Equation of a Straight Line

(i) Let m be the slope of the line, then

m = tan 45° = 1.

Hence, slope of the line is 1.

(ii) Let m be the slope of the line, then

m = tan 30° = $\frac{1}{\sqrt{3}}$.

Hence, slope of the line is $\frac{1}{\sqrt{3}}$.

(i) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ 1 = tan θ
⇒ 1 = tan 45°
∴ tan θ = tan 45°
∴ θ = 45°.

Hence, the inclination is 45°.

(ii) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ $\sqrt{3}$ = tan θ
⇒ $\sqrt{3}$ = tan 60°
∴ tan θ = tan 60°
∴ θ = 60°

Hence, the inclination is 60°.

(iii) Let inclination be θ.

We know that,

m = slope or gradient = tan θ

⇒ $\frac{1}{\sqrt{3}}$ = tan θ

⇒ $\frac{1}{\sqrt{3}}$ = tan 30°

∴ tan θ = tan 30°

∴ θ = 30°.

Hence, the inclination is 30°.

(i) We know that the equation of line parallel to x-axis is y = b, where b is the value of the ordinate.

Since the line is 2 units above so, value of ordinate = b = 2.

∴ Equation of the line ⇒ y = 2 or y - 2 = 0.

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

(ii) We know that the equation of line parallel to x-axis is y = b, where b is the value of the ordinate.

Since the line is 3 units below so, value of ordinate = b = -3.

∴ Equation of the line ⇒ y = -3 or y + 3 = 0.

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

(i) We know that the equation of line parallel to y-axis is x = a, where a is the value of the abscissa.

Since the line is 3 units to the right so, value of abscissa = a = 3.

∴ Equation of the line ⇒ x = 3 or x - 3 = 0.

Hence, the equation of straight line is x - 3 = 0.

(ii) We know that the equation of line parallel to y-axis is x = a, where a is the value of the abscissa.

Since the line is 2 units to the left so, value abscissa = a = -2.

∴ Equation of the line ⇒ x = -2 or x + 2 = 0.

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

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

x = a

Since the line passes through the point (-3, 5), we get

a = -3

∴ Equation of the line ⇒ x = -3 or x + 3 = 0.

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

(i) The equation of the straight line is given by,

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

Given slope = 3 and y-intercept = -5. Putting values in equation we get,

y = 3x - 5.

Hence, the equation of the straight line is y = 3x - 5.

(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 = $-\frac{2}{7}$ and y-intercept = 3. Putting values in equation we get,

$$\begin{aligned}\Rightarrow y = -\frac{2}{7}x + 3 \\[1em] \Rightarrow y = \frac{-2x + 21}{7} \\[1em] \Rightarrow 7y = -2x + 21 \\[1em] \Rightarrow 2x + 7y - 21 = 0.\end{aligned}$$

Hence, the equation of straight line is 2x + 7y - 21 = 0.

(iii) 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 = $-\frac{4}{3}$. Putting values in equation we get,

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

Hence, the equation of straight line is $3\sqrt{3}x - 3y - 4 = 0.$

(iv) The equation of straight line is given by,

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

Given inclination = θ = 30° and y-intercept = 2.
Slope = m = tan θ = tan 30° = $\frac{1}{\sqrt{3}}$

Putting values in equation we get,

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

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

(i) The equation of line is

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

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

m = $\frac{1}{2}$ and c = $-\frac{1}{2}$.

Hence, the slope of the line = $\frac{1}{2}$ and y-intercept = $-\frac{1}{2}$.

(ii) The equation of line is

⇒ 4x - 5y - 9 = 0
⇒ 5y = 4x - 9
⇒ y = $\frac{4x}{5} - \frac{9}{5}.$

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

m = $\frac{4}{5}$ and c = $-\frac{9}{5}$.

Hence, the slope of the line = $\frac{4}{5}$ and y-intercept = $-\frac{9}{5}$.

(iii) The equation of line is

⇒ 3x + 5y + 7 = 0
⇒ 5y = -3x - 7
⇒ y = $-\frac{3x}{5} - \frac{7}{5}.$

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

m = $-\frac{3}{5}$ and c = $-\frac{7}{5}$.

Hence, the slope of the line = $-\frac{3}{5}$ and y-intercept = $-\frac{7}{5}$.

(iv) The equation of line is

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

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

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

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

(v) The equation of line is

⇒ y - 3 = 0
⇒ y = 3
⇒ y = 0.x + 3

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

m = 0 and c = 3.

Hence, the slope of the line = 0 and y-intercept = 3.

(vi) The equation of line is

⇒ x - 3 = 0
⇒ x = 3.

Here, the slope cannot be defined as the line does not meet y-axis.

Hence, the slope of the line is undefined and there is no y-intercept as line does not meet y-axis.

(i) The equation of line is

⇒ 3y - 3x + 7 = 0

⇒ 3y = 3x - 7

⇒ y = $\frac{3\text{x}}{3} - \frac{7}{3}$

⇒ y = x - $\frac{7}{3}$.

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

m = 1.

Hence, the slope of the line PQ is 1.

(ii) We know that m = tan θ

⇒ tan θ = 1
⇒ tan 45° = 1 = tan θ
⇒ θ = 45°

The angle that the line makes with x-axis is 45°.

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

Comparing equations with y = mx + c we get,

m₁ = 1 and m₂ = $\sqrt{3}$.

Let the first line make angle θ₁ and second make θ₂ with positive direction of x-axis.

The inclination that y = x + 1 makes is,
⇒ m₁ = tan θ₁ = 1
⇒ tan θ₁ = 1 = tan 45°
⇒ tan θ₁ = tan 45°
⇒ θ₁ = 45°.

The inclination that y = $\sqrt{3}x$ - 1 makes is,
⇒ m₂ = tan θ₂ = $\sqrt{3}$
⇒ tan θ₂ = $\sqrt{3}$ = tan 60°
⇒ tan θ₂ = tan 60°
⇒ θ₂ = 60°.

From graph we get,

60° is the exterior angle. We know that,

Exterior angle = Sum of two opposite interior angles.

∴ 60° = θ + 45°
⇒ θ = 60° - 45°
⇒ θ = 15°

Hence, y = x + 1 makes 45° and y = $\sqrt{3}x - 1$ makes 60° with the x-axis. The value of θ = 15°.

Since, $\frac{y}{2} = x - p$ passes through (-4, 4) hence, the points must satisfy the equation.

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

Hence, the value of p = -6.

Since, (a, 2a) lies on $\frac{y}{2} = 3x - 6$ hence, the points must satisfy the equation.

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

Hence, the value of a = 3.

Since, (1, 4) and (-2, -5) lie on y = mx + c hence, the points must satisfy the equation.

Putting (1, 4) in the equation,

⇒ 4 = m(1) + c
⇒ 4 = m + c
⇒ m = 4 - c      (Eq 1)

Putting (-2, -5) in the equation,

⇒ -5 = m(-2) + c
⇒ -5 = -2m + c.

Putting value of m from Eq 1 in above equation,

⇒ -5 = -2(4 - c) + c
⇒ -5 = -8 + 2c + c
⇒ -5 + 8 = 3c
⇒ 3 = 3c
⇒ c = 1.

Putting value of c in Eq 1,

⇒ m = 4 - 1
⇒ m = 3.

Hence, the value of m = 3 and c = 1.

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

Since, the line passes through the point (2, -5) hence, the point must satisfy the equation,

⇒ -5 = 2m - 3
⇒ -5 + 3 = 2m
⇒ 2m = -2
⇒ m = -1.

Putting value of m and c in y = mx + c,

⇒ y = -x - 3
⇒ x + y + 3 = 0.

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

The equation of line with slope m and passing through point (x₁, y₁) is given by

y - y₁ = m(x - x₁)

Putting values we get,

$$\begin{aligned}\Rightarrow y - 2 = \frac{2}{5}(x - (-1)) \\[1em] \Rightarrow 5(y - 2) = 2(x + 1) \\[1em] \Rightarrow 5y - 10 = 2x + 2 \\[1em] \Rightarrow 2x - 5y + 2 + 10 = 0 \\[1em] \Rightarrow 2x - 5y + 12 = 0.\end{aligned}$$

Hence, the equation of the line is 2x - 5y + 12 = 0.

Given inclination = θ = 60°.

m = tan θ = tan 60° = $\sqrt{3}$.

Let the equation of line be y = mx + c.

Since, the line passes through point (0, -3) hence, it must satisfy the equation. Putting point and m in the equation,

⇒ -3 = $\sqrt{3} \times 0$ + c
⇒ c = -3.

So, the equation of line whose slope = $\sqrt{3}$ and y-intercept = -3 is,

y = $\sqrt{3}$x - 3 or $\sqrt{3}x - y - 3 = 0.$

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

(i) Gradient of a line = $\frac{y_2 - y_1}{x_2 - x_1}$

Putting values in above formula we get,

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

Hence, the gradient of the line passing through (0, -2), (3, 4) is 2.

(ii) Gradient of a line = $\frac{y_2 - y_1}{x_2 - x_1}$

Putting values in above formula we get,

$$\begin{aligned}\text{Gradient } = \frac{8 - (-7)}{-1 - 3} \\[1em] = \frac{15}{-4} \\[1em] = -\frac{15}{4}.\end{aligned}$$

Hence, the gradient of the line passing through (3, -7), (-1, 8) is $-\frac{15}{4}$.

(i) Gradient of a line = $\frac{y_2 - y_1}{x_2 - x_1}$

Putting values in above formula we get,

$$\begin{aligned}\text{Gradient } = \frac{7 - 4}{3 - 0} \\[1em] = \frac{3}{3} \\[1em] = 1.\end{aligned}$$

Hence, the gradient of EF is 1.

(ii) Equation of EF can be given by,

y - y₁ = m(x - x₁)

Putting values in above equation we get,

⇒ y - 4 = 1(x - 0)
⇒ y - 4 = x
⇒ x - y + 4 = 0.

Hence, the equation of EF is x - y + 4 = 0.

