \frac{4}{3}$

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

Slope = $\frac{1}{3}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to x - 3y = 4 is $\frac{1}{3}$.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $\frac{1}{3}(x - 5)$

⇒ 3(y + 3) = x - 5

⇒ 3y + 9 = x - 5

⇒ x - 3y - 5 - 9 = 0

⇒ x - 3y - 14 = 0.

Hence, equation of the line passing through (5, -3) and parallel to x - 3y = 4 is x - 3y - 14 = 0.

Given,

⇒ 4x + 5y = 6

⇒ 5y = -4x + 6

⇒ y = $-\frac{4}{5}x + \frac{6}{5}$

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

Slope (m₁) = $-\frac{4}{5}$

Let slope of line perpendicular to 4x + 5y = 6 be m₂.

Since, product of slopes of perpendicular lines be -1,

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

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = $\frac{5}{4}$[x - (-2)]

⇒ 4(y - 1) = 5(x + 2)

⇒ 4y - 4 = 5x + 10

⇒ 5x - 4y + 10 + 4 = 0

⇒ 5x - 4y + 14 = 0.

Hence, equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6 is 5x - 4y + 14 = 0.

Let points be A(6, -3) and B(0, 3) and D be the mid-point of AB.

Co-ordinates of D = $\Big(\frac{6 + 0}{2}, \frac{-3 + 3}{2}\Big) = \Big(\frac{6}{2}, \frac{0}{2}\Big)$ = (3, 0).

Slope of AB = $\frac{3 - (-3)}{0 - 6} = \frac{3 + 3}{-6} = \frac{6}{-6}$ = -1.

Let slope of perpendicular bisector be m.

Then

⇒ m × Slope of AB = -1

⇒ m × -1 = -1

⇒ m = 1.

Perpendicular bisector of AB will pass through mid-point of AB i.e. D.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1(x - 3)

⇒ y = x - 3.

Hence, equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3) is y = x - 3.

The rhombus ABCD is shown in the figure below:

By formula,

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

Substituting values we get,

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

Equation of BD by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\frac{1}{3}$[x - (-5)]

⇒ 3(y - 6) = -1(x + 5)

⇒ 3y - 18 = -x - 5

⇒ 3y + x - 18 + 5 = 0

⇒ x + 3y - 13 = 0

⇒ x + 3y = 13.

Since, diagonals of rhombus are perpendicular to each other.

So, product of their slopes will be -1.

Slope of AC × Slope of BD = -1

$\text{Slope of AC} \times -\frac{1}{3}$ = -1

Slope of AC = -1 × -3 = 3.

We know that,

Diagonals of rhombus bisect each other. Let diagonals meet at point O.

∴ Mid-point of AC = Mid-point of BD.

Co-ordinates of mid-point of BD (i.e. O)

= $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) = \Big(\frac{-5 + 1}{2}, \frac{6 + 4}{2}\Big) = \Big(\frac{-4}{2}, \frac{10}{2}\Big)$ = (-2, 5).

∴ Co-ordinates of mid-point of AC = (-2, 5).

Equation of AC by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = 3[x - (-2)]

⇒ y - 5 = 3[x + 2]

⇒ y - 5 = 3x + 6

⇒ 3x - y + 6 + 5 = 0

⇒ 3x - y + 11 = 0

⇒ y = 3x + 11.

Hence, equation of BD is x + 3y = 13 and equation of AC is y = 3x + 11.

(i) Let AD be the median through A.

So, D will be the mid-point of BC.

Co-ordinates of D = $\Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) = \Big(\frac{2 + (-2)}{2}, \frac{2 + 4}{2}\Big) = \Big(\frac{0}{2}, \frac{6}{2}\Big)$ = (0, 3).

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

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = -8(x - 1)

⇒ y + 5 = -8x + 8

⇒ 8x + y = 3.

Hence, equation of median through A is 8x + y = 3.

(ii) Let BF be the altitude.

$$\begin{aligned}\text{Slope of AC } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{4 - (-5)}{-2 - 1} \\[1em] = \frac{9}{-3} \\[1em] = -3.\end{aligned}$$

Since, altitude is at 90°.

So, altitude through B (i.e. BF) will be perpendicular to AC.

Let slope of altitude through B be m.

∴ m × -3 = -1

⇒ m = $\frac{1}{3}$.

Equation of altitude through B by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\frac{1}{3}$(x - 2)

⇒ 3(y - 2) = x - 2

⇒ 3y - 6 = x - 2

⇒ x - 3y - 2 + 6 = 0

⇒ x - 3y + 4 = 0.

Hence, equation of altitude through B is x - 3y + 4 = 0.

(iii) $$\begin{aligned}\text{Slope of AB } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{2 - (-5)}{2 - 1} \\[1em] = \frac{7}{1} \\[1em] = 7.\end{aligned}$$

Since, parallel lines have equal slopes.

So slope of line through C and parallel to AB = 7.

Equation of line through C by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = 7[x - (-2)]

⇒ y - 4 = 7(x + 2)

⇒ y - 4 = 7x + 14

⇒ 7x - y + 14 + 4 = 0

⇒ 7x - y + 18 = 0.

Hence, equation of line through C and parallel to AB is 7x - y + 18 = 0.

(i) Below figure shows the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5:

Given,

⇒ 2y = 3x + 5

⇒ y = $\frac{3}{2}x + \frac{5}{2}$

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

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

Let slope of line AB be m₁.

Since, AB and line 2y = 3x + 5 are perpendicular.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

$\frac{3}{2} \times m_1 = -1$

$m_1 = -\frac{2}{3}$.

By point-slope form,

Equation of AB : y - y₁ = m(x - x₁)

⇒ y - 2 = $-\frac{2}{3}$(x - 3)

⇒ 3(y - 2) = -2(x - 3)

⇒ 3y - 6 = -2x + 6

⇒ 2x + 3y = 6 + 6

⇒ 2x + 3y = 12.

Hence, equation of AB is 2x + 3y = 12.

(ii) Below figure shows AB with its intercepts on x axis and y axis:

At A,

y co-ordinate = 0 as it lies on x-axis.

Substituting y = 0 in equation of AB,

⇒ 2x + 3(0) = 12

⇒ 2x = 12

⇒ x = 6.

A = (x, 0) = (6, 0).

At B,

x co-ordinate = 0 as it lies on y-axis.

Substituting x = 0 in equation of AB,

⇒ 2(0) + 3y = 12

⇒ 3y = 12

⇒ y = 4.

B = (0, y) = (0, 4).

Area of right angle triangle OAB = $\frac{1}{2} \times OA \times OB$

= $\frac{1}{2} \times 6 \times 4$

= 12 sq. units.

Hence, A = (6, 0), B = (0, 4) and area of triangle OAB = 12 sq. units.

At A,

y co-ordinate = 0 as it lies on x-axis.

A = (x, 0)

Substituting y = 0 in equation 4x - 3y + 12 = 0 we get,

⇒ 4x - 3(0) + 12 = 0

⇒ 4x + 12 = 0

⇒ 4x = -12

⇒ x = -3.

