Equation of Straight Line

Find the slope of a line whose inclination is

(i) 45°

(ii) 30°

Answer

(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° = $\dfrac{1}{\sqrt{3}}$.

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

Find the inclination of the line whose gradient is

(i) 1

(ii) $\sqrt{3}$

(iii) $\dfrac{1}{\sqrt{3}}$

Answer

(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

⇒ $\dfrac{1}{\sqrt{3}}$ = tan θ

⇒ $\dfrac{1}{\sqrt{3}}$ = tan 30°

∴ tan θ = tan 30°

∴ θ = 30°.

Hence, the inclination is 30°.

Find the equation of a straight line parallel to x-axis which is at a distance

(i) 2 units above it

(ii) 3 units below it

Answer

(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.

Find the equation of a straight line parallel to y-axis which is at a distance

(i) 3 units to the right

(ii) 2 units to the left.

Answer

(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.

Find the equation of a straight line parallel to y-axis and passing through the point (-3, 5).

Answer

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.

Find the equation of a line whose

(i) slope = 3, y-intercept = -5.

(ii) slope = $-\dfrac{2}{7}$, y-intercept = 3.

(iii) gradient = $\sqrt{3}$, y-intercept = $-\dfrac{4}{3}$

(iv) inclination = 30°, y-intercept = 2.

Answer

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

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

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 = $-\dfrac{4}{3}$. Putting values in equation we get,

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

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° = $\dfrac{1}{\sqrt{3}}$

Putting values in equation we get,

$\Rightarrow y = \dfrac{1}{\sqrt{3}}x + 2 \\[1em] \Rightarrow y = \dfrac{x + 2\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow \sqrt{3}y = x + 2\sqrt{3} \\[1em] \Rightarrow x - \sqrt{3}y + 2\sqrt{3} = 0.$

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

Find the slope and y-intercept of the following lines :

(i) x - 2y - 1 = 0

(ii) 4x - 5y - 9 = 0

(iii) 3x + 5y + 7 = 0

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

(v) y - 3 = 0

(vi) x - 3 = 0

Answer

(i) The equation of line is

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

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

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

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

(ii) The equation of line is

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

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

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

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

(iii) The equation of line is

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

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

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

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

(iv) The equation of line is

$\Rightarrow \dfrac{x}{3} + \dfrac{y}{4} = 1 \\[1em] \Rightarrow \dfrac{4x + 3y}{12} = 1 \\[1em] \Rightarrow 4x + 3y = 12 \\[1em] \Rightarrow 3y = -4x + 12 \\[1em] \Rightarrow y = -\dfrac{4}{3}x + \dfrac{12}{3} \\[1em] \Rightarrow y = -\dfrac{4}{3}x + 4.$

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

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

Hence, the slope of the line = $-\dfrac{4}{3}$ 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.

The equation of the line PQ is 3y - 3x + 7 = 0.

(i) Write down the slope of the line PQ.

(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.

Answer

(i) The equation of line is

⇒ 3y - 3x + 7 = 0

⇒ 3y = 3x - 7

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

⇒ y = x - $\dfrac{7}{3}$.

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°.

The given figure represents the lines y = x + 1 and y = $\sqrt{3}x - 1.$ Write down the angles which the lines make with the positive direction of the x-axis. Hence, determine θ.

Answer

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°.

Find the value of p, given that the line $\dfrac{y}{2} = x - p$ passes through the point (-4, 4).

Answer

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

$\Rightarrow \dfrac{4}{2} = -4 - p \\[1em] \Rightarrow 2 = -4 - p \\[1em] \Rightarrow p = -4 - 2 \\[1em] \Rightarrow p = -6.$

Hence, the value of p = -6.

Given that (a, 2a) lies on the line $\dfrac{y}{2} = 3x - 6$, find the value of a.

Answer

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

$\Rightarrow \dfrac{2a}{2} = 3a - 6 \\[1em] \Rightarrow a = 3a - 6 \\[1em] \Rightarrow 3a - a = 6 \\[1em] \Rightarrow 2a = 6 \\[1em] \Rightarrow a = 3.$

Hence, the value of a = 3.