(iii) The coordinates where EF intersects x-axis will be where y = 0.

Substituting y = 0 in x - y + 4 = 0 ,

⇒ x - 0 + 4 = 0
⇒ x = -4.

Hence, coordinates where EF intersects x-axis are (-4, 0).

Given the equation of line, putting y = 0 we will get intercept made on x-axis

⇒ 2x - 3(0) + 12 = 0
⇒ 2x = -12
⇒ x = -6

In order to find y-intercept, putting x = 0

⇒ 2(0) - 3y + 12 = 0
⇒ 3y = 12
⇒ y = 4.

Hence, the x-intercept is -6 and y-intercept is 4.

The two given points are P(5, 1), Q(1, -1).

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

$$\begin{aligned}= \frac{-1 - 1}{1 - 5} \\[1em] = \frac{-2}{-4} \\[1em] = \frac{1}{2}.\end{aligned}$$

So, the equation of PQ is

⇒ y - y₁ = m(x - x₁)
⇒ y - 1 = $\frac{1}{2}(x - 5)$
⇒ 2(y - 1) = x - 5
⇒ 2y - 2 = x - 5
⇒ x - 2y - 5 + 2 = 0
⇒ x - 2y - 3 = 0.

Now if point R(11, 4) is collinear to points P and Q then it will satisfy the equation x - 2y - 3 = 0,

Putting values in L.H.S of the equation

$$\begin{aligned}= 11 - 2(4) - 3 \\[1em] = 11 - 8 - 3 \\[1em] = 0.\end{aligned}$$

The equation of the line PQ is x - 2y - 3 = 0. Since, L.H.S. = 0 = R.H.S, thus R satisfies the equation. Hence, points P, Q and R are collinear.

Given that ,

A(a, 3), B(2, 1) and C(5, a) are collinear. Hence,

Slope of AB = Slope of BC

$$\begin{aligned}\therefore \frac{1 - 3}{2 - a} = \frac{a - 1}{5 - 2} \qquad \text{....[Eq 1]} \\[1em] \Rightarrow \frac{-2}{2 - a} = \frac{a - 1}{3} \\[1em] \Rightarrow -2 \times 3 = (a - 1) \times (2 - a) \\[1em] \Rightarrow -6 = 2a - a^2 - 2 + a \\[1em] \Rightarrow -6 = 3a - a^2 - 2 \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a + 1 = 0 \text{ or } a - 4 = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.\end{aligned}$$

Let us take points A and B for the equation, by two point form the equation of the line will be,

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

Putting values of points in above formula we get,

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

Putting a = -1 in above equation,

$$\begin{aligned}\Rightarrow y - 3 = \frac{-2}{2 - (-1)}(x - (-1)) \\[1em] \Rightarrow y - 3 = \frac{-2}{3}(x + 1) \\[1em] \Rightarrow 3(y - 3) = -2(x + 1) \\[1em] \Rightarrow 3y - 9 = -2x - 2 \\[1em] \Rightarrow 3y + 2x - 9 + 2 = 0 \\[1em] \Rightarrow 3y + 2x - 7 = 0.\end{aligned}$$

Putting a = 4 in above equation,

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

Hence, the equation of the line is 2x + 3y - 7 = 0 when a = -1 and the equation of the line is x - y - 1 = 0 when a = 4.

Points (h, 4) and ($\frac{1}{2}$, k) lie on the line passing through A(-1, -1) and B(2, 5). The graph is shown below:

From graph we get,

h = $\frac{3}{2}$ and k = 2.

Hence, the value of h = $\frac{3}{2}$ and k = 2.

(i) Let O be the point of intersection of the diagonals.

So, O will be the mid-point of the diagonals so also the mid-point of BD.

By mid-point formula coordinates of O are,

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

Since, O is also the mid-point of AC so,

$$\begin{aligned}\Rightarrow \frac{7}{2} = \frac{x + 4}{2} \text{ and } 2 = \frac{y + 7}{2} \\[1em] \Rightarrow 7 = x + 4 \text{ and } 4 = y + 7 \\[1em] \Rightarrow x = 7 - 4 \text{ and } y = 4 - 7 \\[1em] \Rightarrow x = 3 \text{ and } y = -3.\end{aligned}$$

Hence, the coordinates of A are (3, -3).

(ii) Equation of diagonal BD can be given by two point formula,

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

Putting values in above equation,

$$\begin{aligned}\Rightarrow y - 8 = \frac{-4 - 8}{2 - 5}(x - 5) \\[1em] \Rightarrow y - 8 = \frac{-12}{-3}(x - 5) \\[1em] \Rightarrow y - 8 = 4(x - 5) \\[1em] \Rightarrow y - 8 = 4x - 20 \\[1em] \Rightarrow 4x - y - 20 + 8 = 0 \\[1em] \Rightarrow 4x - y - 12 = 0.\end{aligned}$$

Hence, the equation of the diagonal BD is 4x - y - 12 = 0.

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

Coordinates of D by mid-point formula are,

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

Equation of median AD can be given by two point formula i.e.,

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

Putting values we get,

$$\begin{aligned}\Rightarrow y - 5 = \frac{-1 - 5}{4 - 3}(x - 3) \\[1em] \Rightarrow y - 5 = -6(x - 3) \\[1em] \Rightarrow y - 5 = -6(x - 3) \\[1em] \Rightarrow y - 5 = -6x + 18 \\[1em] \Rightarrow 6x + y - 5 - 18 = 0 \\[1em] \Rightarrow 6x + y - 23 = 0.\end{aligned}$$

Hence, the equation of median through A is 6x + y - 23 = 0.

Since, x-intercept = 4, it means line will intersect x-axis at (4, 0).

Since, line passes through (-2, 3) and (4, 0) so by two-point formula, equation is,

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

Putting values in above equation we get,

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

Hence, the equation of the line is x + 2y - 4 = 0.

Since, x-intercept = 6 and y-intercept = -4, it means line will intersect x-axis at (6, 0) and y-axis at (0, -4).

The equation of line passing through two points is given by two point formula i.e.,

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

Putting values in above equation we get,

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

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

The triangle ABC is shown in the figure below:

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

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

Putting values we get,

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

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

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

Putting values in above equation we get,

$$\begin{aligned}\Rightarrow y - 8 = \frac{4 - 8}{1 - 5}(x - 5) \\[1em] \Rightarrow y - 8 = \frac{-4}{-4}(x - 5) \\[1em] \Rightarrow y - 8 = 1 \times (x - 5) \\[1em] \Rightarrow y - 8 = x - 5 \\[1em] \Rightarrow x - y - 5 + 8 = 0 \\[1em] \Rightarrow x - y + 3 = 0.\end{aligned}$$

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

Since line x - 2y - 11 = 0 intersects y-axis, the point where it will intersect there x-coordinate = 0.

So, putting x = 0 in equation,

⇒ 0 - 2y - 11 = 0
⇒ -2y = 11
⇒ y = -$\frac{11}{2}$

Coordinates = $\Big(0, -\frac{11}{2}\Big)$

So, the line passes through (1, 4) and $\Big(0, -\frac{11}{2}\Big)$.

The equation of the line joining two points is given by,

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

Putting values we get,

$$\begin{aligned}\Rightarrow y - 4 = \frac{-\frac{11}{2} - 4}{0 - 1}(x - 1) \\[1em] \Rightarrow y - 4 = \frac{\frac{-11 - 8}{2}}{-1}(x - 1) \\[1em] \Rightarrow y - 4 = \frac{19}{2}(x - 1) \\[1em] \Rightarrow 2(y - 4) = 19(x - 1) \\[1em] \Rightarrow 2y - 8 = 19x - 19 \\[1em] \Rightarrow 19x - 2y - 19 + 8 = 0 \\[1em] \Rightarrow 19x - 2y - 11 = 0.\end{aligned}$$

Hence, equation of line is 19x - 2y - 11 = 0.

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

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

∴ x = y

$$\begin{aligned}\text{Slope of the line} = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{0 - y}{x - 0} \\[1em] = \frac{-y}{x} \\[1em] = \frac{-x}{x} \\[1em] = -1.\end{aligned}$$

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
⇒ x + y - 5 = 0.

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

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

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

These two should be same. On equating we get,

$$\begin{aligned}\Rightarrow \frac{5 + x}{2} = 3 \text{ and } \frac{10 + y}{2} = 4 \\[1em] \Rightarrow 5 + x = 6 \text{ and } 10 + y = 8 \\[1em] \Rightarrow x = 6 - 5 \text{ and } y = 8 - 10 \\[1em] \Rightarrow x = 1 \text{ and } y = -2.\end{aligned}$$

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

(ii) By distance formula the distance between B(5, 10) and D(1, -2) is given by

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

Putting values we get BD,

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

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.,

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

Putting values we get,

$$\begin{aligned}\Rightarrow y - 6 = \frac{10 - 6}{5 - 3}(x - 3) \\[1em] \Rightarrow y - 6 = \frac{4}{2}(x - 3) \\[1em] \Rightarrow y - 6 = 2(x - 3) \\[1em] \Rightarrow y - 6 = 2x - 6 \\[1em] \Rightarrow 2x - y - 6 + 6 = 0 \\[1em] \Rightarrow 2x - y = 0.\end{aligned}$$

(i) Let the coordinates of A be (x, 0) and B be (0, y).

P(2, -3) is the mid-point of AB. So we have,

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

Hence. the coordinates of A are (4, 0) and B are (0, -6).

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

Putting values we get slope,

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

Hence, the slope of the line AB is $\frac{3}{2}.$

(iii) Equation of AB will be

⇒ y - y₁ = m(x - x₁)
⇒ y - 0 = $\frac{3}{2}$(x - 4)
⇒ 2y = 3x - 12
⇒ 3x - 2y = 12.