A = (x, 0) = (-3, 0).

Given,

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\frac{4}{3}x$ + 4

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

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

Let slope of line perpendicular to 4x - 3y + 12 = 0 be m₁.

∴ Product of their slopes will be equal to -1.

∴ m × m₁ = -1

⇒ $\frac{4}{3} \times m_1 = -1$

⇒ $m_1 = -\frac{3}{4}$.

By point slope form,

Equation of line through A and slope = $-\frac{3}{4}$,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{3}{4}$[x - (-3)]

⇒ 4y = -3(x + 3)

⇒ 4y = -3x - 9

⇒ 3x + 4y + 9 = 0.

Hence, equation of line through A and perpendicular to 4x - 3y + 12 = 0 is 3x + 4y + 9 = 0.

(i) Given,

⇒ 2x - 3y + 18 = 0

⇒ 3y = 2x + 18

⇒ y = $\frac{2}{3}x + 6$

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

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

Since, AP is perpendicular to 2x - 3y + 18 = 0.

∴ Product of slope of AP and 2x - 3y + 18 = 0 will be -1.

Let slope of AP = m₁.

∴ m × m₁ = -1

⇒ $\frac{2}{3} \times m_1 = -1$

⇒ $m_1 = -\frac{3}{2}$.

By point-slope form, equation of AP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 7 = $-\frac{3}{2}$[x - (-5)]

⇒ 2(y - 7) = -3(x + 5)

⇒ 2y - 14 = -3x - 15

⇒ 3x + 2y - 14 + 15 = 0

⇒ 3x + 2y + 1 = 0.

Hence, equation of AP is 3x + 2y + 1 = 0.

(ii) P is the point where AP and 2x - 3y + 18 = 0 meets,

Solving 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0 simultaneously,

⇒ 3x + 2y + 1 = 0

⇒ 2y = -3x - 1

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

Substituting value of y in 2x - 3y + 18 = 0 we get,

$$\begin{aligned}\Rightarrow 2x - 3\Big(\frac{-3x - 1}{2}\Big) + 18 = 0 \\[1em] \Rightarrow 2x - \frac{-9x - 3}{2} + 18 = 0 \\[1em] \Rightarrow 2x + \frac{9x + 3}{2} + 18 = 0 \\[1em] \Rightarrow \frac{4x + 9x + 3 + 36}{2} = 0 \\[1em] \Rightarrow 13x + 39 = 0 \\[1em] \Rightarrow 13x = -39 \\[1em] \Rightarrow x = -3.\end{aligned}$$

Substituting x = -3 in (1) we get,

$$\begin{aligned}y = \frac{-3 \times -3 - 1}{2} \\[1em] = \frac{9 - 1}{2} \\[1em] = \frac{8}{2} \\[1em] = 4.\end{aligned}$$

P = (-3, 4).

Hence, co-ordinates of P = (-3, 4).

By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of AB} = \frac{2 - 0}{2 - 4} \\[1em] = \frac{2}{-2} \\[1em] = -1.\end{aligned}$$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = -1(x - 4)

⇒ y = -x + 4

⇒ x + y = 4.

$$\begin{aligned}\text{Slope of BC} = \frac{6 - 2}{0 - 2} \\[1em] = \frac{4}{-2} \\[1em] = -2.\end{aligned}$$

By point-slope form,

Equation of BC is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = -2(x - 2)

⇒ y - 2 = -2x + 4

⇒ 2x + y = 6.

Let point P be (0, a) and Q be (b, 0).

Substituting value of P in equation of AB we get,

⇒ 0 + a = 4

⇒ a = 4.

Substituting value of Q in equation of BC we get,

⇒ 2b + 0 = 6

⇒ 2b = 6

⇒ b = 3.

Hence, equation of AB is x + y = 4 and BC is 2x + y = 6 and P = (0, 4) and Q = (3, 0).

Putting x and y = 0 in y = 2x we get,

Both sides = 0.

So, line y = 2x passes through origin.

A → L₃.

Putting x = 0 in y - 2x + 2 = 0,

⇒ y - 2(0) + 2 = 0

⇒ y = -2.

Putting y = 0 in y - 2x + 2 = 0

⇒ 0 - 2x + 2 = 0

⇒ 2x = 2

⇒ x = 1.

So, x-intercept is positive and y-intercept is negative.

B → L₄.

Putting x = 0 in 3x + 2y = 6,

⇒ 3(0) + 2y = 6

⇒ 2y = 6

⇒ y = 3.

Putting y = 0 in 3x + 2y = 6

⇒ 3x + 2(0) = 6

⇒ 3x = 6

⇒ x = 2.

So, both intercept are positive.

C → L₂.

Comparing y = 2 with y = mx + c we get,

Slope (m) = 0.

So, the line y = 2 is parallel to x-axis.

D → L₁.

Hence, A → L₃, B → L₄, C → L₂ and D → L₁.

Since, the line 5x - ky + 3 = 0 passes through point (x = y = 3).

∴ Point (3, 3) satisfies the equation 5x - ky + 3 = 0.

∴ 5(3) - 3k + 3 = 0

⇒ 15 - 3k + 3 = 0

⇒ -3k + 18 = 0

⇒ 3k = 18

⇒ k = $\frac{18}{3}$ = 6.

Hence, Option 3 is the correct option.

Given, equation :

$$\begin{aligned}\Rightarrow \sqrt{3}x - 3y + 7 = 0 \\[1em] \Rightarrow 3y = \sqrt{3}x + 7 \\[1em] \Rightarrow y = \frac{\sqrt{3}}{3}x + \frac{7}{3} \\[1em] \Rightarrow y = \frac{1}{\sqrt{3}}x + \frac{7}{3}.\end{aligned}$$

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

⇒ m = $\frac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, Option 1 is the correct option.

We know that,

It is true that if slopes of both the lines are same, they are parallel to each other and in order to find whether two lines are parallel or not we find and compare their slopes.

Hence, Option 3 is the correct option.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A. Answer

Given,

1st equation :

y = x + 4

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

m = 1

Slope of 1st line = 1.

2nd equation :

y = $\sqrt{3}x + 5$

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

m = $\sqrt{3}$

Slope of 2nd line = $\sqrt{3}$.

Equating equations (1) and (2), we get :

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

Rationalising,

$$\begin{aligned}\Rightarrow x = \frac{1}{1 - \sqrt{3}} \times \frac{1 + \sqrt{3}}{1 + \sqrt{3}} \\[1em] \Rightarrow x = \frac{1 + \sqrt{3}}{1^2 - (\sqrt{3})^2} \\[1em] \Rightarrow x = \frac{1 + \sqrt{3}}{1 - 3} \\[1em] \Rightarrow x = \frac{1 + \sqrt{3}}{-2} \\[1em] \Rightarrow x = \frac{-(1 + \sqrt{3})}{2}\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y = \frac{-(1 + \sqrt{3})}{2} + 4 \\[1em] \Rightarrow y = \frac{-1 - \sqrt{3} + 8}{2} \\[1em] \Rightarrow y = \frac{7 - \sqrt{3}}{2}.\end{aligned}$$

∴ y = x + 4 and y = $\sqrt{3}x + 5$ are two intersecting lines and also the inclinations of both the given lines are not equal.

