Equation of a Straight Line — Multiple Choice Questions

The inclination of every line parallel to the x-axis is:

1. 0°

2. 90°

3. 180°

4. none of these

Answer

The angle that a horizontal line makes with the x-axis is 0°.

The angle between the line and the x-axis is 0°.

Hence, option 1 is the correct option.

The inclination of y-axis is:

1. 0°

2. 45°

3. 90°

4. 180°

Answer

A vertical line makes a right angle with the horizontal x-axis.

Therefore, the angle of inclination of the y-axis is 90°.

Hence, option 3 is the correct option.

The slope of a line whose inclination is 30°, is:

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

2. $\sqrt{3}$

3. 1

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

Answer

The slope (m) of a line is related to its inclination (θ) by the formula:

m = tan θ

m = tan 30° = $\dfrac{1}{\sqrt3}$.

Hence, option 4 is the correct option.

The slope of a horizontal line is:

1. 0

2. 1

3. 2

4. not defined

Answer

A horizontal line is a line parallel to the x-axis.

Thus θ = 0°

We know that,

m = tan θ

m = tan 0° = 0

Hence, option 1 is the correct option.

The slope of a vertical line is:

1. 0

2. 1

3. 90

4. not defined

Answer

A vertical line is perpendicular to x-axis.

Thus, θ = 90°.

We know that,

m = tan θ

m = tan 90° = not defined

Hence, option 4 is the correct option.

The slope of a line passing through two given points A(x₁, y₁) and B(x₂, y₂) is given by:

1. $m = \dfrac{x_2 - x_1}{y_2 - y_1}$

2. $m = \dfrac{y_2 \times y_1}{x_2 \times x_1}$

3. $m = \dfrac{x_2 \times x_1}{y_2 \times y_1}$

4. $m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Answer

The correct formula for slope is: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Hence, option 4 is the correct option.

Which of the following lines does not have x-intercept?

1. A horizontal line

2. A vertical line

3. An oblique line

4. None of these

Answer

A horizontal line has equation : y = c.

It does not crosses x-axis at any point, thus x-intercept = 0.

Hence, option 1 is the correct option.

Which of the following lines does not have y-intercept?

1. A horizontal line

2. A vertical line

3. y-axis

4. A transverse line

Answer

A y-intercept is the point where a line crosses the y-axis.

A vertical line: A vertical line has the equation x = a.

If a = 0 , it has an infinite number of y-intercepts.

If a ≠ 0 , the line is parallel to the y-axis and never crosses it. Therefore, it has no y-intercept.

Hence, option 2 is the correct option.

The equation of x-axis is:

1. x = 0

2. y = 0

3. x = a

4. y = a

Answer

The equation of x-axis is :

y = 0

Every point on the x-axis has a y-coordinate of 0, regardless of its x-coordinate.

Hence, option 2 is the correct option.

The equation of y-axis is :

1. x = 0

2. y = 0

3. x = a

4. y = a

Answer

The equation of y-axis is:

x = 0

Every point on the y-axis has an x-coordinate of 0, regardless of its y-coordinate.

Hence, option 1 is the correct option.

The equation of a line parallel to x-axis and at a distance of 5 units below it, is:

1. x – 5 = 0

2. x + 5 = 0

3. y – 5 = 0

4. y + 5 = 0

Answer

Any line parallel to the x-axis is a horizontal line and has the equation of form y = k.

The distance is 5 units below x-axis, thus the y-coordinate (k) of every point on the line is -5.

y = -5

y + 5 = 0.

Hence, option 4 is the correct option.

The equation of a line parallel to y-axis and at a distance of 8 units to the right of it, is:

1. x – 8 = 0

2. x + 8 = 0

3. y – 5 = 0

4. y + 5 = 0

Answer

Any line parallel to the y-axis is a vertical line and has the equation form x = k.

The distance is 8 units to the right, thus the x-coordinate (k) of every point on the line is +8.

x = 8

x - 8 = 0

Hence, option 1 is the correct option.

The equation of a line with slope m and y-intercept c is given by:

1. x = my + c

2. cy = mx

3. y = mx + c

4. none of these

Answer

y = mx + c, is the standard form of a linear equation known as the slope-intercept form.

Hence, option 3 is the correct option.

The slope m of a line whose inclination is α, is given by :

1. sin α

2. cos α

3. tan α

4. cot α

Answer

The slope (m) of a line is defined as the tangent of its angle of inclination.

m = tan α

Hence, option 3 is the correct option.