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

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

By section formula the coordinates of P are,

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

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

$$\begin{aligned}\Rightarrow 3 = \frac{3x}{5} \text{ and } 2 = \frac{2y}{5} \\[1em] \Rightarrow 3x = 15 \text{ and } 2y = 10 \\[1em] \Rightarrow x = 5 \text{ and } y = 5.\end{aligned}$$

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

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

Putting value in above equation we get slope,

$$\begin{aligned}= \frac{5 - 0}{0 - 5} \\[1em] = -\frac{5}{5} \\[1em] = -1.\end{aligned}$$

Hence, the slope of the line is -1.

(i) Slope of the line PQ = tan 45° = 1.

(ii) Equation of line PQ can be given by point slope form i.e.,

⇒ y - y₁ = m(x - x₁)
⇒ y - 3 = 1(x - 5)
⇒ y - 3 = x - 5
⇒ x - y - 5 + 3 = 0
⇒ x - y - 2 = 0.

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

(iii) The line touches y-axis at Q there x-coordinate will be 0 so putting x = 0 in equation of line,

⇒ 0 - y - 2 = 0
⇒ y = -2.

Hence, coordinates of Q are (0, -2).

(i) By section formula coordinates of P are,

$$\begin{aligned}\Rightarrow \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow \Big(\frac{1 \times 17 + 2 \times -4}{1 + 2}, \frac{1 \times 10 + 2 \times 1}{1 + 2}\Big) \\[1em] \Rightarrow \Big(\frac{17 - 8}{3}, \frac{10 + 2}{3}\Big) \\[1em] \Rightarrow \Big(\frac{9}{3}, \frac{12}{3}\Big) \\[1em] \Rightarrow (3, 4).\end{aligned}$$

Hence, the coordinates of P are (3, 4).

(ii) By distance formula

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

$$\begin{aligned}\therefore \text{OP} = \sqrt{(0 - 3)^2 + (0 - 4)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5.\end{aligned}$$

Hence, the length of OP is 5 units.

(iii) Let AB be divided by the y-axis in the ratio m : n.

By section formula,

$$\begin{aligned}\Rightarrow 0 = \frac{m \times 17 + n \times -4}{m + n} \\[1em] \Rightarrow 17m - 4n = 0 \\[1em] \Rightarrow 17m = 4n \\[1em] \Rightarrow \frac{m}{n} = \frac{4}{17} \\[1em] \Rightarrow m : n = 4 : 17.\end{aligned}$$

Thus, the ratio in which the y-axis divide the line AB is 4 : 17.

These lines x = -1, x = 2, y = -2 and y = 6 form a rectangle when they intersect at A, B, C and D.

From graph we get coordinates of A, B, C and D as (-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.

Equation of AC can be given by two point formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - (-2) = \frac{6 - (-2)}{2 - (-1)}(x - (-1)) \\[1em] \Rightarrow y + 2 = \frac{8}{3}(x + 1) \\[1em] \Rightarrow 3(y + 2) = 8(x + 1) \\[1em] \Rightarrow 3y + 6 = 8x + 8 \\[1em] \Rightarrow 8x - 3y + 2 = 0.\end{aligned}$$

Equation of BD can also be given by two point formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - (-2) = \frac{6 - (-2)}{-1 - 2}(x - 2) \\[1em] \Rightarrow y + 2 = \frac{8}{-3}(x - 2) \\[1em] \Rightarrow -3(y + 2) = 8(x - 2) \\[1em] \Rightarrow -3y - 6 = 8x - 16 \\[1em] \Rightarrow 8x + 3y - 10 = 0.\end{aligned}$$

Hence, the equations of the diagonals of the rectangle are 8x - 3y + 2 = 0 and 8x + 3y - 10 = 0.

5x + 7y = 3 ....(i)

2x - 3y = 7 ....(ii)

Multiply (i) by 3 and (ii) by 7,

15x + 21y = 9 ....(iii)

14x - 21y = 49 ....(iv)

Adding (iii) and (iv) we get,

⇒ 29x = 58
⇒ x = 2.

Substituting x = 2 in (i), we get

⇒ 5(2) + 7y = 3
⇒ 10 + 7y = 3
⇒ 7y = 3 - 10
⇒ 7y = -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.,

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

Putting values in above equation we get,

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

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

Lines are y = 3x - 5 and 2y = 4x + 7 or y = 2x + $\frac{7}{2}$.

Comparing y = 3x - 5 and y = 2x + $\frac{7}{2}$ with y = mx + c we get,

slopes = 3 and 2.

Since, slope of both the lines are neither equal nor their products is -1. Thus, the lines are neither parallel nor perpendicular.

∴ Option (iii) is correct

Hence, the straight lines y = 3x - 5 and 2y = 4x + 7 are neither parallel nor perpendicular.

Converting 6x + 5y - 7 = 0 in the form y = mx + c we get,

⇒ 6x + 5y - 7 = 0

⇒ 5y = -6x + 7

⇒ y = $-\frac{6}{5}x + \frac{7}{5}$

Comparing, we get slope of this line = m₁ = $-\frac{6}{5}$.

Converting 2px + 5y + 1 = 0 in the form y = mx + c we get,

⇒ 2px + 5y + 1 = 0

⇒ 5y = -2px - 1

⇒ y = $-\frac{2p}{5}x - \frac{1}{5}$

Comparing, we get slope of this line = m₂ = $-\frac{2p}{5}$

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

m₁ = m₂

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

Hence, the value of p = 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,

m₁.m₂ = -1

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

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

Given,

Lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other. Then the product of their slopes is -1.

Converting 3x + by + 5 = 0 in the form y = mx + c.

⇒ by = -3x - 5

⇒ y = $-\frac{3}{\text{b}}\text{x} - \frac{5}{\text{b}}$.

Comparing with y = mx + c we get,

Slope of first line = m₁ = $-\frac{3}{\text{b}}$.

Converting ax - 5y + 7 = 0 in the form y = mx + c.

⇒ 5y = ax + 7

⇒ y = $\frac{\text{a}}{5}\text{x} + \frac{7}{5}$.

Comparing with y = mx + c we get,

Slope of second line = m₂ = $\frac{\text{a}}{5}$.

For perpendicular lines, product of their slopes is -1.

∴ m₁.m₂ = -1.

$$\begin{aligned}\Rightarrow -\frac{3}{b} \times \frac{a}{5} = -1 \\[1em] \Rightarrow -\frac{3a}{5b} = -1 \\[1em] \Rightarrow -3a = -5b \\[1em] \Rightarrow 3a = 5b.\end{aligned}$$

Hence, the relation between a and b is given by 3a = 5b.

Equation of line through (-2, 3) and (4, 1) can be given by two-point form i.e.,

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

Putting values in above formula we get,

$$\begin{aligned}\Rightarrow y - 3 = \frac{1 - 3}{4 - (-2)}(x - (-2)) \\[1em] \Rightarrow y - 3 = \frac{-2}{6}(x + 2) \\[1em] \Rightarrow y - 3 = \frac{-1}{3}(x + 2) \\[1em] \Rightarrow y - 3 = -\frac{1}{3}x - \frac{2}{3} \\[1em] \Rightarrow y = -\frac{1}{3}x - \frac{2}{3} + 3 \\[1em] \Rightarrow y = -\frac{1}{3}x - \frac{2 + 9}{3} \\[1em] \Rightarrow y = -\frac{1}{3}x - \frac{11}{3}.\end{aligned}$$

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

slope = m₁ = $-\frac{1}{3}$

The other equation is 3x = y + 1 or y = 3x - 1, comparing this with y = mx + c we get,

slope = m₂ = 3.

Product of slopes,

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

Since, the product of slopes is -1 hence, the lines are perpendicular to each other.

Mid-point of (-2, 3) and (4, 1) can be given by mid-point formula i.e.,

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

Line 3x = y + 1 bisects the line joining (-2, 3) and (4, 1) if the mid-point i.e., (1, 2) satisfies the equation.

Putting (1, 2) in 3x = y + 1.

L.H.S. = 3x = 3(1) = 3.

R.H.S. = y + 1 = 2 + 1 = 3.

Since, L.H.S. = R.H.S. hence, (1, 2) satisfies 3x = y + 1.

Hence, the line 3x = y + 1 is perpendicular to the line joining (-2, 3) and (4, 1) and also bisects it.

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

Slope of line passing through A and B is m₁,

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

The equation of other line is,

⇒ 2x - 4y = 5

⇒ 4y = 2x - 5

⇒ y = $\frac{2}{4}x - \frac{5}{4}$

⇒ y = $\frac{1}{2}x - \frac{5}{4}$

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

slope = m₂ = $\frac{1}{2}$

As lines are perpendicular to each other, we have

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

Hence, the value of b is -9.

Given lines are :

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

x- ay + 7 = 0 ....(ii)

bx + 2y + 5 = 0 ....(iii)

It's said that these lines form three consecutive sides of a rectangle.

So,

Lines (i) and (ii) must be perpendicular and also (ii) and (iii) will be perpendicular.

Slope of line (i) is

⇒ 3x + y = 4

⇒ y = -3x + 4.

Comparing with y = mx + c we get,

slope = m₁ = -3.

Slope of line (ii) is

⇒ x - ay + 7 = 0

⇒ ay = x + 7

⇒ y = $\frac{1}{a}x + \frac{7}{a}$

Comparing with y = mx + c we get,

slope = m₂ = $\frac{1}{a}$.

Slope of line (iii) is

⇒ bx + 2y + 5 = 0

⇒ 2y = -bx - 5

⇒ y = $-\frac{b}{2}x - \frac{5}{2}$

Comparing with y = mx + c we get,

slope = m₃ = $-\frac{b}{2}$.

Since, lines (i) and (ii) are perpendicular so,

⇒ m₁ × m₂ = -1

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

Since, lines (ii) and (iii) are perpendicular so,

⇒ m₂ × m₃ = -1

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

Thus, the value of a is 3 and the value of b is 6.