Hence, Option 3 is the correct option.

Given,

1st equation :

⇒ 4x + 3y = 84

⇒ 3y = -4x + 84

⇒ y = $-\frac{4}{3}x + \frac{84}{3}$

⇒ y = $-\frac{4}{3}x$ + 28

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

Slope of first equation = $-\frac{4}{3}$

2nd equation :

⇒ 3x + ky + 7 = 0

⇒ ky = -3x - 7

⇒ y = $-\frac{3}{k}x - \frac{7}{k}$

Slope of second equation = $-\frac{3}{k}$

We know that,

Product of slope of two perpendicular lines = -1.

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

Hence, Option 2 is the correct option.

We know that,

x-coordinate at y-axis = 0.

Let point P be (0, a).

Since,

Line 5x + 3y = 15 meets y-axis at point P.

∴ Point P(0, a) satisfies the equation 5x + 3y = 15.

⇒ 5(0) + 3a = 15

⇒ 0 + 3a = 15

⇒ 3a = 15

⇒ a = $\frac{15}{3}$ = 5.

∴ P = (0, a) = (0, 5).

Hence, Option 1 is the correct option.

By formula,

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

Slope of AC = $\frac{-8 - (-1)}{-4 - 3} = \frac{-8 + 1}{-7} = \frac{-7}{-7}$ = 1.

We know that,

Diagonals of rhombus are perpendicular to each other.

Let slope of diagonal of BD be x.

We know that,

Product of slope of two perpendicular lines = -1.

∴ Slope of AC × Slope of BD = -1

⇒ 1 × Slope of BD = -1

⇒ Slope of BD = -1.

Hence, Option 4 is the correct option.

Given,

Line OP bisects angle O.

Since, ∠O = 90°.

∴ OP makes 45° with x-axis.

From figure,

OP lies in the third quadrant.

Since, tan is positive in third quadrant.

Slope = tan 45° = 1.

Hence, Option 1 is the correct option.

We know that,

y-coordinate at x-axis = 0.

Let point A be (a, 0).

Since,

Line 2x - 3y = 12 meets x-axis at point A.

∴ Point A(a, 0) satisfies the equation 2x - 3y = 12.

⇒ 2a - 3(0) = 12

⇒ 2a - 0 = 12

⇒ 2a = 12

⇒ a = $\frac{12}{2}$ = 6.

∴ A = (a, 0) = (6, 0).

We know that,

x-coordinate at y-axis = 0.

Let point B be (0, b).

Since,

Line 2x - 3y = 12 meets y-axis at point B.

∴ Point B(0, b) satisfies the equation 2x - 3y = 12.

⇒ 2(0) - 3b = 12

⇒ 0 - 3b = 12

⇒ -3b = 12

⇒ b = $\frac{12}{-3}$ = -4.

∴ B = (0, b) = (0, -4).

Hence, Option 1 is the correct option.

Given, two lines: 3x - 2y = 15 and 2x + ky + 8 = 0

Converting the equation of first line in slope-intercept form (y = mx + c), we get :

⇒ 3x - 2y = 15

⇒ -2y = -3x + 15

⇒ y = $\frac{3}{2}x - \frac{15}{2}$

∴ Slope of the line (m₁) = $\frac{3}{2}$

Converting the equation of second line in slope-intercept form (y = mx + c), we get :

⇒ 2x + ky + 8 = 0

⇒ ky = -2x - 8

⇒ y = $-\frac{2}{k}x - \frac{8}{k}$

∴ Slope of the line (m₂) = $-\frac{2}{k}$

We know that,

The two lines are perpendicular if product of their slopes is -1.

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

So, assertion (A) is true.

If the inclination of two lines are α and β, then Slope of first line (m₁) = tan α and Slope of second line (m₂) = tan β.

⇒ m₁ x m₂ = -1

⇒ tan α x tan β = -1

⇒ tan α = $-\frac{1}{\text{tan } β}$

⇒ tan α = -cot β

So, reason (R) is true, but it is not the correct reason for assertion (A).

Hence, option 4 is the correct option.

Given, 2x + 8y = 15

Converting in slope-intercept form (y = mx + c), we get :

⇒ 8y = -2x + 15

⇒ y = $-\frac{2}{8}x + \frac{15}{8}$

⇒ y = $-\frac{1}{4}x + \frac{15}{8}$

∴ Slope of the line (m) = $-\frac{1}{4}$

By point-slope formula,

⇒ (y - y₁) = m(x - x₁)

Equation of line parallel to the line 2x + 8y = 15 and passing through origin (0, 0) :

⇒ (y - 0) = $-\frac{1}{4}$ (x - 0)

⇒ y = $-\frac{1}{4}$ x

⇒ 4y = -x

⇒ x + 4y = 0

So, assertion (A) is true.

Given, equation of line : ax + by + c = 0

Converting in slope-intercept form (y = mx + c), we get :

⇒ by = -ax - c

⇒ y = $-\frac{a}{b}x - \frac{c}{b}$

∴ Slope of the line (m) = $-\frac{a}{b}$

By point-slope formula,

⇒ (y - y₁) = m(x - x₁)

Equation of line parallel to ax + by + c = 0 and passing through origin (0, 0) :

⇒ (y - 0) = $-\frac{a}{b}$ (x - 0)

⇒ y = $-\frac{a}{b}$ x

⇒ by = -ax

⇒ ax + by = 0

So, reason (R) is true, and it is the correct reason for assertion (A).

Hence, option 3 is the correct option.

Given, P(x, 2), A(-2, 3) and B(-5, 4).

If three points are collinear, it means they lie on same line.

∴ Slope will be equal.

∴ Slope of PA = Slope of PB = Slope of AB

So, statement 1 is true.

Using formula,

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

Slope of AB = Slope of PA

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

So, statement 2 is true.

Hence, option 1 is the correct option.

Given equation of line 2x - by + 7 = 0 and ax - 2y - 7 = 0

Converting the equation of first line in slope-intercept form (y = mx + c), we get :

⇒ 2x - by + 7 = 0

⇒ by = 2x + 7

⇒ y = $\frac{2}{b}x + \frac{7}{b}$

∴ Slope of the line (m₁) = $\frac{2}{b}$

Converting the equation of second line in slope-intercept form y = mx + c, we get :

⇒ ax - 2y - 7 = 0

⇒ 2y = ax - 7

⇒ y = $\frac{a}{2}x - \frac{7}{2}$

∴ Slope of the line (m₂) = $\frac{a}{2}$

So, statement 2 is true.