The graph of the equation y = mx + c passes through the points (1, 4) and (-2, -5). Determine the values of m and c.

Answer

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.

Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.

Answer

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.

Find the equation of the straight line passing through (-1, 2) and whose slope is $\dfrac{2}{5}.$

Answer

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,

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

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

Find the equation of a straight line whose inclination is 60° and which passes through the point (0, -3).

Answer

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

Find the gradient of a line passing through the following pairs of points :

(i) (0, -2), (3, 4)

(ii) (3, -7), (-1, 8).

Answer

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

Putting values in above formula we get,

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

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

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

Putting values in above formula we get,

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

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

The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :

(i) the gradient of EF.

(ii) the equation of EF.

(iii) the coordinates of the point where the line EF intersects the x-axis.

Answer

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

Putting values in above formula we get,

$\text{Gradient } = \dfrac{7 - 4}{3 - 0} \\[1em] = \dfrac{3}{3} \\[1em] = 1.$

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).

Find the intercepts made by the line 2x - 3y + 12 = 0 on the coordinate axes.

Answer

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.

Find the equation of the line passing through the points P(5, 1) and Q(1, -1). Hence, show that the points P, Q and R(11, 4) are collinear.

Answer

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

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

$= \dfrac{-1 - 1}{1 - 5} \\[1em] = \dfrac{-2}{-4} \\[1em] = \dfrac{1}{2}.$

So, the equation of PQ is

⇒ y - y₁ = m(x - x₁)
⇒ y - 1 = $\dfrac{1}{2}(x - 5)$
⇒ 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

$= 11 - 2(4) - 3 \\[1em] = 11 - 8 - 3 \\[1em] = 0.$

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.

Find the value of 'a' for which the following points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.

Answer

Given that ,

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

Slope of AB = Slope of BC

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

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

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

Putting values of points in above formula we get,

$\Rightarrow y - 3 = \dfrac{1 - 3}{2 - a}(x - a) \\[1em]$

Putting a = -1 in above equation,

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

Putting a = 4 in above equation,

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

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.

Use a graph paper for this question. The graph of a linear equation in x and y, passes through A (-1, -1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and ($\dfrac{1}{2}$, k).

Answer

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

From graph we get,

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

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

ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find

(i) the coordinates of A.

(ii) the equation of the diagonal BD.

Answer

(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,

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

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

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

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

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

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

Putting values in above equation,

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

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

In △ABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.

Answer

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

Coordinates of D by mid-point formula are,

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

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

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

Putting values we get,

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

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

Find the equation of a line passing through the point (-2, 3) and having x-intercept 4 units.

Answer

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,

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

Putting values in above equation we get,

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

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

Find the equation of the line whose x-intercept is 6 and y-intercept is -4.

Answer

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.,

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

Putting values in above equation we get,

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

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

A(2, 5), B(-1, 2) and C(5, 8) are the vertices of a triangle ABC, 'M' is a point on AB such that AM : MB = 1 : 2. Find the coordinates of 'M'. Hence, find the equation of the line passing through C and M.

Answer

The triangle ABC is shown in the figure below:

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

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

Putting values we get,

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

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

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

Putting values in above equation we get,

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

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

Find the equation of the line passing through the point (1, 4) and intersecting the line x - 2y - 11 = 0 on the y-axis.

Answer

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 = -$\dfrac{11}{2}$

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

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

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

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

Putting values we get,

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

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

Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.

Answer

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

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

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.

Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2) find :

(i) the coordinates of the fourth vertex D.

(ii) length of diagonal BD.

(iii) equation of side AB of the parallelogram ABCD.

Answer

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

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

These two should be same. On equating we get,

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

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,

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

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.,

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

Putting values we get,

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

A and B are the two points on the x-axis and y-axis respectively. P(2, -3) is the mid-point of AB.

Find :

(i) the coordinates of A and B.

(ii) the slope of the line AB.

(iii) the equation of the line AB.