Given lines,

⇒ 5x - 3y + 2 = 0 and 6x - py + 7 = 0

⇒ 3y = 5x + 2 and py = 6x + 7

⇒ y = $\frac{5}{3}x + \frac{2}{3} \text{ and } y = \frac{6}{p}x + \frac{7}{p}$

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

Slope of 1st line = $\frac{5}{3}$

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

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

$$\begin{aligned}\therefore \frac{5}{3} \times \frac{6}{p} = -1 \\[1em] \Rightarrow 5 \times \frac{2}{p} = -1 \\[1em] \Rightarrow \frac{10}{p} = -1 \\[1em] \Rightarrow p = -10 \\[1em]\end{aligned}$$

Given,

⇒ 6x - py + 7 = 0

⇒ 6x - (-10)y + 7 = 0

⇒ 6x + 10y + 7 = 0

⇒ 10y = -6x - 7

⇒ y = $-\frac{6}{10}x - \frac{7}{10}$

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

Slope = $-\frac{6}{10}$

Since, parallel lines have equal slope.

∴ Slope of line parallel to line 6x + 10y + 7 = 0 is $-\frac{6}{10}$

By point-slope form,

⇒ y - y₁ = m(x - x₁)

Equation of line passing through (-2, -1) and slope $-\frac{6}{10}$ is

$$\begin{aligned}\Rightarrow y - (-1) = -\frac{6}{10}[x - (-2)] \\[1em] \Rightarrow 10(y + 1) = -6(x + 2) \\[1em] \Rightarrow 10y + 10 = -6x - 12 \\[1em] \Rightarrow 10y + 10 + 6x + 12 = 0 \\[1em] \Rightarrow 6x + 10y + 22 = 0 \\[1em] \Rightarrow 2(3x + 5y + 11) = 0 \\[1em] \Rightarrow 3x + 5y + 11 = 0\end{aligned}$$

Hence, p = -10 and the equation of line is 3x + 5y + 11 = 0.

Given equation of line,

⇒ 2x - 3y - 7 = 0,

Converting it in the form y = mx + c,

⇒ 3y = 2x - 7

⇒ y = $\frac{2}{3}x - \frac{7}{3}$.

Comparing with y = mx + c, slope = $\frac{2}{3}$.

Since,the other line is parallel so, its slope will also be equal to $\frac{2}{3}$. Given, y-intercept is 4 or c = 4.

Putting values of slope and y-intercept in y = mx + c, we will get the equation of line as,

⇒ y = $\frac{2}{3}x + 4$

⇒ y = $\frac{2x + 12}{3}$

⇒ 3y = 2x + 12

⇒ 2x - 3y + 12 = 0.

At the point where the line intersects the x-axis, the y-coordinate there will be zero. So, putting y = 0 in 2x - 3y + 12 = 0.

⇒ 2x - 3(0) + 12 = 0
⇒ 2x = -12
⇒ x = -6.

∴ Coordinates = (-6, 0).

Hence, the equation of the line is 2x - 3y + 12 = 0 and it intersects the x-axis at (-6, 0).

Given equation of line,

⇒ 2x + 5y + 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x - 7

⇒ y = $-\frac{2}{5}x - \frac{7}{5}$.

Comparing with y = mx + c we get,

m = $-\frac{2}{5}$

Let slope of other line be m', since lines are perpendicular so,

⇒ m × m' = -1

$$\begin{aligned}\Rightarrow -\frac{2}{5} \times m' = -1 \\[1em] \Rightarrow m' = \frac{5}{2}.\end{aligned}$$

Given, y-intercept = -3, putting values of slope and y-intercept in y = mx + c we get,

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

Hence, the equation of the line is 5x - 2y - 6 = 0.

Given equation of line,

⇒ 3x - 4y + 12 = 0

Converting it in the form y = mx + c,

⇒ 4y = 3x + 12

⇒ y = $\frac{3}{4}x + 3$.

Comparing with y = mx + c we get,

Slope (m₁) = $\frac{3}{4}$.

Let the slope of the line perpendicular to the given line be m₂. So,

m₁ × m₂ = -1

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

The other line is 2x - y + 5 = 0 or y = 2x + 5.

Comparing with y= mx + c we get, c = 5.

So, the new line has slope = $-\frac{4}{3}$ and y-intercept = 5.

Putting these values in y = mx + c,

$$\begin{aligned}\Rightarrow y = -\frac{4}{3}x + 5 \\[1em] \Rightarrow y = \frac{-4x + 15}{3} \\[1em] \Rightarrow 3y = -4x + 15 \\[1em] \Rightarrow 4x + 3y = 15 \\[1em] \Rightarrow 4x + 3y - 15 = 0.\end{aligned}$$

Hence, the equation of the line is 4x + 3y - 15 = 0.

Given equation of line,

⇒ 3x + 5y + 15 = 0

Converting it in the form y = mx + c,

⇒ 5y = -3x - 15

⇒ y = $-\frac{3}{5}x - 3$

Comparing with y = mx + c we get,

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

The slope of the line parallel to the given line will also be $-\frac{3}{5}$.

Given, the new line has slope = $-\frac{3}{5}$ and passes through (0, 4).

So, equation can be given by,

⇒ y - y₁ = m(x - x₁)

$$\begin{aligned}\Rightarrow y - 4 = -\frac{3}{5}(x - 0) \\[1em] \Rightarrow 5(y - 4) = -3x \\[1em] \Rightarrow 5y - 20 = -3x \\[1em] \Rightarrow 3x + 5y - 20 = 0,\end{aligned}$$

Hence, the equation of the line is 3x + 5y - 20 = 0.

(i) When the line meets x-axis, its y-coordinate = 0.

So, putting y = 0 in 4x - 3y + 12 = 0, we get

⇒ 4x - 3(0) + 12 = 0
⇒ 4x = -12
⇒ x = -3.

Hence, the line meets the x-axis at A(-3, 0).

(ii) Converting 4x - 3y + 12 = 0, in the form y = mx + c.

⇒ 4x - 3y + 12 = 0

⇒ 3y = 4x + 12

⇒ y = $\frac{4}{3}x + 4$

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

Slope (m₁) = $\frac{4}{3}$

Let the slope of the line perpendicular to the given line be m₂.

∴ m₁ × m₂ = -1

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

Equation of the line having slope = $-\frac{3}{4}$ and passing through (-3, 0) can be given by,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 0 = -\frac{3}{4}(x - (-3)) \\[1em] \Rightarrow 4y = -3(x + 3) \\[1em] \Rightarrow 4y = -3x - 9 \\[1em] \Rightarrow 4y + 3x + 9 = 0.\end{aligned}$$

Hence, the equation of the line is 3x + 4y + 9 = 0.

Given equation of line,

⇒ 2x + 5y - 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x + 7

⇒ y = $-\frac{2}{5}x + \frac{7}{5}$

So, the slope is $-\frac{2}{5}$.

Since, slope of parallel lines are equal. So, slope of parallel line will be $-\frac{2}{5}$

By mid-point formula, the mid-point of the line segment joining the points (2, 7) and (-4, 1) is

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

Equation of the line having slope = $-\frac{2}{5}$ and passing through (-1, 4) can be given by,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = -\frac{2}{5}(x - (-1)) \\[1em] \Rightarrow 5(y - 4) = -2(x + 1) \\[1em] \Rightarrow 5y - 20 = -2x - 2 \\[1em] \Rightarrow 5y + 2x = 20 - 2 \\[1em] \Rightarrow 2x + 5y = 18 \\[1em] \Rightarrow 2x + 5y - 18 = 0.\end{aligned}$$

Hence, the equation of the line is 2x + 5y - 18 = 0.

Given equation of line,

⇒ 3x + 2y - 8 = 0

Converting it in the form y = mx + c,

⇒ 2y = -3x + 8

⇒ y = $-\frac{3}{2}x + 4$

Comparing with y = mx + c we get,

Slope (m₁) = $-\frac{3}{2}$

Now, the coordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be

$$\begin{aligned}\Rightarrow \Big(\frac{5 + 2}{2}, \frac{-2 + 2}{2}\Big) \\[1em] \Rightarrow \Big(\frac{7}{2}, 0\Big).\end{aligned}$$

Let's consider the slope of the line perpendicular to the given line be m₂.

Then,

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

The equation of the new line with slope m₂ and passing through $(\frac{7}{2}, 0)$ can be given by point-slope form i.e.,

y - y₁ = m(x - x₁)

Putting values we get,

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

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

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,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 0 = \frac{3}{7}(x - 2) \\[1em] \Rightarrow 7y = 3x - 6 \\[1em] \Rightarrow 3x - 7y - 6 = 0.\end{aligned}$$

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

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 is -2.

By point-slope formula,

Equation of line :

⇒ y - y₁ = m(x - x₁)

Substituting value we get :

Equation of line parallel to line 2x + y - 7 = 0 and passing through the point of intersection of the lines x + y - 4 = 0 and 2x - y = 8 is :

⇒ y - 0 = -2(x - 4)

⇒ y = -2x + 8

⇒ 2x + y = 8.

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

(i) Given, 3x + 4y - 7 = 0

Converting the equation in the form of y = mx + c,

⇒ 4y = -3x + 7

⇒ y = $-\frac{3}{4}x + \frac{7}{4}$

Comparing the equation with y = mx + c, we get slope (m₁) = $-\frac{3}{4}$.

(ii) Let the slope of the line perpendicular to the given line be m₂.

Then,

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

Now to find the point of intersection of

x - y + 2 = 0 ... (i)
3x + y - 10 = 0 ... (ii)

On adding (i) and (ii), we get

⇒ x - y + 2 + 3x + y - 10 = 0
⇒ 4x - 8 = 0
⇒ 4x = 8
⇒ x = 2.

Putting x = 2 in (i), we get

⇒ 2 - y + 2 = 0
⇒ y = 4.

Hence, the point of intersection of the lines is (2, 4).

The equation of the line with slope $\frac{4}{3}$ and passing through (2, 4) can be given by point-slope form,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = \frac{4}{3}(x - 2) \\[1em] \Rightarrow 3(y - 4) = 4(x - 2) \\[1em] \Rightarrow 3y - 12 = 4x - 8 \\[1em] \Rightarrow 4x - 3y + 4 = 0.\end{aligned}$$

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

Converting 4x - 3y - 5 = 0 in the form of y = mx + c.

