Equation of a Straight Line — Exercise 12.1

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.