Answer

(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,

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

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

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

Putting values we get slope,

$= \dfrac{(-6 - 0)}{(0 - 4)} \\[1em] = \dfrac{-6}{-4} \\[1em] = \dfrac{3}{2}.$

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

(iii) Equation of AB will be

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

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

M and N are two points on the x-axis and y-axis respectively. P(3, 2) divides the line segment MN in the ratio 2 : 3. Find :

(i) the coordinates of M and N.

(ii) slope of the line MN.

Answer

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

By section formula the coordinates of P are,

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

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

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

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

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

Putting value in above equation we get slope,

$= \dfrac{5 - 0}{0 - 5} \\[1em] = -\dfrac{5}{5} \\[1em] = -1.$

Hence, the slope of the line is -1.

The line through P(5, 3) intersects y-axis at Q.

(i) Write the slope of the line.

(ii) Write the equation of the line.

(iii) Find the co-ordinates of Q.

Answer

(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) Write down the coordinates of point P that divides the line joining A(-4, 1) and B(17, 10) in the ratio 1 : 2.

(ii) Calculate the distance OP, where O is the origin.

(iii) In what ratio does the y-axis divide the line AB?

Answer

(i) By section formula coordinates of P are,

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

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

(ii) By distance formula

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

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

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,

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

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

Find the equations of the diagonals of a rectangle whose sides are x = -1, x = 2, y = -2 and y = 6.

Answer

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.,

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

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

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

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

Find the equation of the straight line passing through the origin and through the point of intersection of the lines 5x + 7y = 3 and 2x - 3y = 7.

Answer

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.,

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

Putting values in above equation we get,

$\Rightarrow y - (-1) = \dfrac{0 - (-1)}{0 - 2}(x - 2) \\[1em] \Rightarrow y + 1 = \dfrac{1}{-2}(x - 2) \\[1em] \Rightarrow -2(y + 1) = x - 2 \\[1em] \Rightarrow -2y - 2 = x - 2 \\[1em] \Rightarrow x + 2y - 2 + 2 = 0 \\[1em] \Rightarrow x + 2y = 0.$

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

State which one of the following is true :
The straight lines y = 3x - 5 and 2y = 4x + 7 are

(i) parallel

(ii) perpendicular

(iii) neither parallel nor perpendicular.

Answer

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

Comparing y = 3x - 5 and y = 2x + $\dfrac{7}{2}$ 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.

Hence, (iii) is the correct option.

If 6x + 5y - 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

Answer

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

⇒ 6x + 5y - 7 = 0

⇒ 5y = -6x + 7

⇒ y = $-\dfrac{6}{5}x + \dfrac{7}{5}$

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

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

⇒ 2px + 5y + 1 = 0

⇒ 5y = -2px - 1

⇒ y = $-\dfrac{2p}{5}x - \dfrac{1}{5}$

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

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

m₁ = m₂

$\Rightarrow -\dfrac{6}{5} = -\dfrac{2p}{5} \\[1em] \Rightarrow 2p = 6 \\[1em] \Rightarrow p = 3.$

Hence, the value of p = 3.

If the straight lines 3x - 5y + 7 = 0 and 4x + ay + 9 = 0 are perpendicular to one another, find the value of a.

Answer

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

⇒ 3x - 5y + 7 = 0

⇒ 5y = 3x + 7

⇒ y = $\dfrac{3}{5}x + \dfrac{7}{5}$

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

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

⇒ 4x + ay + 9 = 0

⇒ ay = -4x - 9

⇒ y = $-\dfrac{4}{a}x - \dfrac{9}{a}$

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

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

m₁.m₂ = -1

$\Rightarrow \dfrac{3}{5} \times -\dfrac{4}{a} = -1 \\[1em] \Rightarrow -\dfrac{12}{5a} = -1 \\[1em] \Rightarrow a = \dfrac{12}{5}.$

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

If the lines 3x + by + 5 = 0 and ax - 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Answer

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 = $-\dfrac{3}{\text{b}}\text{x} - \dfrac{5}{\text{b}}$.

Comparing with y = mx + c we get,

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

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

⇒ 5y = ax + 7

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

Comparing with y = mx + c we get,

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

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

∴ m₁.m₂ = -1.