⇒ 4x - 3y - 5 = 0

⇒ 3y = 4x - 5

⇒ y = $\frac{4}{3}x - \frac{5}{3}$

Slope of the line (m₁) = $\frac{4}{3}$.

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

Then, m₁ × m₂ = -1.

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

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

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-2) = -\frac{3}{4}(x - 1) \\[1em] \Rightarrow 4(y + 2) = -3(x - 1) \\[1em] \Rightarrow 4y + 8 = -3x + 3 \\[1em] \Rightarrow 3x + 4y + 5 = 0\end{aligned}$$

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

4x - 3y - 5 = 0 ....(i)
3x + 4y + 5 = 0 ....(ii)

On multiplying (i) by 4 and (ii) by 3 we get,

16x - 12y - 20 = 0 ....(iii)
9x + 12y + 15 = 0 ....(iv)

Adding (iii) and (iv) we get,

⇒ 16x - 12y - 20 + 9x + 12y + 15 = 0

⇒ 25x - 5 = 0

⇒ x = $\frac{5}{25}$

⇒ x = $\frac{1}{5}$.

Putting value of x in (i), we have

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

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

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

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

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

So, slope (m₁) of (0, 0) and (2, 3) is,

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

So, slope (m₂) of (2, -2) and (6, 4) is,

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

Since, m₁ = $\frac{3}{2}$ = m₂.

Hence, the lines are parallel to each other.

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,

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

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

$$\begin{aligned}= \frac{24 - 12}{4 - 8} \\[1em] = \frac{12}{-4} \\[1em] = -3.\end{aligned}$$

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

Hence, the lines are perpendicular to each other.

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 the line joining A(1, 3) and B(3, -1) is,

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

Slope (m₂) of line joining B(3, -1) and C(-5, -5) is,

$$\begin{aligned}= \frac{-5 - (-1)}{-5 - 3} \\[1em] = \frac{-4}{-8} \\[1em] = \frac{1}{2}.\end{aligned}$$

Since, m₁ × m₂ = $-2 \times \frac{1}{2} = -1$.

Thus AB and BC are perpendicular to each other.

Hence, △ABC is a right-angled triangle.

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 of the line joining the points (0, -2) and (4, 5) is,

$$\begin{aligned}= \frac{5 + 2}{4 - 0} \\[1em] = \frac{7}{4}.\end{aligned}$$

∴ Slope of line parallel to the line joining (0, -2) and (4, 5) = $\frac{7}{4}$

The equation of the line having slope $\frac{7}{4}$ and passing through (-1, 3) can be given by point-slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \frac{7}{4}(x - (-1)) \\[1em] \Rightarrow 4(y - 3) = 7(x + 1) \\[1em] \Rightarrow 4y - 12 = 7x + 7 \\[1em] \Rightarrow 7x - 4y + 19 = 0.\end{aligned}$$

Hence, the equation of the line through the point (-1, 3) and parallel to the line joining the points (0, -2) and (4, 5) is 7x - 4y + 19 = 0.

(i) Centroid of the triangle is given by,

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

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

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

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

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.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 1 = -\frac{5}{4}(x - 2) \\[1em] \Rightarrow 4(y - 1) = -5(x - 2) \\[1em] \Rightarrow 4y - 4 = -5x + 10 \\[1em] \Rightarrow 4y + 5x = 14 \\[1em] \Rightarrow 5x + 4y - 14 = 0.\end{aligned}$$

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

The slope (m₁) of the line joining (-3, 2) and (9, 1) i.e. two points is $\frac{y_2 - y_1}{x_2 - x_1}$ so,

$$\begin{aligned}\text{m}_1 = \frac{1 - 2}{9 - (-3)} \\[1em] = -\frac{1}{12}.\end{aligned}$$

Let the slope of the line perpendicular to the above line be m₂.

Then, m₁ × m₂ = -1.

$$\begin{aligned}\Rightarrow -\frac{1}{12} \times m_2 = -1 \\[1em] \Rightarrow m_2 = 12.\end{aligned}$$

So, the equation of the line passing through (0, -3) and slope 12 can be given by point-slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-3) = 12(x - 0) \\[1em] \Rightarrow y + 3 = 12x \\[1em] \Rightarrow 12x - y - 3 = 0.\end{aligned}$$

Hence, the equation of the required line is 12x - y - 3 = 0.

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

$$\begin{aligned}\text{m}_1 = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-6 - 2}{6 - (-1)} \\[1em] = -\frac{8}{7}.\end{aligned}$$

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

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

So, m₁ × m₂ = -1.

$$\begin{aligned}\Rightarrow -\frac{8}{7} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \frac{7}{8}.\end{aligned}$$

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

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 8 = \frac{7}{8}(x - 3) \\[1em] \Rightarrow 8(y - 8) = 7(x - 3) \\[1em] \Rightarrow 8y - 64 = 7x - 21 \\[1em] \Rightarrow 7x - 8y - 21 + 64 = 0 \\[1em] \Rightarrow 7x - 8y + 43 = 0.\end{aligned}$$

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

Given, vertices of a triangle are A(10, 4), B(4, -9) and C(-2, -1).

Now,

Slope of line BC (m₁),

$$\begin{aligned}\text{m}_1 = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-1 - (-9)}{-2 - 4} \\[1em] = -\frac{8}{6} \\[1em] = -\frac{4}{3}.\end{aligned}$$

Let the slope of the altitude from A(10, 4) to BC be m₂.

Then, m₁ × m₂ = -1.

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

Equation of the line having the slope = $\frac{3}{4}$ and passing through A(10, 4) can be given by point-slope formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = \frac{3}{4}(x - 10) \\[1em] \Rightarrow 4(y - 4) = 3(x - 10) \\[1em] \Rightarrow 4y - 16 = 3x - 30 \\[1em] \Rightarrow 3x - 4y + 16 - 30 = 0 \\[1em] \Rightarrow 3x - 4y - 14 = 0.\end{aligned}$$

Hence, the equation of the required line is 3x - 4y - 14 = 0.

Triangle ABC with vertices A(2, -4), B(3, 3) and C(-1, 5) is shown below:

(i) Let D be the mid-point of BC. So, AD will be the median.

Coordinates of D by mid-point formula will be,

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

The equation of AD can be given by two-point formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - (-4) = \frac{4 - (-4)}{1 -2}(x - 2) \\[1em] \Rightarrow y + 4 = -8(x - 2) \\[1em] \Rightarrow y + 4 = -8x + 16 \\[1em] \Rightarrow 8x + y - 12 = 0.\end{aligned}$$

Hence, the equation of the median of the triangle through A is 8x + y - 12 = 0.

(ii) Let E be a point on AC such that BE is perpendicular to AC.

Slope (m₁) of AC is,

$$\begin{aligned}\text{m}_1 = \frac{5 - (-4)}{-1 - 2} \\[1em] = -\frac{9}{3} \\[1em] = -3.\end{aligned}$$

Let slope of BE be m₂. Since, BE is perpendicular to AC so,

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

So, the equation of BE by point-slope form will be

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \frac{1}{3}(x - 3) \\[1em] \Rightarrow 3(y - 3) = x - 3 \\[1em] \Rightarrow 3y - 9 = x - 3 \\[1em] \Rightarrow x - 3y + 6 = 0.\end{aligned}$$

Hence, the equation of the required line is x - 3y + 6 = 0.

Slope of the line joining the points (1, 2) and (5, -6) is,

$$\begin{aligned}\text{m}_1 = \frac{-6 - 2}{5 - 1} \\[1em] = -\frac{8}{4} \\[1em] = -2.\end{aligned}$$

Let m₂ be the slope of the right bisector of the above line. Then,

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

The mid-point of the line segment joining (1, 2) and (5, -6) will be

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

Equation of the line having the slope = $\frac{1}{2}$ and passing through (3, -2) can be given by point-slope formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-2) = \frac{1}{2}(x - 3) \\[1em] \Rightarrow 2(y + 2) = x - 3 \\[1em] \Rightarrow 2y + 4 = x - 3 \\[1em] \Rightarrow x - 2y - 3 - 4 = 0 \\[1em] \Rightarrow x - 2y - 7 = 0.\end{aligned}$$

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

(i) Slope (m₁) of AB is,

$$\begin{aligned}\text{m}_1 = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{9 - (-3)}{1 - 7} \\[1em] = -\frac{12}{6} \\[1em] = -2.\end{aligned}$$

Hence, the slope of AB is -2.

(ii) Let PQ be the perpendicular bisector of AB intersecting it at M. Now, the coordinates of M will be

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

Let the slope of the line PQ be m₂. Since, PQ is perpendicular to AB then product of their slopes will be equal to -1,

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

Thus, by point-slope form equation of PQ is,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = \frac{1}{2}(x - 4) \\[1em] \Rightarrow 2(y - 3) = x - 4 \\[1em] \Rightarrow 2y - 6 = x - 4 \\[1em] \Rightarrow x - 2y + 2 = 0.\end{aligned}$$

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

(iii) As (-2, p) lies on the above line. The point will satisfy the line equation x - 2y + 2 = 0.

⇒ -2 - 2p + 2 = 0
⇒ 2p = 0
⇒ p = 0.

Hence, the value of p is 0.

Slope of BD is given by

m₁ = $\frac{8 - 3}{6 - 1} = \frac{5}{5}$ = 1.

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

So, the slope of AC (m₂) will be,

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

Coordinates of mid-point of BD and AC will be same as diagonals of a square meet at their mid-point.

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

By point-slope formula equation of AC is

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

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

Slope of AC is given by,

$m_1 = \frac{2 - 6}{-1 - 3} = \frac{-4}{-4}$ = 1.

We know that diagonal of a rhombus bisect each other at right angles. So, the diagonal BD is perpendicular to diagonal AC.