The equation of a line with slope m and passing through a point P(a, b), is given by :

1. (y – a) = m(x – b)

2. (x – b) = m(y – a)

3. (x – a) = m(y – b)

4. (y – b) = m(x – a)

Answer

By point-slope form,

Equation of line :

y - y₁ = m (x - x₁)

Thus, equation of line passing through (a, b) and slope m is :

y - b = m(x - a)

Hence, option 4 is the correct option.

The equation of a line passing through two points A(x₁, y₁) and B(x₂, y₂) is given by:

1. (x – x₁) = $\dfrac{y_2 - y_1}{x_2 - x_1}$ (y – y₁)

2. (y – y₁) = $\dfrac{x_2 - x_1}{y_2 - y_1}$ (x – x₁)

3. (y – y₁) = $\dfrac{y_2 - y_1}{x_2 - x_1}$ (x – x₁)

4. (y – y₁) = $\dfrac{y_1 - y_2}{x_2 - x_1}$ (x – x₁)

Answer

By two-point formula,

Equation of line :

(y – y₁) = $\dfrac{y_2 - y_1}{x_2 - x_1}$ (x – x₁)

Hence, option 3 is the correct option.

The equation of the line passing through origin and parallel to the line 3x + 4y + 7 = 0 is:

1. 3x + 4y + 5 = 0

2. 4x - 3y - 5 = 0

3. 4x - 3y = 0

4. 3x + 4y = 0

Answer

Given,

⇒ 3x + 4y + 7 = 0

⇒ 4y = -3x - 7

⇒ y = $-\dfrac{3}{4}x - \dfrac{7}{4}$

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

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

By point-slope formula,

Equation of line : y - y₁ = m(x - x₁)

Thus, equation of line parallel to 3x + 4y + 7 = 0 and passing through origin (0, 0) is :

⇒ y - 0 = $-\dfrac{3}{4}$ (x - 0)

⇒ y = $-\dfrac{3x}{4}$

⇒ 4y = -3x

⇒ 4y + 3x = 0.

Hence, option 4 is the correct option.

The gradient of the line passing through the points A(–3, 4) and B(2, –6) is:

1. –$\dfrac{1}{2}$

2. –2

3. 2

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

Answer

We know that,

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] m_{AB} = \dfrac{-6 - 4}{2-(-3)} \\[1em] = \dfrac{-10}{5} = -2.$

Hence, option 2 is the correct option.

The equation of a straight line whose inclination with x-axis is 30° and whose y-intercept is –4, is:

1. $\sqrt{3}x - y - 4\sqrt{3} = 0$

2. $x + \sqrt{3}y - 4\sqrt{3} = 0$

3. $x - \sqrt{3}y - 4\sqrt{3} = 0$

4. $\sqrt{3}x + y - 4\sqrt{3} = 0$

Answer

The slope m is determined by the inclination θ = 30°.

m = tan θ

m = tan 30°

m = $\dfrac{1}{\sqrt3}$

Slope-intercept form:

y = mx + c

$\Rightarrow y = \dfrac{1}{\sqrt3}x - 4 \\[1em] \Rightarrow \sqrt3y = x - 4\sqrt3 \\[1em] \Rightarrow x - \sqrt3y - 4\sqrt3 = 0.$

Hence, option 3 is the correct option.

The slope of the straight line passing through the points A(3, –2) and B(3, –4) is:

1. 0

2. 1

3. –2

4. not defined

Answer

We know that,

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

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

Slope is not defined.

Hence, option 4 is the correct option.

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

1. 0°

2. 30°

3. 45°

4. 60°

Answer

Comparing equation, y = $\sqrt{3}$ x – 5 with y = mx + c, we get :

⇒ m = $\sqrt3$

⇒ tan θ = $\sqrt{3}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, option 4 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. 0

2. –1

3. 1

4. 2

Answer

We know that,

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

Substitute values we get:

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

Hence, option 3 is the correct option.

What can be said regarding a straight line if its slope is negative?

1. Inclination, θ is an acute angle.

2. Inclination, θ is an obtuse angle.

3. Either the line is x-axis or it is parallel to the x-axis.

4. Either the line is y-axis or it is parallel to the y-axis.

Answer

If the slope is negative, then tan θ must be negative.

tan θ < 0

This happens only when the angle θ made with the positive x-axis is between 90° and 180°.