$\Rightarrow -\dfrac{3}{b} \times \dfrac{a}{5} = -1 \\[1em] \Rightarrow -\dfrac{3a}{5b} = -1 \\[1em] \Rightarrow -3a = -5b \\[1em] \Rightarrow 3a = 5b.$

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

Is the line through (-2, 3) and (4, 1) perpendicular to the line 3x = y + 1? Does the line 3x = y + 1 bisect the join of (-2, 3) and (4, 1)?

Answer

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

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

Putting values in above formula we get,

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

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

slope = m₁ = $-\dfrac{1}{3}$

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,

$= m_1 \times m_2 \\[1em] = -\dfrac{1}{3} \times 3 \\[1em] = -1.$

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.,

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

The line through A(-2, 3) and B(4, b) is perpendicular to the line 2x - 4y = 5. Find the value of b.

Answer

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

Slope of line passing through A and B is m₁,

$= \dfrac{b - 3}{4 - (-2)} \\[1em] = \dfrac{b - 3}{6}.$

The equation of other line is,

⇒ 2x - 4y = 5

⇒ 4y = 2x - 5

⇒ y = $\dfrac{2}{4}x - \dfrac{5}{4}$

⇒ y = $\dfrac{1}{2}x - \dfrac{5}{4}$

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

slope = m₂ = $\dfrac{1}{2}$

As lines are perpendicular to each other, we have

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

Hence, the value of b is -9.

If the lines 3x + y = 4, x - ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the values of a and b.

Answer

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 = $\dfrac{1}{a}x + \dfrac{7}{a}$

Comparing with y = mx + c we get,

slope = m₂ = $\dfrac{1}{a}$.

Slope of line (iii) is

⇒ bx + 2y + 5 = 0

⇒ 2y = -bx - 5

⇒ y = $-\dfrac{b}{2}x - \dfrac{5}{2}$

Comparing with y = mx + c we get,

slope = m₃ = $-\dfrac{b}{2}$.

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

⇒ m₁ × m₂ = -1

$\Rightarrow -3 \times \dfrac{1}{a} = -1 \\[1em] \Rightarrow a = \dfrac{-3}{-1} \\[1em] \Rightarrow a = 3.$

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

⇒ m₂ × m₃ = -1

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

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

Find the value of 'p' if the lines 5x - 3y + 2 = 0 and 6x - py + 7 = 0 are perpendicular to each other. Hence find the equation of a line passing through (-2, -1) and parallel to 6x - py + 7 = 0.

Answer

Given lines,

⇒ 5x - 3y + 2 = 0 and 6x - py + 7 = 0

⇒ 3y = 5x + 2 and py = 6x + 7

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

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

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

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

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

$\therefore \dfrac{5}{3} \times \dfrac{6}{p} = -1 \\[1em] \Rightarrow 5 \times \dfrac{2}{p} = -1 \\[1em] \Rightarrow \dfrac{10}{p} = -1 \\[1em] \Rightarrow p = -10 \\[1em]$

Given,

⇒ 6x - py + 7 = 0

⇒ 6x - (-10)y + 7 = 0

⇒ 6x + 10y + 7 = 0

⇒ 10y = -6x - 7

⇒ y = $-\dfrac{6}{10}x - \dfrac{7}{10}$

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

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

Since, parallel lines have equal slope.

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

By point-slope form,

⇒ y - y₁ = m(x - x₁)

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

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

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

Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x - 3y - 7 = 0. Find the coordinates of the point where it cuts the x-axis.

Answer

Given equation of line,

⇒ 2x - 3y - 7 = 0,

Converting it in the form y = mx + c,

⇒ 3y = 2x - 7

⇒ y = $\dfrac{2}{3}x - \dfrac{7}{3}$.

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

Since,the other line is parallel so, its slope will also be equal to $\dfrac{2}{3}$. 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 = $\dfrac{2}{3}x + 4$

⇒ y = $\dfrac{2x + 12}{3}$

⇒ 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).

Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept -3.