Let the slope of BD be m₂. Then,

$$\begin{aligned}\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow 1 \times m_2 = -1 \\[1em] \Rightarrow m_ 2 = -1.\end{aligned}$$

Coordinates of mid-point of AC and BD are same which are

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

Equation of the line having the slope = -1 and passing through (1, 4) can be given by point-slope formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 4 = -1(x - 1) \\[1em] \Rightarrow y - 4 = -x + 1 \\[1em] \Rightarrow x + y - 4 - 1 = 0 \\[1em] \Rightarrow x + y - 5 = 0.\end{aligned}$$

Hence, the equation of BD is x + y - 5 = 0.

The given line is x - 2y - 7 = 0 .....(i)

⇒ 2y = x - 7

⇒ y = $\frac{1}{2}x - \frac{7}{2}$.

The slope of the line (i) = m₁ = $\frac{1}{2}.$

Let the point (1, 2) be P.

From P draw a perpendicular to the line (i) and produce it to point P' such that P'M = MP, then P' is the image of P in line (i) and line (i) is the right bisector of the segment PP'.

Let P' be (a, b).

Then slope of PP' = m₂ = $\frac{b - 2}{a - 1}$.

Since, line (i) is perpendicular to PP' so,

$$\begin{aligned}\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow \frac{1}{2} \times \frac{b - 2}{a - 1} = -1 \\[1em] \Rightarrow \frac{b - 2}{2a - 2} = -1 \\[1em] \Rightarrow b - 2 = -2a + 2 \\[1em] \Rightarrow 2a + b = 4 \space .....(\text{iii})\end{aligned}$$

Also mid-point of PP' is M$\Big(\frac{a + 1}{2}, \frac{b + 2}{2}\Big)$.

Since, (i) is the right bisector of the segment PP', M lies on (i)

$$\begin{aligned}\Rightarrow \frac{a + 1}{2} - 2\Big(\frac{b + 2}{2}\Big) - 7 = 0 \\[1em] \Rightarrow \frac{a + 1}{2} - b - 2 - 7 = 0 \\[1em] \Rightarrow \frac{a + 1}{2} - b = 9 \\[1em] \Rightarrow \frac{a + 1 - 2b}{2} = 9 \\[1em] \Rightarrow a + 1 - 2b = 18 \\[1em] \Rightarrow a - 2b = 17 \space ....(\text{iv})\end{aligned}$$

Multiplying equation (iv) by 2 and subtracting from (iii) we get,

$$\begin{aligned}\Rightarrow 2a + b - 2(a - 2b) = 4 - 34 \\[1em] \Rightarrow 2a + b - 2a + 4b = -30 \\[1em] \Rightarrow 5b = -30 \\[1em] \Rightarrow b = -6.\end{aligned}$$

Putting value of b in Eq (iii),

⇒ 2a - 6 = 4
⇒ 2a = 10
⇒ a = 5.

P' = (a, b) = (5, -6).

Hence, the coordinates of image are (5, -6).

Given, equation of line,

⇒ x - 4y - 6 = 0

⇒ 4y = x - 6

⇒ y = $\frac{1}{4}x - \frac{6}{4}.$

Comparing with y = mx + c we get, slope = $\frac{1}{4}$.

Since, given line and PQ are perpendicular so their products will be equal to -1. Let slope of PQ be m₁,

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

Hence, slope of PQ = -4.

Now equation of PQ can be found by point slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 3 = -4(x - 1) \\[1em] \Rightarrow y - 3 = -4x + 4 \\[1em] \Rightarrow 4x + y - 7 = 0,\end{aligned}$$

Since, line x - 4y - 6 = 0 is perpendicular bisector of 4x + y - 7 = 0 hence solving them simultaneously to find point of intersection,

⇒ x - 4y = 6 ......(i)
⇒ 4x + y = 7 ......(ii)

Multiplying (ii) with 4 and adding with (i) we get,

⇒ 16x + 4y + x - 4y = 28 + 6
⇒ 17x = 34
⇒ x = 2.

Putting value of x = 2 in (i),

⇒ 2 - 4y = 6
⇒ -4y = 4
⇒ y = -1.

Hence, the point of intersection which is the mid-point of PQ is (2, -1).

Let coordinates of Q be (a, b).

By mid-point formula, coordinates of mid-point of PQ are

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

Equating with mid-point of PQ (2, -1) we get,

$$\begin{aligned}2 = \frac{1 + a}{2} \text{ and } -1 = \frac{3 + b}{2} \\[1em] 4 = 1 + a \text{ and } -2 = 3 + b \\[1em] a = 4 - 1 \text{ and } b = -2 - 3 \\[1em] a = 3 \text{ and } b = -5\end{aligned}$$

Hence, the coordinates of Q are (3, -5).

The square OABC is plotted on the graph below:

$$\begin{aligned}\text{OA} = \sqrt{(3 - 0)^2 + (0 - 0)^2} \\[1em] = \sqrt{3^2 + 0^2} \\[1em] = \sqrt{9} \\[1em] = 3. \\[1em] \text{AB} = \sqrt{(3 - p)^2 + (0 - q)^2} \\[1em] = \sqrt{(3 - p)^2 + q^2}\end{aligned}$$

Since, OA = AB (as sides of square are equal)

$$\begin{aligned}\therefore \sqrt{(p - 3)^2 + q^2} = 3 \\[1em] \Rightarrow (p - 3)^2 + q^2 = 9 \\[1em] \Rightarrow p^2 + 9 - 6p + q^2 = 9 \\[1em] \Rightarrow p^2 + q^2 - 6p = 0 \space .....(\text{i})\end{aligned}$$

By pythagoras theorem, OB² = OA² + AB².

$$\begin{aligned}\Rightarrow \Big(\sqrt{(p - 0)^2 + (q - 0)^2}\Big)^2 = 3^2 + \Big(\sqrt{(3 - p)^2 + q^2}\Big)^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + (3 - p)^2 + q^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + 9 + p^2 - 6p + q^2 \\[1em] \Rightarrow p^2 - p^2 + q^2 - q^2 + 6p = 18 \\[1em] \Rightarrow 6p = 18 \\[1em] \Rightarrow p = 3.\end{aligned}$$

Substituting value of p in (i),

$$\begin{aligned}\Rightarrow 3^2 + q^2 - 6(3) = 0 \\[1em] \Rightarrow q^2 + 9 - 18 = 0 \\[1em] \Rightarrow q^2 - 9 = 0 \\[1em] \Rightarrow q^2 - (3)^2 = 0 \\[1em] \Rightarrow (q - 3)(q + 3) = 0 \\[1em] \Rightarrow q - 3 = 0 \text{ or } q + 3 = 0 \\[1em] \Rightarrow q = 3 \text{ or } q = -3 \\[1em] \Rightarrow q = 3, -3.\end{aligned}$$

But q = -3 is not possible as the square is in 1st quadrant and the coordinates are positive in 1st quadrant.

∴ p = 3 and q = 3.

AB is parallel to y-axis,

∴ Equation of AB will be x = 3 or x - 3 = 0.

BC is parallel to x-axis,

∴ Equation BC will be y = 3 or y - 3 = 0.

Hence, the value of p = 3 and q = 3. Equation of AB is x - 3 = 0 and BC is y - 3 = 0.

We know that slope of y-axis is not defined. Since, slope of parallel lines are equal.

∴ Slope of line parallel to y-axis is not defined.

Hence, Option 4 is the correct option.

Slope of the line which makes an angle of 30° with positive direction of x-axis = tan 30° = $\frac{1}{\sqrt{3}}.$

Hence, Option 2 is the correct option.

Slope of the line passing through (x₁, y₁) and (x₂, y₂) is given by,

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{2 - (-4)}{-6 - 0} \\[1em] = \frac{6}{-6} \\[1em] = -1.\end{aligned}$$

Hence, Option 3 is the correct option.

Slope of the line passing through (x₁, y₁) and (x₂, y₂) is given by,

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-2 - (-2)}{-7 - 3} \\[1em] = \frac{-2 + 2}{-10} \\[1em] = -\frac{0}{10} \\[1em] = 0.\end{aligned}$$

Hence, Option 1 is the correct option.

Slope of the line passing through (x₁, y₁) and (x₂, y₂) is given by,

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-4 - (-2)}{3 - 3} \\[1em] = \frac{-4 + 2}{0} \\[1em] = -\frac{-2}{0} \\[1em]\end{aligned}$$

= not defined.

Hence, Option 4 is the correct option.

Given, y = $\sqrt{3}x - 5$ comparing with y = mx + c we get,

m = $\sqrt{3}$.

Slope is given by m = tan θ or,

$$\begin{aligned}\Rightarrow \sqrt{3} = \text{tan θ} \\[1em] \Rightarrow \text{tan } 60° = \text{tan θ} \\[1em] \Rightarrow \text{θ} = 60°.\end{aligned}$$

Hence, Option 2 is the correct option.

Slope of the line passing through (x₁, y₁) and (x₂, y₂) is given by,

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

Given, slope of the line passing through the points (2, 5) and (k, 3) is 2.

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

Hence, Option 3 is the correct option.

Slope of a line parallel to the line passing through the points (0, 6) and (7, 3) = Slope of the line passing through the points (0, 6) and (7, 3) which is given by

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{3 - 6}{7 - 0} \\[1em] = -\frac{3}{7} \\[1em]\end{aligned}$$

Hence, Option 2 is the correct option.

Slope (m₁) of line joining the points (2, 5) and (-3, 6) is given by

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x1} \\[1em] = \frac{6 - 5}{-3 - 2} \\[1em] = -\frac{1}{5}.\end{aligned}$$

Let slope of perpendicular line be m₂. Then,

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

∴ Slope of line perpendicular to this line = 5.

Hence, Option 4 is the correct option.

Given, 2x + 3y - 7 = 0.

⇒ 3y = -2x + 7

⇒ y = $-\frac{2}{3}x + \frac{7}{3}$.

Comparing with y = mx + c we get,

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

Since, parallel lines have equal slopes so the slope of line parallel to 2x + 3y - 7 = 0 is $-\frac{2}{3}$.

Hence, Option 1 is the correct option.

Given, 3x = 4y + 11.