We know that,

The two lines are perpendicular if product of their slopes is -1.

$$\begin{aligned}\Rightarrow m_1 \times m_2 = -1\\[1em] \Rightarrow \frac{2}{b} \times \frac{a}{2} = -1\\[1em] \Rightarrow \frac{2}{b} = -1 \times \frac{2}{a}\\[1em] \Rightarrow \frac{2}{b} = -\frac{2}{a}\\[1em]\end{aligned}$$

So, statement 1 is false.

Hence, option 4 is the correct option.

Given points, A (8, 0) and B (16, -8)

By section formula, the co-ordinates of the point P which divides AB in the ratio 3 : 5 is given by

$$\begin{aligned}P = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\frac{3 \times 16 + 5 \times 8}{3 + 5}, \frac{3 \times -8 + 5 \times 0}{3 + 5}\Big) \\[1em] = \Big(\frac{48 + 40}{8}, \frac{-24 + 0}{8}\Big) \\[1em] = \Big(\frac{88}{8}, \frac{-24}{8}\Big) \\[1em] = (11, -3).\end{aligned}$$

Given line equation is,

⇒ 3x + 5y = 7

⇒ 5y = -3x + 7

⇒ y = $-\frac{3}{5}x + \frac{7}{5}$

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

Slope = $-\frac{3}{5}$

The line parallel to the line 3x + 5y = 7 will have the same slope.

Hence, the slope of the required line = $-\frac{3}{5}$

By point-slope form, equation of the required line,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-3) = $-\frac{3}{5}$(x - 11)

⇒ 5(y + 3) = -3(x - 11)

⇒ 5y + 15 = -3x + 33

⇒ 3x + 5y = 33 - 15

⇒ 3x + 5y = 18.

Hence, P = (11, -3) and equation of the required line is 3x + 5y = 18.

Given points, A(3, -4) and B(-2, 1)

By section formula, the co-ordinates of the point P which divides AB in the ratio 1: 3 is given by

$$\begin{aligned}P = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\frac{1 \times -2 + 3 \times 3}{1 + 3}, \frac{1 \times 1 + 3 \times -4}{1 + 3}\Big) \\[1em] = \Big(\frac{-2 + 9}{4}, \frac{1 + -12}{4}\Big) \\[1em] = \Big(\frac{7}{4}, -\frac{11}{4}\Big)\end{aligned}$$

Given line equation is,

5x – 3y = 4

3y = 5x - 4

y = $\frac{5}{3}x - \frac{4}{3}$

So, the slope of this line (m) = $\frac{5}{3}$

Let slope of perpendicular line be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \frac{5}{3} = -1$

⇒ m₁ = $-\frac{3}{5}$.

Slope of the required line = $-\frac{3}{5}$.

By point-slope form,

Equation of line through P and slope = $-\frac{3}{5}$ is,

⇒ y – y₁ = m(x – x₁)

$$\begin{aligned}\Rightarrow y - \Big(-\frac{11}{4}\Big) = -\frac{3}{5}\Big(x - \frac{7}{4}\Big) \\[1em] \Rightarrow \frac{4y + 11}{4} = -\frac{3}{5} \times \frac{4x - 7}{4} \\[1em] \Rightarrow 4y + 11 = -\frac{3}{5} \times (4x - 7) \\[1em] \Rightarrow 5(4y + 11) = -3(4x - 7) \\[1em] \Rightarrow 20y + 55 = -12x + 21 \\[1em] \Rightarrow 12x + 20y + 55 - 21 = 0 \\[1em] \Rightarrow 12x + 20y + 34 = 0 \\[1em] \Rightarrow 2(6x + 10y + 17) = 0 \\[1em] \Rightarrow 6x + 10y + 17 = 0.\end{aligned}$$

Hence, P = $\Big(\frac{7}{4}, -\frac{11}{4}\Big)$ the equation of required line is 6x + 10y + 17 = 0.

As the point P lies on y-axis, the x co-ordinate of point P will be = 0.

Putting x = 0 in the equation 5x + 3y + 15 = 0, we get

⇒ 5(0) + 3y + 15 = 0

⇒ 3y + 15 = 0

⇒ 3y = -15

⇒ y = -5.

Hence, the co-ordinates of the point P are (0, -5).

Given line equation,

x - 3y + 4 = 0

3y = x + 4

y = $\frac{1}{3}x + \frac{4}{3}$

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

Slope of this line (m) = $\frac{1}{3}$

Product of slope of perpendicular lines = -1.

Let slope of line perpendicular to x - 3y + 4 = 0 be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \frac{1}{3} = -1$

⇒ m₁ = -3.

Slope of the required line = -3.

By point-slope form,

Equation of line through P and slope = -3 is,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = -3(x - 0)

⇒ y + 5 = -3x

⇒ 3x + y + 5 = 0.

Hence, P = (0, -5) and equation of required line is 3x + y + 5 = 0.

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

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

By point-slope form,

Equation of the line PQ,

⇒ y – y₁ = m(x – x₁)

⇒ y – 4 = -1[x - (-1)]

⇒ y – 4 = -[x + 1]

⇒ y - 4 = -x - 1

⇒ x + y = -1 + 4

⇒ x + y = 3.

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

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

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

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

Given, A (3, 2), B (6, -2) and C (2, -5)

By distance formula,

Distance between two points = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$$\begin{aligned}\text{AB} = \sqrt{(6 - 3)^2 + (-2 - 2)^2} \\[1em] = \sqrt{(3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \\[1.5em] \text{BC} = \sqrt{(2 - 6)^2 + (-5 - (-2))^2} \\[1em] = \sqrt{(-4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5\end{aligned}$$

Thus, AC = BC

By formula,

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

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

Slope of BC = $\frac{-5 - (-2)}{2 - 6} = \frac{-3}{-4} = \frac{3}{4}$

Slope of AB x Slope of BC = $\frac{-4}{3} \times \frac{3}{4}$ = -1

Hence, AB ⊥ BC

Therefore, A, B, C can be the vertices of a square.

(i) Slope of CD = Slope of AB = $-\frac{4}{3}$ (As they are parallel)

So, the equation of CD is

⇒ y – y₁ = m(x – x₁)

⇒ y - (-5) = $-\frac{4}{3}$(x – 2)

⇒ 3(y + 5) = -4(x - 2)

⇒ 3y + 15 = -4x + 8

⇒ 4x + 3y + 7 = 0 .........(1)

Slope of AD = Slope of BC = $\frac{3}{4}$ (As these lines are parallel)

So, the equation of the line AD is

⇒ y – y₁ = m(x – x₁)

⇒ y - 2 = $\frac{3}{4}$(x - 3)

⇒ 4(y – 2) = 3(x – 3)

⇒ 4y - 8 = 3x - 9

⇒ 3x - 4y = -8 + 9

⇒ 3x - 4y = 1 ......... (2)

Now, D is the point of intersection of CD and AD.