Answer

Given equation of line,

⇒ 2x + 5y + 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x - 7

⇒ y = $-\dfrac{2}{5}x - \dfrac{7}{5}$.

Comparing with y = mx + c we get,

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

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

⇒ m × m' = -1

$\Rightarrow -\dfrac{2}{5} \times m' = -1 \\[1em] \Rightarrow m' = \dfrac{5}{2}.$

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

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

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

Find the equation of a straight line perpendicular to the line 3x - 4y + 12 = 0 and having same y-intercept as 2x - y + 5 = 0.

Answer

Given equation of line,

⇒ 3x - 4y + 12 = 0

Converting it in the form y = mx + c,

⇒ 4y = 3x + 12

⇒ y = $\dfrac{3}{4}x + 3$.

Comparing with y = mx + c we get,

Slope (m₁) = $\dfrac{3}{4}$.

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

m₁ × m₂ = -1

$\Rightarrow \dfrac{3}{4} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -\dfrac{4}{3}.$

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 = $-\dfrac{4}{3}$ and y-intercept = 5.

Putting these values in y = mx + c,

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

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

Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.

Answer

Given equation of line,

⇒ 3x + 5y + 15 = 0

Converting it in the form y = mx + c,

⇒ 5y = -3x - 15

⇒ y = $-\dfrac{3}{5}x - 3$

Comparing with y = mx + c we get,

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

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

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

So, equation can be given by,

⇒ y - y₁ = m(x - x₁)

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

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

(i) The line 4x - 3y + 12 = 0 meets the x-axis at A. Write down the coordinates of A.

(ii) Determine the equation of the line passing through A and perpendicular to 4x - 3y + 12 = 0.

Answer

(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 = $\dfrac{4}{3}x + 4$

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

Slope (m₁) = $\dfrac{4}{3}$

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

∴ m₁ × m₂ = -1

$\Rightarrow \dfrac{4}{3} \times m_2 = -1 \\[1em] \Rightarrow m_2 = -\dfrac{3}{4}.$

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

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

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

Find the equation of the line that is parallel to 2x + 5y - 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and (-4, 1).

Answer

Given equation of line,

⇒ 2x + 5y - 7 = 0

Converting it in the form y = mx + c,

⇒ 5y = -2x + 7

⇒ y = $-\dfrac{2}{5}x + \dfrac{7}{5}$

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

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

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

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

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

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

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

Find the equation of the line that is perpendicular to 3x + 2y - 8 = 0 and passes through the mid-point of the line segment joining the points (5, -2) and (2, 2).

Answer

Given equation of line,

⇒ 3x + 2y - 8 = 0

Converting it in the form y = mx + c,

⇒ 2y = -3x + 8

⇒ y = $-\dfrac{3}{2}x + 4$

Comparing with y = mx + c we get,

Slope (m₁) = $-\dfrac{3}{2}$

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

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

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

Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{3}{2} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{2}{3}.$

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

y - y₁ = m(x - x₁)

Putting values we get,

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

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

Find the equation of a straight line passing through the intersection of 2x + 5y - 4 = 0 with x-axis and parallel to the line 3x - 7y + 8 = 0.

Answer

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 = $\dfrac{3}{7}x + \dfrac{8}{7}$

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

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

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

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

Line AB is perpendicular to line CD. Coordinates of B, C and D are (4, 0), (0, -1) and (4, 3) respectively. Find

(i) the slope of CD

(ii) the equation of line AB

Answer

(i) By formula,

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

Substituting values we get :

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

Find the equation of a line parallel to the line 2x + y - 7 = 0 and passing through the point of intersection of the lines x + y - 4 = 0 and 2x - y = 8.

Answer

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.

The equation of a line is 3x + 4y - 7 = 0. Find

(i) slope of the line.

(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.

Answer

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

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

⇒ 4y = -3x + 7

⇒ y = $-\dfrac{3}{4}x + \dfrac{7}{4}$

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

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

Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{3}{4} \times m_2 = -1 \\[1em] \Rightarrow m_2 = \dfrac{4}{3}.$

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 $\dfrac{4}{3}$ and passing through (2, 4) can be given by point-slope form,

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

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