And angles between 90° and 180° are called obtuse angles.

Hence, option 2 is the correct option.

The equation of a straight line passing through the point (–6, 8) and parallel to the x-axis is:

1. x – 6 = 0

2. x + 6 = 0

3. y – 8 = 0

4. y + 8 = 0

Answer

The equation of horizontal line is of the form y = c.

Since the line must pass through the point (-6, 8), its y-coordinate must be equal to 8.

y = 8

y - 8 = 0.

Hence, option 3 is the correct option.

The slope of the line, 3x – $2\sqrt2y$ – 4 = 0 is:

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

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

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

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

Answer

Solving,

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

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

m = $\dfrac{3}{2\sqrt2}$

Rationalizing the Denominator,

$m = \dfrac{3}{2\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\[1em] = \dfrac{3\sqrt{2}}{2 \times 2} \\[1em] = \dfrac{3\sqrt{2}}{4}.$

Hence, option 3 is the correct option.

The equation of the line passing through the points A(4, 3) and B(–2, 6) is :

1. x + 2y – 10 = 0

2. x – 2y – 6 = 0

3. x – 3y + 8 = 0

4. x + 2y – 6 = 0

Answer

We know that,

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] m_{AB} = \dfrac{6 - 3}{-2 - 4} \\[1em] = \dfrac{3}{-6} \\[1em] = -\dfrac{1}{2}.$

By point-slope form :

⇒ y - y₁ = m(x - x₁)

⇒ y - 3 = $-\dfrac{1}{2}$ (x - 4)

⇒ 2(y - 3) = -1(x - 4)

⇒ 2y - 6 = -x + 4

⇒ x + 2y - 6 - 4 = 0

⇒ x + 2y - 10 = 0.

Hence, option 1 is the correct option.

The equation of a line passing through the point (5, –3) and having the y-intercept of 8 units below the x-axis is:

1. x + y – 8 = 0

2. x – y – 8 = 0

3. 2x + y – 4 = 0

4. x – 2y – 8 = 0

Answer

The line intersects the y-axis 8 units below the x-axis. This means the line intersect y-axis at (0, -8).

Thus, y-intercept of lie (c) = -8

Thus, line passes through (0, -8) and (5, -3).

We know that,

$m = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-3 - (-8)}{5 - 0} \\[1em] = \dfrac{5}{5} \\[1em] = 1.$

Substitute m = 1 and c = -8 into y = mx + c, we get :

⇒ y = 1.x + (-8)

⇒ y = x - 8

⇒ x - y - 8 = 0.

Hence, option 2 is the correct option.

The value of m such that the points A(5, –2), B(8, –3) and C(m, –12) are collinear, is:

1. 29

2. 33

3. 35

4. 41

Answer

For three points A, B, and C to be collinear.

m_AB = m_BC

We know that,

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

Using A(5, -2) and B(8, -3) :

$m_{AB} = \dfrac{-3 - (-2)}{8 - 5} \\[1em] = \dfrac{-1}{3}.$

Using B(8, -3) and C(m, -12):

$m_{BC} = \dfrac{-12 - (-3)}{m - 8} \\[1em] = \dfrac{-9}{m - 8}$

m_AB = m_BC

⇒ $\dfrac{-1}{3} = \dfrac{-9}{m - 8}$

⇒ -1(m - 8) = -9 × 3

⇒ -m + 8 = -27

⇒ -m = -27 - 8

⇒ -m = -35

⇒ m = 35.

Hence, option 3 is the correct option.

The equation of the straight line passing through the point (9, –9) and parallel to the y-axis is:

1. y – 9 = 0

2. y + 9 = 0

3. x + 9 = 0

4. x – 9 = 0

Answer

The equation of line parallel to y-axis is :

x = c

Since the line must pass through the point (9, -9), its x-coordinate must be x = 9.

x - 9 = 0

Hence, option 4 is the correct option.

Two non-vertical lines with slopes m₁ and m₂ are parallel to each other, if:

1. m₁m₂ = 1

2. m₁m₂ = –1

3. m₁ = –m₂

4. m₁ = m₂

Answer

Two distinct non-vertical lines are parallel if and only if their slopes are equal.

m₁ = m₂

Hence, option 4 is the correct option.

Two non-vertical lines with slopes m₁ and m₂ are perpendicular to each other, if:

1. m₁.m₂ = 1

2. m₁.m₂ = –1

3. m₁ = –m₂

4. m₁ = m₂

Answer

Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

m₁.m₂ = –1

Hence, option 2 is the correct option.