Solving (1) and (2),

Multiplying equation (1) by 4 and (2) by 3 and adding them we get,

⇒ 16x + 12y + 28 + 9x – 12y = 0 + 3

⇒ 25x = 3 - 28

⇒ 25x = -25

⇒ x = -1

Putting value of x in (1), we get

⇒ 4(-1) + 3y + 7 = 0

⇒ 3y - 4 + 7 = 0

⇒ 3y + 3 = 0

⇒ 3y = -3

⇒ y = -1

Hence, the co-ordinates of point D are (-1, -1).

(ii) From the equation (2)

The equation of the line AD is,

⇒ 3x – 4y = 1

⇒ 4y = 3x - 1.

Slope of AC = $\frac{-5 - 2}{2 - 3} = \frac{-7}{-1} = 7.$

Since, diagonals of a square are perpendicular to each other. So, product of their slopes = -1.

∴ Slope of AC × Slope of BD = -1

⇒ $7 \times$ Slope of BD = -1

⇒ Slope of BD = -$\frac{1}{7}$.

The equation of the diagonal BD is

⇒ y – y₁ = m(x – x₁)

⇒ y - (-2) = $-\frac{1}{7}$(x - 6)

⇒ 7(y + 2) = –1(x – 6)

⇒ 7y + 14 = -x + 6

⇒ x + 7y + 14 - 6 = 0

⇒ x + 7y + 8 = 0.

Hence, equation of line BD is x + 7y + 8 = 0 and AD = 4y = 3x - 1.

The given line equation is

⇒ x = 3y + 2 ..........(1)

⇒ 3y = x - 2

⇒ y = $\frac{1}{3}x - \frac{2}{3}$

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

Slope (m) = $\frac{1}{3}$

Let slope of perpendicular line be m₁.

⇒ m₁ × m = -1 (As product of slope of perpendicular lines = -1).

⇒ $m_1 \times \frac{1}{3}$ = -1

⇒ m₁ = -3.

Equation of the line passing through origin and with slope = -3, by point-slope form is:

⇒ y – y₁ = m(x – x₁)

⇒ y – 0 = -3(x – 0)

⇒ y = -3x

⇒ 3x + y = 0 ...........(2)

Next,

Point X is the intersection of the lines (1) and (2).

Substituting value of x from (1) in (2),

⇒ 3(3y + 2) + y = 0

⇒ 9y + 6 + y = 0

⇒ 10y = -6

⇒ y = $-\frac{6}{10} = -\frac{3}{5}$.

⇒ x = 3y + 2 = $3 \times -\frac{3}{5} + 2 = \frac{-9 + 10}{5} = \frac{1}{5}$.

Hence, the co-ordinates of the point X are $\Big(\frac{1}{5}, -\frac{3}{5}\Big)$.

Let P = (3, 2).

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).

Since, P is the mid-point of AB, we have:

$$\begin{aligned}\Rightarrow P = \Big(\frac{x + 0}{2}, \frac{0 + y}{2}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\frac{x}{2}, \frac{y}{2}\Big) \\[1em] \Rightarrow \frac{x}{2} = 3 \text{ and } \frac{y}{2} = 2\\[1em] \Rightarrow x = 6 \text{ and } y = 4.\end{aligned}$$

Thus, A = (6, 0) and B = (0, 4)

Slope of line AB = $\frac{4 - 0}{0 - 6} = \frac{4}{-6} = -\frac{2}{3}$.

So, the required equation of the line AB is given by

⇒ y – y₁ = m(x – x₁)

⇒ y – 0 = $-\frac{2}{3}$(x – 6)

⇒ 3y = -2x + 12

⇒ 2x + 3y = 12

Hence, equation of the required line is 2x + 3y = 12.

Given line equations are,

7x + 6y = 71 .......(1)

and

5x - 8y = -23 .........(2)

Multiplying (1) by 4 and (2) by 3 we get,

⇒ 28x + 24y = 284 ......... (3)

⇒ 15x – 24y = -69 .........(4)

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

⇒ 28x + 15x + 24y - 24y = 284 + (-69)

⇒ 43x = 215

⇒ x = $\frac{215}{43}$

⇒ x = 5.

From (2), we get

⇒ 8y = 5x + 23

⇒ 8y = 5(5) + 23

⇒ 8y = 25 + 23

⇒ 8y = 48

⇒ y = $\frac{48}{8}$

⇒ y = 6.

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

Given, 4x – 2y = 1

⇒ 2y = 4x – 1

⇒ y = 2x – $\frac{1}{2}$.

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

$$\begin{aligned}\text{Slope of AB }(\text{m}_1) = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{1 - 3}{4 - (-2)} \\[1em] = \frac{1 - 3}{4 + 2} \\[1em] = \frac{-2}{6} \\[1em] = -\frac{1}{3}\end{aligned}$$

By point-slope form, the equation of line AB is

⇒ y – y₁ = m(x – x₁)

⇒ y – 3 = $-\frac{1}{3}$[x - (-2)]

⇒ 3(y – 3) = -1(x + 2)

⇒ 3y - 9 = -x - 2

⇒ 3y + x = -2 + 9

⇒ x + 3y = 7 .........(1)

Given,

⇒ 3x = y + 1

⇒ y = 3x - 1

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

⇒ Slope (m₂) = 3

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

Hence, the line through points A and B is perpendicular to the 3x = y + 1.

Given line is 3x = y + 1 ........(2)

Let P be the mid-point of AB,

The co-ordinates of the mid-point of AB (i.e. P) are

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

Now, Let’s check if point P satisfies the line equation (2)

⇒ 3(1) = 2 + 1

⇒ 3 = 3

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.

Given line equation,

⇒ x cos 30° + y sin 30° = 2

⇒ $\frac{\sqrt{3}}{2}x + \frac{y}{2} = 2$

⇒ y + $\sqrt{3}x$ = 4

⇒ y = -$\sqrt{3}x + 4$.

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

Slope (m₁) = -$\sqrt{3}$.

Since, slope of parallel lines are equal,

Slope of a line which is parallel to this given line = -$\sqrt{3}$.

By slope-point form, the equation of the required line is,

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -$\sqrt{3}$(x - 4)

⇒ y - 3 = $-\sqrt{3}x + 4\sqrt{3}$

⇒ $\sqrt{3}x + y = 4\sqrt{3} + 3$.

Hence, the equation of the required line is $\sqrt{3}x + y = 4\sqrt{3} + 3$.

Given line equation,

⇒ 2x - y + 7 = 0 ..........(1)

⇒ y = 2x + 7

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

Slope (m₁) = 2

Given line equation,

⇒ (k - 2)x + (k + 3)y - 5 = 0 ..........(2)

⇒ (k + 3)y = -(k - 2)x + 5

⇒ y = $-\frac{k - 2}{k + 3}x + \frac{5}{k + 3}$

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

Slope (m₂) = $-\frac{k - 2}{k + 3}$

(i) If lines (1) and (2) are perpendicular then product of their slopes = -1.