⇒ 4y = 3x - 11

⇒ y = $\frac{3}{4}x - \frac{11}{4}$.

Comparing with y = mx + c we get,

Slope (m₁) = $\frac{3}{4}$.

Let the slope of perpendicular line be m₂. Since, lines are perpendicular so,

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

Hence, Option 4 is the correct option.

Given,

⇒ 2x + 3y = 5 and kx - 6y = 7

⇒ 3y = -2x + 5 and 6y = kx - 7

⇒ y = $-\frac{2}{3}x + \frac{5}{3}$ and y = $\frac{k}{6}x - \frac{7}{6}$

Comparing both the equations with y = mx + c,

Slope of first line = m₁ = $-\frac{2}{3}$

Slope of second line = m₂ = $\frac{k}{6}$

Since, both the lines are parallel so,

m₁ = m₂

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

Hence, Option 2 is the correct option.

Given,

3x - 4y + 7 = 0 and
2x + ky + 5 = 0

⇒ 4y = 3x + 7 and ky = -2x - 5

⇒ y = $\frac{3}{4}x + \frac{7}{4}$ and y = $-\frac{2}{k}x - \frac{5}{k}$

Comparing both the equations with y = mx + c,

Slope of first line = m₁ = $\frac{3}{4}$

Slope of second line = m₂ = $-\frac{2}{k}$

Since, both the lines are perpendicular so,

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

Hence, Option 1 is the correct option.

Substituting x = 0 and y = 0 in L.H.S. of the equation 2x - 3y = 0, we get :

⇒ 2 × 0 - 3 × 0

⇒ 0 - 0

⇒ 0.

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

∴ Line 2x - 3y = 0 represents a line passing through origin.

Hence, Option 2 is the correct option.

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC.

$$\begin{aligned}\Rightarrow \frac{-2 - y}{3 - x} = \frac{-5 - (-2)}{4 - 3} \\[1em] \Rightarrow \frac{-2 - y}{3 - x} = \frac{-5 + 2}{1} \\[1em] \Rightarrow \frac{-2 - y}{3 - x} = -3 \\[1em] \Rightarrow -2 - y = -3(3 - x) \\[1em] \Rightarrow -2 - y = -9 + 3x \\[1em] \Rightarrow y = -2 + 9 - 3x \\[1em] \Rightarrow y = 7 - 3x.\end{aligned}$$

Hence, Option 4 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 2 is the correct option.

Line y = -2 is a line parallel to x-axis at a distance of 2 units from it.

For any point on x-axis the y-coordinate equals to zero, which is not the case with line y = -2.

So, it never intersects the x-axis.

So, reason (R) is true.

The x-intercept is the distance from the origin to the point where the line intersects the x-axis. Since this line never intersects the x-axis, so its x-intercept is undefined.

So, assertion (A) is false.

Thus, Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

By point-slope form,

⇒ y - y₁ = m(x - x₁)

Equation of line passing through (-1, 0) and slope = 1 is :

⇒ y - 0 = 1[x - (-1)]

⇒ y = x + 1

⇒ x - y + 1 = 0

In order to find x-intercept, substitute y = 0 in the equation of line :

⇒ x - 0 + 1 = 0

⇒ x = -1

In order to find y-intercept, substitute x = 0 in the equation of line :

⇒ 0 - y + 1 = 0

⇒ y = 1

Thus, x-intercept and y-intercept are not equal.

So, assertion (A) is false.

The triangle are formed with the axes has vertices : (0, 0), (-1, 0) and (0, 1).

By distance formula,

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

$$\begin{aligned}\text{Distance between (0, 0) and (-1, 0)} = \sqrt{(-1 - 0)^2 + (0 - 0)^2}\\[1em] = \sqrt{(-1)^2}\\[1em] = \sqrt{1}\\[1em] = 1 \\[1em] \text{Distance between (0, 0) and (0, 1)} = \sqrt{(0 - 0)^2 + (0 - 1)^2}\\[1em] = \sqrt{(-1)^2}\\[1em] = \sqrt{1}\\[1em] = 1\end{aligned}$$

Since,two sides are equal in length. Therefore, it is an isosceles triangle.

So, reason (R) is true.

Thus, Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Given lines,

⇒ y = 2x + 8 and y = $\frac{1}{2}x - 7$

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

Slope of 1st line = 2

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

The slopes of 1st line and 2nd line are reciprocal of each other.

So, reason (R) is true.

⇒ Slope of 1st line × Slope of 2nd line

⇒ 2 × $\frac{1}{2}$

⇒ 1.

Since, product ≠ -1, so lines are not perpendicular as product of slope of perpendicular lines = -1.

So, assertion (A) is false.

Thus, Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Both lines intersect at the same point on the y-axis. So, they have the same y-intercept.

So, assertion (A) is true.

Given,

x-intercept are equal but they are of different signs (one is positive, one is negative).

tan θ = $\frac{y}{x}$

So, the inclination of both the lines will be of opposite signs.

So, reason (R) is false.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

We know that,

The ordinate (y-coordinate) of every point on x-axis is zero.

So, reason (R) is true.

Substituting y = 0, in the equation of line 3x + 3y + 5 = 0, we get :

⇒ 3x + 3 x 0 + 5 = 0

⇒ 3x + 5 = 0

⇒ x = $-\frac{5}{3}$

Point of intersection = $\Big(-\frac{5}{3}, 0\Big)$

So, assertion (A) is false.

Thus, Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

Given, θ = 60° and c = -4.

We know that m = tan θ = tan 60° = $\sqrt{3}$.

Putting values of m and c in equation y = mx + c.

$\Rightarrow y = \sqrt{3}x - 4.$

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

Given, 3y + 2x = 12

⇒ 3y = -2x + 12

⇒ y = $-\frac{2}{3}x + 4$

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

m = $-\frac{2}{3}$ and c = 4.

Hence, the gradient and intercept on the y-axis of the line 3y + 2x = 12 is $-\frac{2}{3}$ and 4 respectively.

Comparing the equation y = $\sqrt{3}x + 1$ with y = mx + c we get,

m = $\sqrt{3}$.

We know that m = tan θ.

∴ tan θ = $\sqrt{3}$
⇒ tan θ = tan 60°
⇒ θ = 60°.

Hence, the inclination of the line y = $\sqrt{3}$x + 1 is 60°.

Equation of line passing through (2, -4) and (-3, 1) i.e. two points can be given by two-point formula i.e.

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

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

m = -1 and c = -2.

Hence, the value of m is -1 and c is -2.

Equation of line passing through (1, 4) and (3, -2) i.e two points can be given by two-point formula i.e.

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \therefore y - 4 = \frac{-2 - 4}{3 - 1}(x - 1) \\[1em] \Rightarrow y - 4 = \frac{-6}{2}(x - 1) \\[1em] \Rightarrow y - 4 = -3(x - 1) \\[1em] \Rightarrow y - 4 = -3x + 3 \\[1em] \Rightarrow y = -3x + 7.\end{aligned}$$

Since (p, -5) lies on the line it will satisfy it. Putting the values,

⇒ -5 = -3p + 7
⇒ 3p = 7 + 5
⇒ 3p = 12
⇒ p = 4.

Hence, the value of p is 4.

Slope of the line joining P and Q i.e two points is given by,

$$\begin{aligned}\Rightarrow \frac{y_2 - y_1}{x_2 - x_1} \\[1em] \therefore \frac{3 - 0}{7 - 4} \\[1em] = \frac{3}{3} \\[1em] = 1.\end{aligned}$$

We know that, m = tan θ. Since m = 1.

$$\begin{aligned}\Rightarrow \text{tan θ} = \text{tan 45°} \\[1em] \Rightarrow \text{θ} = 45°.\end{aligned}$$

Hence, the inclination of the line is 45°.

Equation of the lines are
⇒ 2x + y = 5 ...(i)
⇒ x - 2y = 5 ...(ii)

Multiplying (i) by 2, we get
⇒ 4x + 2y = 10 ....(iii)

Adding (iii) and (ii) we get,
⇒ 4x + 2y + x - 2y = 10 + 5
⇒ 5x = 15
⇒ x = 3.

Substituting the values of x in (i)
⇒ 2(3) + y = 5
⇒ 6 + y = 5
⇒ y = -1.

∴ Coordinates of point of intersection are (3, -1).

Hence, line passes through (3, -1). So it will satisfy y = mx + c.
⇒ -1 = 3m + c

Given, y-intercept is $-\frac{3}{7}$ so, c = $-\frac{3}{7}.$

⇒ -1 = 3m + $\Big(-\frac{3}{7}\Big)$

⇒ -7 = 21m - 3

⇒ -7 + 3 = 21m

⇒ -4 = 21m

⇒ m = $-\frac{4}{21}.$

Putting value of m and c in y = mx + c,

$$\begin{aligned}\Rightarrow y = -\frac{4}{21}x + \Big(-\frac{3}{7}\Big) \\[1em] \Rightarrow y = \frac{-4x - 3 \times 3}{21} \\[1em] \Rightarrow 21y = -4x - 9 \\[1em] \Rightarrow 4x + 21y + 9 = 0.\end{aligned}$$

Hence, the equation of the line is 4x + 21y + 9 = 0.

Given, $\frac{x}{3} + \frac{y}{4} = 7$ and 3x + ky = 11.

Converting $\frac{x}{3} + \frac{y}{4} = 7$ in the form of y = mx + c.

$$\begin{aligned}\Rightarrow \frac{4x + 3y}{12} = 7 \\[1em] \Rightarrow 4x + 3y = 84 \\[1em] \Rightarrow 3y = -4x + 84 \\[1em] \Rightarrow y = -\frac{4}{3}x + 28.\end{aligned}$$

Comparing above equation with y = mx + c we get slope (m₁),

$m_1 = -\frac{4}{3}$.

Converting 3x + ky = 11 in the form of y = mx + c.

$$\begin{aligned}\Rightarrow 3x + ky = 11 \\[1em] \Rightarrow ky = -3x + 11 \\[1em] \Rightarrow y = -\frac{3}{k}x + \frac{11}{k}.\end{aligned}$$

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

$m_2 = -\frac{3}{k}$.