The slope of a line parallel to the line passing through the points A(3, –7) and B(5, –7) is:

1. 2

2. 1

3. –1

4. 0

Answer

Given, points A(3, –7) and B(5, –7)

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

Substitute values we get,

$m_{AB} = \dfrac{-7 - (-7)}{5 - 3} \\[1em] = \dfrac{-7 + 7}{2} \\[1em] = \dfrac{0}{2} = 0.$

As, slope of parallel lines are equal.

Thus, slope of line parallel to AB = 0.

Hence, option 4 is the correct option.

The slope of a line perpendicular to the line passing through the points P(3, –4) and Q(1, –8) is:

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

2. 2

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

4. –2

Answer

Slope of the line passing through P(3, –4) and Q(1, –8).

$m_{PQ} = \dfrac{y_2 - y_1}{x_2 - x_1} \\[1em] = \dfrac{-8 - (-4)}{1 - 3} \\[1em] = \dfrac{-8 + 4}{-2} \\[1em] = \dfrac{-4}{-2} = 2.$

Since the required line is perpendicular to line PQ, the product of their slopes must be -1.

Let slope of required line be m.

⇒ m_PQ × m = -1

⇒ 2 × m = -1

⇒ m = $\dfrac{-1}{2}$

Hence, option 1 is the correct option.

The slope of a line parallel to the line, 3x – 5y + 8 = 0, is:

1. $\dfrac{-5}{3}$

2. $\dfrac{5}{3}$

3. $–\dfrac{3}{5}$

4. $\dfrac{3}{5}$

Answer

Solving,

⇒ 3x - 5y + 8 = 0

⇒ 3x + 8 = 5y

⇒ $y = \dfrac{3}{5}x + \dfrac{8}{5}$.

The slope of the given line is $m = \dfrac{3}{5}$.

Since the required line is parallel, its slope must be the same as m.

Let slope of required line be m₁,

m₁ = m = $\dfrac{3}{5}$.

Hence, option 4 is the correct option.

The slope of a line perpendicular to the line, 3x = 4y – 10, is :

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

2. $–\dfrac{4}{3}$

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

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

Answer

Solving,

⇒ 3x = 4y - 10

⇒ 4y = 3x + 10

$\Rightarrow y = \dfrac{3}{4}x + \dfrac{10}{4}$.

The slope of the given line is $m = \dfrac{3}{4}$.

Since the required line is perpendicular, the product of the slopes must be -1.

Let slope of required line be a, then :

⇒ m × a = -1

⇒ $\dfrac{3}{4}$ × a = -1

⇒ a = $\dfrac{-4}{3}$.

Hence, option 2 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. $–\dfrac{1}{4}$

3. 4

4. –5

Answer

For two lines to be parallel, their slopes must be equal.

Line 1 : 2x + 3y = 5

First, convert the equation 2x + 3y = 5 into the slope-intercept form, y = mx + c, to find its slope, m.

$3y = -2x + 5 \\[1em] y = -\dfrac{2}{3}x + \dfrac{5}{3} \\[1em] m_1 = -\dfrac{2}{3}$

Line 2: kx - 6y = 7

First, convert the equation kx - 6y = 7 into the slope-intercept form, y = mx + c, to find its slope, m.

$-6y = -kx + 7 \\[1em] y = \dfrac{-k}{-6}x + \dfrac{7}{-6} = \dfrac{k}{6}x - \dfrac{7}{6} \\[1em] m_2 = \dfrac{k}{6}$

The slopes of parallel lines are equal:

$\Rightarrow m_1 = m_2 \\[1em] \Rightarrow -\dfrac{2}{3} = \dfrac{k}{6} \\[1em] \Rightarrow k = -\dfrac{2 \times 6}{3} \\[1em] \Rightarrow k = -\dfrac{12}{3} \\[1em] \Rightarrow k = -4.$

Hence, option 1 is the correct option.

If the lines x – my + 3 = 0 and 2x + 3y – 7 = 0 are perpendicular to each other, then the value of m is:

1. $–\dfrac{2}{3}$

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

3. $–\dfrac{3}{2}$

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

Answer

For two lines to be perpendicular, the product of their slopes must be -1.