⇒ m₁ x m₂ = -1

$$\begin{aligned}\Rightarrow 2 \times -\frac{k - 2}{k + 3} = -1 \\[1em] \Rightarrow -2(k - 2) = -1(k + 3) \\[1em] \Rightarrow -2k + 4 = -k - 3 \\[1em] \Rightarrow -k + 2k = 4 + 3 \\[1em] \Rightarrow k = 7.\end{aligned}$$

Hence, k = 7.

(ii) If lines 1 and 2 are parallel then their slopes will be equal.

⇒ $-\frac{k - 2}{k + 3}$ = 2

⇒ -(k - 2) = 2(k + 3)

⇒ -k + 2 = 2k + 6

⇒ 2k + k = 2 - 6

⇒ 3k = -4

⇒ k = $-\frac{4}{3}$.

Hence, k = $-\frac{4}{3}$.

$$\begin{aligned}\text{Slope of BC } = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{7 - (-2)}{11 - (-1)} \\[1em] = \frac{7 + 2}{11 + 1} \\[1em] = \frac{9}{12} \\[1em] = \frac{3}{4}\end{aligned}$$

Then the equation of the line BC is

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = $\frac{3}{4}$[x - (-1)]

⇒ 4(y + 2) = 3(x + 1)

⇒ 4y + 8 = 3x + 3

⇒ 3x - 4y = 8 - 3

⇒ 3x - 4y = 5 ......... (1)

(i) Let slope of line perpendicular to BC = m₁

As product of slope of perpendicular lines is -1

∴ m x m₁ = -1

⇒ $\frac{3}{4}$ x m₁ = -1

⇒ m₁ = $-\frac{4}{3}$

So, the required equation of the line through A (0, 5) and perpendicular to BC is given by

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = $-\frac{4}{3}$(x - 0)

⇒ 3y - 15 = -4x

⇒ 4x + 3y = 15 ......... (2)

Hence, equation of the required line is 4x + 3y = 15.

(ii) The required point P will be the point of intersection of lines (1) and (2).

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

4x + 3y = 15 ......... (2)

Multiplying equation (1) by 3 and (2) by 4 and adding we get,

⇒ 9x - 12y + 16x + 12y = 15 + 60

⇒ 25x = 75

⇒ x = $\frac{75}{25}$

⇒ x = 3

Substituting the value of x in equation (1) we get,

3(3) - 4y = 5

⇒ 4y = 3(3) – 5

⇒ 4y = 9 - 5

⇒ 4y = 4

⇒ y = 1

Hence, the co-ordinates of the required point P is (3, 1).

(i) From figure,

The co-ordinates of A = (2, 3), B = (-1, 2), C = (3, 0).

(ii) Slope of BC = $\frac{0 - 2}{3 - (-1)} = \frac{-2}{4} = -\frac{1}{2}$.

Slope of the line which is parallel to BC = $-\frac{1}{2}$ (As parallel lines have equal slope).

Hence, the required equation of the line through A and parallel to BC is given by

⇒ y – y₁ = m(x – x₁)

⇒ y – 3 = $-\frac{1}{2}$(x – 2)

⇒ 2(y – 3) = -1(x - 2)

⇒ 2y - 6 = -x + 2

⇒ x + 2y = 2 + 6

⇒ x + 2y = 8.

Hence, equation of line through A and parallel to BC is x + 2y = 8.

ΔABC with vertices A (8, -6), B (-4, 2) and C (0, -10) is shown below:

By Mid-point formula,

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

Co-ordinates of P = $\Big[\frac{8 + (-4)}{2}, \frac{-6 + 2}{2}\Big] = \Big(\frac{4}{2}, \frac{-4}{2}\Big) = (2, -2).$

Co-ordinates of Q = $\Big[\frac{8 + 0}{2}, \frac{-6 + (-10)}{2}\Big] = \Big(\frac{8}{2}, \frac{-16}{2}\Big)$ = (4, -8).

By formula,

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

Substituting values we get,

Slope of PQ = $\frac{-8 - (-2)}{4 - 2} = \frac{-6}{2}$ = -3.

Slope of BC = $\frac{-10 - 2}{0 - (-4)} = \frac{-12}{4}$ = -3.

Since, slope of PQ = slope of BC.

∴ PQ || BC.

From figure,

PBCQ is a trapezium.

(i) Let’s assume the co-ordinates of point A, lying on x-axis be (x, 0) and the co-ordinates of point B (lying on y-axis) be (0, y).

Given,

P = (-4, -2) and AP : PB = 1 : 2

By section formula, we get

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

Hence, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) By formula,

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

Slope of AB = $\frac{-6 - 0}{0 - (-6)} = \frac{-6}{6}$ = -1.

Let slope of perpendicular line be m.

⇒ m × -1 = -1

⇒ -m = -1

⇒ m = 1.

Therefore, the required equation of the line passing through P and perpendicular to AB is given by

⇒ y – y₁ = m(x – x₁)

⇒ y - (-2) = 1[x - (-4)]

⇒ y + 2 = x + 4

⇒ y = x + 2.

Hence, the equation of line through P and perpendicular to AB is y = x + 2.

Let line intersect x-axis at point A.

So, A = (-2, 0)

Let line cuts off an intercept of 3 units from positive side of y-axis at point B.

So, co-ordinates of B are (0, 3).

By formula,

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

Slope of AB = $\frac{3 - 0}{0 - (-2)} = \frac{3}{2}$.

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\frac{3}{2}$[x - (-2)]

⇒ 2y = 3(x + 2)

⇒ 2y = 3x + 6

⇒ 3x - 2y + 6 = 0

⇒ 2y = 3x + 6.

Hence, equation of AB is 2y = 3x + 6.

Let the line intersect x-axis at point A.

So, A = (4, 0).

Let B = (2, 3).

By formula,

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

Slope of AB = $\frac{3 - 0}{2 - 4} = -\frac{3}{2}$.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $-\frac{3}{2}$(x - 4)

⇒ 2y = -3(x - 4)

⇒ 2y = -3x + 12

⇒ 3x + 2y = 12.

Hence, the equation of required line is 3x + 2y = 12.

Given,

Equation of AB :

⇒ y = x + 1

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

Slope of AB = 1.

Let θ₁ be the inclination of AB then,

⇒ tan θ₁ = 1

⇒ tan θ₁ = tan 45°

⇒ θ₁ = 45°.

Equation of CD :

⇒ y = $\sqrt{3}$x - 1

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

Slope of CD = $\sqrt{3}$.

Let θ₂ be the inclination of AB then,

⇒ tan θ₂ = $\sqrt{3}$

⇒ tan θ₂ = tan 60°

⇒ θ₂ = 60°.

From figure,

∠FGE = 180° - θ₂ = 180° - 60° = 120°.

In △EFG,

∠E + ∠F + ∠G = 180

⇒ θ₁ + θ + 120° = 180°

⇒ 45° + θ + 120° = 180°

⇒ θ + 165° = 180°

⇒ θ = 180° - 165° = 15°.

Hence, inclination of AB is 45°, inclination of CD is 60° and θ = 15°.

