Equation of a Straight Line — Multiple Choice Questions

The slope of a line parallel to y-axis is

1. 0

2. 1

3. -1

4. not defined

Answer

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.

The slope of a line which makes an angle of 30° with the positive direction of x-axis is

1. 1

2. $\dfrac{1}{\sqrt{3}}$

3. $\sqrt{3}$

4. $-\dfrac{1}{\sqrt{3}}$

Answer

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

Hence, Option 2 is the correct option.

The slope of the line passing through the points (0, -4) and (-6, 2) is

1. 0

2. 1

3. -1

4. 6

Answer

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

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

Hence, Option 3 is the correct option.

The slope of the line passing through the points (3, -2) and (-7, -2) is

1. 0

2. 1

3. -$\dfrac{1}{10}$

4. not defined

Answer

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

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

Hence, Option 1 is the correct option.

The slope of the line passing through the points (3, -2) and (3, -4) is

1. -2

2. 0

3. 1

4. not defined

Answer

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

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

= not defined.

Hence, Option 4 is the correct option.

The inclination of the line y = $\sqrt{3}x - 5$ is

1. 30°

2. 60°

3. 45°

4. 0°

Answer

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

m = $\sqrt{3}$.

Slope is given by m = tan θ or,

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

Hence, Option 2 is the correct option.

If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is

1. -2

2. -1

3. 1

4. 2

Answer

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

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

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

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

Hence, Option 3 is the correct option.

The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is

1. $\dfrac{3}{7}$

2. -$\dfrac{3}{7}$

3. $\dfrac{7}{3}$

4. -$\dfrac{7}{3}$

Answer

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

$= \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{3 - 6}{7 - 0} \\[1em] = -\dfrac{3}{7} \\[1em]$

Hence, Option 2 is the correct option.

The slope of a line perpendicular to the line passing through the points (2, 5) and (-3, 6) is

1. $-\dfrac{1}{5}$

2. $\dfrac{1}{5}$

3. -5

4. 5

Answer

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

$= \dfrac{y_2 - y_1}{x_2 - x1} \\[1em] = \dfrac{6 - 5}{-3 - 2} \\[1em] = -\dfrac{1}{5}.$

Let slope of perpendicular line be m₂. Then,

$\Rightarrow m_1 \times m_2 = -1 \\[1em] \Rightarrow -\dfrac{1}{5} \times m_2 = -1 \\[1em] \Rightarrow m_2 = 5.$

∴ Slope of line perpendicular to this line = 5.

Hence, Option 4 is the correct option.

The slope of a line parallel to the line 2x + 3y - 7 = 0 is

1. $-\dfrac{2}{3}$

2. $\dfrac{2}{3}$

3. $-\dfrac{3}{2}$

4. $\dfrac{3}{2}$

Answer

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

⇒ 3y = -2x + 7

⇒ y = $-\dfrac{2}{3}x + \dfrac{7}{3}$.

Comparing with y = mx + c we get,

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

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

Hence, Option 1 is the correct option.

The slope of a line perpendicular to the line 3x = 4y + 11 is

1. $\dfrac{3}{4}$

2. $-\dfrac{3}{4}$

3. $\dfrac{4}{3}$

4. $-\dfrac{4}{3}$

Answer

Given, 3x = 4y + 11.

⇒ 4y = 3x - 11

⇒ y = $\dfrac{3}{4}x - \dfrac{11}{4}$.

Comparing with y = mx + c we get,

Slope (m₁) = $\dfrac{3}{4}$.

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

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

Hence, Option 4 is the correct option.

If the lines 2x + 3y = 5 and kx - 6y = 7 are parallel, then the value of k is

1. 4

2. -4

3. $\dfrac{1}{4}$

4. -$\dfrac{1}{4}$

Answer

Given,

⇒ 2x + 3y = 5 and kx - 6y = 7

⇒ 3y = -2x + 5 and 6y = kx - 7

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

Comparing both the equations with y = mx + c,

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

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

Since, both the lines are parallel so,

m₁ = m₂

$\Rightarrow -\dfrac{2}{3} = \dfrac{k}{6} \\[1em] \Rightarrow k = -\dfrac{2}{3} \times 6 \\[1em] \Rightarrow k = -4.$

Hence, Option 2 is the correct option.

If the line 3x - 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is

1. $\dfrac{3}{2}$

2. $-\dfrac{3}{2}$

3. $\dfrac{2}{3}$

4. $-\dfrac{2}{3}$

Answer

Given,

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

⇒ 4y = 3x + 7 and ky = -2x - 5

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

Comparing both the equations with y = mx + c,

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

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

Since, both the lines are perpendicular so,

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

Hence, Option 1 is the correct option.

Which of the following equations represents a line passing through origin ?

1. 3x - 2y + 5 = 0

2. 2x - 3y = 0

3. x = 5

4. y = -6

Answer

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.

Points A(x, y), B(3, -2) and C(4, -5) are collinear. The value of y in terms of x is ∶

1. 3x - 11

2. 11 - 3x

3. 3x - 7

4. 7 - 3x

Answer

Since, points A, B and C are collinear.

∴ Slope of AB = Slope of BC.

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

Hence, Option 4 is the correct option.

Which of the following equation represents a line equally inclined to the axes ?

1. 2x - 3y + 7 = 0

2. x - y = 7

3. x = 7

4. y = -7

Answer

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.