Line 1: x - my + 3 = 0

First, convert the equation x - my + 3 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$-my = -x - 3 \\[1em] y = \dfrac{-1}{-m}x + \dfrac{-3}{-m} = \dfrac{1}{m}x + \dfrac{3}{m} \\[1em] m_1 = \dfrac{1}{m}.$

Line 2: 2x + 3y - 7 = 0

First, convert the equation 2x + 3y - 7 = 0 into the slope-intercept form, y = mx + c, to find its slope, m.

$3y = -2x + 7 \\[1em] y = -\dfrac{2}{3}x + \dfrac{7}{3} \\[1em] m_2 = -\dfrac{2}{3}$

The product of slopes of perpendicular lines is equal to -1:

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

Hence, option 2 is the correct option.

The equation of the straight line passing through the point (1, 2) and parallel to the line y = 3x + 1, is:

1. x – 3y + 1 = 0

2. 3x + y + 1 = 0

3. 3x – y – 1 = 0

4. 3x – y + 1 = 0

Answer

The given line is $y = 3x + 1$, so its slope is m = 3.

Since the required line is parallel, its slope is also m = 3.

Using the point-slope form y - y₁ = m(x - x₁) with the point (x₁, y₁) = (1, 2) and m=3:

y - 2 = 3(x - 1)

y - 2 = 3x - 3

Rearranging the equation to the standard form (Ax + By + C = 0):

0 = 3x - y - 3 + 2

3x - y - 1 = 0

Hence, option 3 is the correct option.

The slope of the line, ax + by + c = 0, is:

1. $-\dfrac{a}{b}$

2. $-\dfrac{b}{a}$

3. $\dfrac{c}{b}$

4. $\dfrac{b}{a}$

Answer

Convert the general linear equation ax + by + c = 0 into the slope-intercept form, y = mx + c, to find the slope, m.

by = -ax - c

$y = \Big(\dfrac{-a}{b}\Big)x - \dfrac{c}{b}$

The slope, m, is the coefficient of x.

m = $-\dfrac{a}{b}$

Hence, option 1 is the correct option.

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

1. y = -7

2. x = 7

3. x - y = 7

4. 2x - 3y + 7 = 0

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 3 is the correct option.

In the given figure line l₁ is a parallel to line l₂. If line l₃ is perpendicular to line l₁, then the slopes of lines l₂ and l₃ respectively are :

1. 1, 1

2. -1, -1

3. 1, -1

4. -1, 1

Answer

Slope of line l₁ = tan 45° = 1.

We know that,

Slope of parallel lines are equal.

Slope of line l₂ = Slope of line l₁ = 1.

We know that,

Product of slope of perpendicular lines is equal to -1.

⇒ Slope of line l₂ × Slope of line l₃ = -1

⇒ 1 × Slope of line l₃ = -1

⇒ Slope of line l₃ = -1.

Hence, Option 3 is the correct option.

Which of the following lines cut the positive x-axis and positive y-axis at equal distances from the origin?

1. 3x + 3y = 6

2. 5x + 10y = 10

3. -x + y = 1

4. 10x + 5y = 5

Answer

Substituting x = 0 in first equation,

⇒ 3(0) + 3y = 6

⇒ 3y = 6

⇒ y = $\dfrac{6}{3}$

⇒ y = 2.

The line touches y-axis at point (0, 2).

Substituting y = 0 in first equation,

⇒ 3x + 3(0) = 6

⇒ 3x = 6

⇒ x = $\dfrac{6}{3}$

⇒ x = 2.

The line touches y-axis at point (2, 0).

∴ Line 3x + 3y = 6 cuts positive x-axis and positive y-axis at equal distance i.e. 2 units form the origin.

Hence, Option 1 is the correct option.

In the given diagram, O is the origin and P is the mid-point of AB. The equation of OP is :

1. y = x

2. 2y = x

3. y = 2x

4. y = -x

Answer

From graph,

A = (4, 0) and B = (0, 2).

Given,

P is the mid-point of AB.

P = $\Big(\dfrac{4 + 0}{2}, \dfrac{0 + 2}{2}\Big) = \Big(\dfrac{4}{2}, \dfrac{2}{2}\Big)$ = (2, 1).

By two-point form,

⇒ y - y₁ = $\dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$

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

⇒ y - 1 = $\dfrac{1}{2}(x - 2)$

⇒ 2(y - 1) = x - 2

⇒ 2y - 2 = x - 2

⇒ 2y = x - 2 + 2

⇒ 2y = x.

Hence, Option 2 is the correct option.