Given two lines are perpendicular,

∴ m₁ × m₂ = -1.

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

Hence, the value of k is -4.

Given equation of line,

x - 2y + 8 = 0.

Converting in the form of y = mx + c,

⇒ 2y = x + 8

⇒ y = $\frac{1}{2}x + 4$

Comparing we get,

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

Line parallel to x - 2y + 8 = 0 will have the same slope.

Equation of the line having slope $\frac{1}{2}$ and passing through (1, 2) can be given by point-slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 2 = \frac{1}{2}(x - 1) \\[1em] \Rightarrow 2(y - 2) = x - 1 \\[1em] \Rightarrow 2y - 4 = x - 1 \\[1em] \Rightarrow x - 2y + 3 = 0.\end{aligned}$$

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

Given equation of the line is,

⇒ 3y = 5 - x

⇒ y = $-\frac{1}{3}x + \frac{5}{3}.$

Comparing with y = mx + c we get,

Slope (m₁) = $-\frac{1}{3}$.

Let slope of the line perpendicular to the given line be m₂.

∴ m₁ × m₂ = -1.

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

Equation of the line having slope 3 and passing through (-3, 2) can be given by point-slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 2 = 3(x - (-3)) \\[1em] \Rightarrow y - 2 = 3(x + 3) \\[1em] \Rightarrow y - 2 = 3x + 9 \\[1em] \Rightarrow 3x - y + 11 = 0.\end{aligned}$$

Hence, the equation of the required line is 3x - y + 11 = 0.

Let the slope of line joining A(1, 2) and B(6, 7) be s₁. Slope of two points is given by,

$$\begin{aligned}s_1 = \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{7 - 2}{6 - 1} \\[1em] = \frac{5}{5} \\[1em] = 1.\end{aligned}$$

Let slope of perpendicular line be s₂. So,

$$\begin{aligned}\Rightarrow s_1 \times s_2 = -1 \\[1em] \Rightarrow 1 \times s_2 = -1 \\[1em] \Rightarrow s_2 = -1.\end{aligned}$$

Given, the new line passes through the point which divides the line segment AB in the ratio 3 : 2. By section formula coordinates are,

$$\begin{aligned}= \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] = \Big(\frac{3 \times 6 + 2 \times 1}{3 + 2}, \frac{3 \times 7 + 2 \times 2}{3 + 2} \Big) \\[1em] = \Big(\frac{18 + 2}{5}, \frac{21 + 4}{5}\Big) \\[1em] = \Big(\frac{20}{5}, \frac{25}{5}\Big) \\[1em] = (4, 5).\end{aligned}$$

Equation of the line having slope -1 and passing through (4, 5) can be given by point-slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - 5 = -1(x - 4) \\[1em] \Rightarrow y - 5 = -x + 4 \\[1em] \Rightarrow x + y - 9 = 0.\end{aligned}$$

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

Rhombus ABCD with A(7, 3) and C(0, -4) as the two opposite vertices is shown in the figure below:

Slope of the line AC (m₁),

$$\begin{aligned}= \frac{y_2 - y_1}{x_2 - x_1} \\[1em] = \frac{-4 - 3}{0 - 7} \\[1em] = \frac{-7}{-7} \\[1em] = 1.\end{aligned}$$

Diagonals of rhombus bisect each other at right angles.

∴ BD is perpendicular to AC. Let slope of BD be m₂.

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

Let O be the mid-point of diagonals. It's coordinates are given by,

$$\begin{aligned}= \Big(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\Big) \\[1em] = \Big(\frac{7 + 0}{2}, \frac{3 + (-4)}{2}\Big) \\[1em] = \Big(\frac{7}{2}, -\frac{1}{2}\Big).\end{aligned}$$

Equation of BD can be given by point slope form i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - \Big(-\frac{1}{2}\Big) = -1(x - \Big(\frac{7}{2}\Big)) \\[1em] \Rightarrow y + \frac{1}{2} = -x + \frac{7}{2} \\[1em] \Rightarrow \frac{2y + 1}{2} = \frac{-2x + 7}{2} \\[1em] \Rightarrow 2y + 1 = -2x + 7 \\[1em] \Rightarrow 2y + 2x - 6 = 0 \\[1em] \Rightarrow 2(y + x - 3) = 0 \\[1em] \Rightarrow x + y - 3 = 0.\end{aligned}$$

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

(a) From figure,

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

(b) 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 :

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

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

(c) 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

⇒ y = x + 2.

Hence, required equation is y = x + 2.

A lies on x-axis and B lies on y-axis. Let coordinates of A be (x, 0) and B be (0, y) and P(2, 1) divides BA in the ratio 3 : 1.

By section formula,

$$\begin{aligned}\text{x-coordinate } = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \\[1em] \Rightarrow 2 = \frac{3x + 0}{3 + 1} \\[1em] \Rightarrow 2 = \frac{3x}{4} \\[1em] \Rightarrow x = \frac{8}{3}.\end{aligned}$$

Similarly,

$$\begin{aligned}\text{y-coordinate } = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \\[1em] \Rightarrow 1 = \frac{3 \times 0 + 1 \times y}{3 + 1} \\[1em] \Rightarrow 1 = \frac{y}{4} \\[1em] \Rightarrow y = 4.\end{aligned}$$

Hence, coordinates of A are $\Big(\frac{8}{3}, 0\Big)$ and of B are (0, 4).

(ii) By two point formula equation of AB will be,

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - 0 = \frac{4 - 0}{0 - \frac{8}{3}}(x - \frac{8}{3}) \\[1em] \Rightarrow y = \frac{4 \times 3}{-8}\Big(\frac{3x - 8}{3}\Big) \\[1em] \Rightarrow y = \frac{3x - 8}{-2} \\[1em] \Rightarrow -2y = 3x - 8 \\[1em] \Rightarrow 3x + 2y - 8 = 0.\end{aligned}$$

Hence, the equation of the required line is 3x + 2y = 8.

Let the line make intercept a and b with the x-axis and y-axis respectively. Let line intersect x-axis at A and y-axis at B.

Coordinates of A will be (a, 0) and B will be (0, b).

Given, sum of intercepts = 7.

∴ a + b = 7 or b = 7 - a.

Equation of AB can be given by two point formula i.e.,

$$\begin{aligned}\Rightarrow y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \\[1em] \Rightarrow y - 0 = \frac{b - 0}{0 - a}(x - a) \\[1em] \Rightarrow y = \frac{b}{-a}(x - a) \\[1em] \Rightarrow -ay = bx - ba \\[1em] \Rightarrow bx + ay - ab = 0 \space .....(i)\end{aligned}$$

Since, line passes through (-3, 8) it will satisfy the above equation. Also, putting b = 7 - a.

$$\begin{aligned}\Rightarrow (7 - a)(-3) + 8a - a(7 - a) = 0 \\[1em] \Rightarrow -21 + 3a + 8a - 7a + a^2 = 0 \\[1em] \Rightarrow a^2 + 4a - 21 = 0 \\[1em] \Rightarrow a^2 + 7a - 3a - 21 = 0 \\[1em] \Rightarrow a(a + 7) - 3(a + 7) = 0 \\[1em] \Rightarrow (a - 3)(a + 7) = 0 \\[1em] \Rightarrow a = 3 \text{ or } a = -7.\end{aligned}$$

Since, only positive intercepts are made hence a ≠ -7.

b = 7 - a = 7 - 3 = 4.

Putting value of b and a in (i) we get,

⇒ 4x + 3y - 12 = 0.

Hence, the equation of the required line is 4x + 3y = 12.

Diagonals AC and BD of the square ABCD bisect each other at right angle at O.

∴ O is the mid-point of AC and BD.

Equation of BD is 3x - 7y + 6 = 0

⇒ 7y = 3x + 6

⇒ y = $\frac{3}{7}x + \frac{6}{7}$.

∴ Slope of BD = m₁ = $\frac{3}{7}$

Let slope of AC be m₂. Since, BD and AC are perpendicular,

$$\begin{aligned}\therefore m_1 \times m_2 = -1 \\[1em] \Rightarrow \frac{3}{7}m_2 = -1 \\[1em] \Rightarrow m_2 = -\frac{7}{3}.\end{aligned}$$

Equation of AC will be

$$\begin{aligned}\Rightarrow y - y_1 = m(x - x_1) \\[1em] \Rightarrow y - (-2) = -\frac{7}{3}(x - 3) \\[1em] \Rightarrow 3(y + 2) = -7(x - 3) \\[1em] \Rightarrow 3y + 6 = -7x + 21 \\[1em] \Rightarrow 7x + 3y = 15\end{aligned}$$

Now we will find the coordinates of O, the points of intersection of AC and BD

⇒ 3x - 7y = -6 ....(i)
⇒ 7x + 3y = 15 ....(ii)

Multiplying (i) by 3 and (ii) by 7, we get,

⇒ 9x - 21y = -18 ....(iii)
⇒ 49x + 21y = 105 ...(iv)

Adding (iii) and (iv) we get,

⇒ 9x + 49x - 21y + 21y = -18 + 105

⇒ 58x = 87

⇒ x = $\frac{87}{58} = \frac{3}{2}.$

Putting value of x in (i) we get,

$$\begin{aligned}\Rightarrow 3 \times \frac{3}{2} - 7y = -6 \\[1em] \Rightarrow \frac{9}{2} - 7y = -6 \\[1em] \Rightarrow \frac{9}{2} + 6 = 7y \\[1em] \Rightarrow \frac{9 + 12}{2} = 7y \\[1em] \Rightarrow \frac{21}{2} = 7y \\[1em] \Rightarrow y = \frac{3}{2}.\end{aligned}$$

Hence, the equation of AC is 7x + 3y - 15 = 0 and coordinates of the center are $\Big(\frac{3}{2}, \frac{3}{2}\Big)$.

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

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.

(a) 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.

(b) 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.

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

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.