By section-formula,

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

Substituting values we get,

$$\begin{aligned}P = \Big(\frac{2 \times 3 + 3 \times -2}{2 + 3}, \frac{2 \times -4 + 3 \times 6}{2 + 3}\Big) \\[1em] = \Big(\frac{6 + (-6)}{5}, \frac{-8 + 18}{5}\Big) \\[1em] = \Big(\frac{0}{5}, \frac{10}{5}\Big) \\[1em] = (0, 2).\end{aligned}$$

By point-slope form,

Equation of line with slope = $\frac{3}{2}$ and passing through (0, 2) is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 2 = $\frac{3}{2}$(x - 0)

⇒ 2(y - 2) = 3x

⇒ 2y - 4 = 3x

⇒ 2y = 3x + 4.

Hence, equation of required line is 2y = 3x + 4.

Let point A = (6, 4) and B = (7, -5).

Let the point P be (a, -23).

By formula,

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

Slope of AB = $\frac{-5 - 4}{7 - 6} = \frac{-9}{1}$ = -9.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 4 = -9(x - 6)

⇒ y - 4 = -9x + 54

⇒ 9x + y = 54 + 4

⇒ 9x + y = 58.

Since, P lies on AB so it will satisfy the equation.

Substituting values of P in equation we get,

⇒ 9(a) + (-23) = 58

⇒ 9a - 23 = 58

⇒ 9a = 58 + 23

⇒ 9a = 81

⇒ a = $\frac{81}{9}$

⇒ a = 9.

∴ P = (a, -23) = (9, -23).

Hence, co-ordinates of the required point are (9, -23).

(i) By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of AB } = \frac{9 - (-3)}{1 - 7} \\[1em] = \frac{12}{-6} \\[1em] = -2.\end{aligned}$$

Hence, slope of AB = -2.

(ii) Let O be the mid-point of AB.

By mid-point formula,

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

O = $\Big(\frac{7 + 1}{2}, \frac{-3 + 9}{2}\Big) = \Big(\frac{8}{2}, \frac{6}{2}\Big)$ = (4, 3).

⇒ Slope of AB × Slope of perpendicular bisector = -1 (As product of slopes of perpendicular lines = -1)

⇒ -2 × Slope of perpendicular bisector = -1

⇒ Slope of perpendicular bisector = $\frac{1}{2}$.

By point-slope form,

Equation of perpendicular bisector is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $\frac{1}{2}$(x - 4)

⇒ 2(y - 3) = 1(x - 4)

⇒ 2y - 6 = x - 4

⇒ x - 2y - 4 + 6 = 0

⇒ x - 2y + 2 = 0.

Hence, the equation of perpendicular bisector of AB is x - 2y + 2 = 0.

(iii) Since, (-2, p) lies on perpendicular bisector. So, it will satisfy the equation.

⇒ -2 - 2p + 2 = 0

⇒ -2p = 0

⇒ p = 0.

Hence, p = 0.

(i) Let point A be (x, 0) and B be (0, y). Since, P is the mid-point of AB.

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

∴ A = (x, 0) = (4, 0) and B = (0, y) = (0, -6).

Hence, co-ordinates of A = (4, 0) and B = (0, -6).

(ii) By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of AB }= \frac{-6 - 0}{0 - 4} \\[1em] = \frac{-6}{-4} \\[1em] = \frac{3}{2} \\[1em] = 1\frac{1}{2}\end{aligned}$$

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

(iii) By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\frac{3}{2}$(x - 4)

⇒ 2y = 3(x - 4)

⇒ 2y = 3x - 12

⇒ 3x - 2y = 12

⇒ 2y = 3x - 12

Hence, equation of AB is 2y = 3x - 12.

(i) Since, ABCD is a parallelogram.

So, AB || CD and AD || BC.

Since, AB || CD so slope of AB = slope of CD.

$$\begin{aligned}\therefore \frac{8 - y}{5 - x} = \frac{7 - (-4)}{4 - 2} \\[1em] \Rightarrow \frac{8 - y}{5 - x} = \frac{11}{2} \\[1em] \Rightarrow 2(8 - y) = 11(5 - x) \\[1em] \Rightarrow 16 - 2y = 55 - 11x \\[1em] \Rightarrow 11x - 2y = 55 - 16 \\[1em] \Rightarrow 11x - 2y = 39 ..........(1)\end{aligned}$$

Since, AD || BC so slope of AD = slope of BC

$$\begin{aligned}\therefore \frac{-4 - y}{2 - x} = \frac{7 - 8}{4 - 5} \\[1em] \Rightarrow \frac{-4 - y}{2 - x} = \frac{-1}{-1} \\[1em] \Rightarrow \frac{-4 - y}{2 - x} = 1 \\[1em] \Rightarrow -4 - y = 2 - x \\[1em] \Rightarrow x - y - 4 - 2 = 0 \\[1em] \Rightarrow x - y - 6 = 0 \\[1em] \Rightarrow x = y + 6 .........(2)\end{aligned}$$

Substituting value of x from (2) in (1) we get,

⇒ 11(y + 6) - 2y = 39

⇒ 11y + 66 - 2y = 39

⇒ 9y = 39 - 66

⇒ 9y = -27

⇒ y = $\frac{-27}{9}$

⇒ y = -3.

Substituting value of y in (2) we get,

⇒ x = -3 + 6 = 3.

∴ A = (3, -3).

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

(ii) By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of BD }= \frac{-4 - 8}{2 - 5} \\[1em] = \frac{-12}{-3} = 4.\end{aligned}$$

By-point slope form,

Equation of BD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 8 = 4(x - 5)

⇒ y - 8 = 4x - 20

⇒ 4x - y = -8 + 20

⇒ 4x - y = 12.

Hence, equation of BD is 4x - y = 12.

(i) The angle between x and y axis = 90°.

Given,

L₂ is the bisector of angle O. So, inclination of line L₂ is 45°.

Slope of L₂ = tan 45° = 1.

Hence, slope of L₂ = 1.

(ii) Equation of line L₂, passing through origin (0, 0) and slope = 1 is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = 1(x - 0)

⇒ y = x.

From figure,

P is the point of intersection of y = 4 and y = x.

Solving,

y = 4 ........(1)

y = x ........(2)

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

x = 4.

So, point of intersection = (4, 4).

Hence, P = (4, 4).

(iii) From part (ii) we get equation of L₂ as,

y = x

Hence, equation of L₂ is y = x.

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,

Slope (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.

(i) Slope of PQ = $\frac{-2 - 4}{5 - (-1)} = -\frac{6}{6}$ = -1.

By point-slope form,

Equation of PQ is :

⇒ 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.

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

(ii) Given,

PQ intersects x-axis at A and y-axis at B.

At x-axis, y co-ordinate = 0.

Substituting y = 0 in equation we get,

⇒ x + 0 = 3

⇒ x = 3.

∴ A = (3, 0)

At y-axis, x co-ordinate = 0.

Substituting x = 0 in equation we get,

⇒ 0 + y = 3

⇒ y = 3.

∴ B = (0, 3)

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

(iii) By formula,

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

Substituting values we get,

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

Hence, co-ordinates of M = $\Big(\frac{3}{2}, \frac{3}{2}\Big)$.

(i) By formula,

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

Substituting values we get,

$$\begin{aligned}\text{Slope of AB} = \frac{8 - 6}{-6 - 0} \\[1em] = \frac{2}{-6} \\[1em] = -\frac{1}{3}.\end{aligned}$$

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\frac{1}{3}$(x - 0)

⇒ 3(y - 6) = -1(x)

⇒ 3y - 18 = -x

⇒ x + 3y = 18.

Hence, equation of line AB is x + 3y = 18.

(ii) From figure,

CD is perpendicular to AB.

Slope of AB (m₁) = $-\frac{1}{3}$

Let slope of CD be m₂.

⇒ m₁ × m₂ = -1

⇒ $-\frac{1}{3} \times m_2 = -1$

⇒ m₂ = 3.

Equation of CD is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 10 = 3(x - 2)

⇒ y - 10 = 3x - 6

⇒ 3x - y - 6 + 10 = 0

⇒ 3x - y + 4 = 0.

Hence, equation of CD is 3x - y + 4 = 0.

(iii) From figure,

E lies on x-axis. Let co-ordinates of E be (a, 0).

Since, E lies on line AB it will satisfy its equation.

Substituting value of E in AB we get,

⇒ a + 3(0) = 18

⇒ a = 18.

E = (18, 0)

D lies on x-axis. Let co-ordinates of D be (b, 0).

Since, D lies on CD it will satisfy its equation.

Substituting value of D in CD we get,

3b - 0 + 4 = 0

3b = -4

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

D = $(-\frac{4}{3}, 0)$.

Hence, co-ordinates of E = (18, 0) and D = $(-\frac{4}{3}, 0)$.

By formula,

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

Substituting values we get,

Slope of AB = $\frac{2 - 3}{0 - (-1)} = \frac{-1}{1}$ = -1.

By point-slope form,

Equation of AB is :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = -1[x - (-1)]

⇒ y - 3 = -1(x + 1)

⇒ y - 3 = -x - 1

⇒ x + y = -1 + 3

⇒ x + y = 2.

Slope of BC = $\frac{1 - 2}{1 - 0} = \frac{-1}{1}$ = -1.

Since, slope of AB = slope of BC.

Hence, proved that A, B and C are collinear and equation of AB is x + y = 2.

(i) From graph,

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

(ii) By distance formula,

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

Substituting values we get,

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

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

(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,

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

k : 1 = 2 : 1.

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

(iv) Slope of AC = $\frac{-4 - 0}{-2 - 4} = \frac{-4}{-6} = \frac{2}{3}$.

By point-slope form,

Equation of AB 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.

(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,

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

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

(ii) By formula,

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

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

Let slope of perpendicular line be m₁.

Since, slope of product of perpendicular lines = -1.

⇒ m₁ × Slope of AB = -1

⇒ m₁ $\times \frac{1}{2} = -1$

⇒ m₁ = -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 = -2(x - 4)

⇒ y + 1 = -2x + 8

⇒ 2x + y = 7.

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

Let the point of intersection of the line 2x + 5y - 4 = 0 and the x-axis be (x₁, 0).

Substituting the value of points in equation,

⇒ 2x₁ + 5 × 0 - 4 = 0

⇒ 2x₁ = 4

⇒ x₁ = 2.

Coordinates of the point of intersection will be (2, 0).

Given new line is parallel to 3x - 7y + 8 = 0.

Converting it in the form y = mx + c,

3x - 7y + 8 = 0

⇒ 7y = 3x + 8 ​ ⇒ y = $\frac{3}{7}x + \frac{8}{7}$

Comparing equation with y = mx + c, we get slope = $\frac{3}{7}$.

The equation of the line with slope $\frac{3}{7}$ and passing through (2, 0) can be given by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 0 = $\frac{3}{7}(x - 2)$

⇒ 7y = 3(x - 2)

⇒ 7y = 3x - 6

⇒ 3x - 7y - 6 = 0.

Hence, the equation of the new line is 3x - 7y - 6 = 0.

(i) Given,

The line 3x - 4y - 12 = 0 meets x-axis at point A and y-axis at point B.

Substituting x = 0, we get :

⇒ 3.0 - 4y - 12 = 0

⇒ 4y = -12

⇒ y = -3.

B = (0, -3)

Substituting y = 0, we get :

⇒ 3x - 4.0 - 12 = 0

⇒ 3x = 12

⇒ x = $\frac{12}{3}$ = 4.

A = (4, 0)

By section formula,

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

Substituting values we get :

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

Hence, coordinates of P = $\Big(\frac{4}{3}, -2\Big)$.

(ii) Slope of AB = $\frac{-3 - 0}{0 - 4} = \frac{-3}{-4} = \frac{3}{4}$.

We know that,

Product of slope of perpendicular lines = -1. Let slope of line perpendicular to AB be m.

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

By point slope form :

Equation of line :

y - y₁ = m(x - x₁)

Equation of line passing through P and perpendicular to AB is :

$$\begin{aligned}\Rightarrow y - (-2) = -\frac{4}{3}\Big(x - \frac{4}{3}\Big) \\[1em] \Rightarrow y + 2 = -\frac{4}{3}\Big(x - \frac{4}{3}\Big) \\[1em] \Rightarrow 3(y + 2) = -4\Big(x - \frac{4}{3}\Big) \\[1em] \Rightarrow 3y + 6 = -4x + \frac{16}{3} \\[1em] \Rightarrow 3(3y + 6) = -12x + 16 \\[1em] \Rightarrow 9y + 18 = -12x + 16 \\[1em] \Rightarrow 12x + 9y + 2 = 0.\end{aligned}$$

Hence, equation of line passing through P and perpendicular to AB is 12x + 9y + 2 = 0.

Given,

⇒ y + (2p + 1)x + 3 = 0

⇒ y = -(2p + 1)x - 3

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

⇒ Slope (m₁) = -(2p + 1)

Given,

⇒ 8y - (2p - 1)x = 5

⇒ 8y = (2p - 1)x + 5

⇒ y = $\frac{2p - 1}{8}.x + \frac{5}{8}$

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

⇒ Slope (m₂) = $\frac{2p - 1}{8}$

We know that,

Product of slopes of perpendicular lines = -1.

$$\begin{aligned}\Rightarrow -(2p + 1) \times \frac{2p - 1}{8} = -1 \\[1em] \Rightarrow \frac{(2p)^2 - 1^2}{8} = 1 \\[1em] \Rightarrow 4p^2 - 1 = 8 \\[1em] \Rightarrow 4p^2 = 9 \\[1em] \Rightarrow p^2 = \frac{9}{4} \\[1em] \Rightarrow p = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow p = \pm \frac{3}{2}.\end{aligned}$$

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