Parallel and Perpendicular Lines

What you'll learn

How to read the gradient from a straight-line equation.
How to write equations of parallel lines.
How to write equations of perpendicular lines.
How to use two points, or an unknown value like $k$ , to find a line’s gradient.

1. Start with $y = m x + c$

Most questions in this topic are about comparing gradients. So your first job is often to get the equation into the form $y = m x + c$ .

Definition

Gradient and y-intercept

The gradient tells you the steepness and direction of a straight line. In $y = m x + c$ , the gradient is $m$ .

A line in the form y = mx + c, showing the gradient as rise divided by run and the y-intercept c.

The y-intercept is where the line crosses the y-axis. In $y = m x + c$ , the y-intercept is $c$ .

“Make $y$ the subject” means rearrange the equation so that $y$ is on its own.

Example

Finding the gradient

Start with the equation $6 y = 3 x - 12$ .
Divide every term by 6 to make $y$ the subject: $y=12x−2y=\frac{1}{2}x-2$ .
Compare it with $y = m x + c$ .
The gradient is $m=12m=\frac{1}{2}$ and the y-intercept is $c = - 2$ .

Common Mistake

Reading the wrong coefficient

In $2 x + 5 y = 10$ , the coefficient of $x$ is not the gradient yet. Rearrange first until $y$ is on its own.

2. Parallel lines

Definition

Parallel lines

Parallel lines are straight lines that never meet. In $y = m x + c$ form, parallel lines have the same gradient.

Parallel lines have the same steepness but different y-intercepts.

The y-intercept can change, but the gradient must stay the same.

Example

Parallel line through the y-axis

The line $y = - 3 x + 8$ has gradient $m = - 3$ .
A parallel line has the same gradient, so start with $y = - 3 x + c$ .
Passing through (0, 5) means the y-intercept is 5, so $c = 5$ .
The equation is $y = - 3 x + 5$ .

Tip

When it says “a line”

If no point is given, there are many correct answers. For example, a line parallel to $y = 3 x + 1$ could be $y = 3 x - 4$ .

3. Perpendicular lines

Definition

Perpendicular lines

Perpendicular lines meet at a right angle, 90°. For usual straight lines in $y = m x + c$ form, their gradients are negative reciprocals, meaning you flip the fraction and change the sign. Their gradients satisfy $m_1m_2=-1$ .

Perpendicular lines meet at 90°, and their gradients are negative reciprocals.

For example:

$m = 4$ becomes $m=−14m=-\frac{1}{4}$ .
$m=−23m=-\frac{2}{3}$ becomes $m=32m=\frac{3}{2}$ .

Example

Perpendicular line through the y-axis

The line $y=14x−2y=\frac{1}{4}x-2$ has gradient $m=14m=\frac{1}{4}$ .
A perpendicular line has gradient $m = - 4$ .
Passing through (0, -6) means the y-intercept is -6, so $c = - 6$ .
The equation is $y = - 4 x - 6$ .

Common Mistake

Only doing half the reciprocal rule

For perpendicular gradients, you must do both parts: flip the fraction and change the sign.

Common Mistake

Horizontal and vertical lines

The negative-reciprocal rule assumes both lines can be written as $y = m x + c$ . A horizontal line like $y = 4$ is perpendicular to a vertical line like $x = - 1$ .

A horizontal line and a vertical line are perpendicular, even though the vertical line is not in y = mx + c form.

4. If the equation is not already in $y = m x + c$

Sometimes the given line is written in a different form, such as $3 x + 6 y = 12$ . Always rearrange first, then use the parallel or perpendicular rule.

Example

Rearrange first, then use the rule

Rearrange $3 x + 6 y = 12$ by subtracting $3 x$ : $6 y = - 3 x + 12$ .
Divide by 6: $y=−12x+2y=-\frac{1}{2}x+2$ .
The given line has gradient $m=−12m=-\frac{1}{2}$ .
A perpendicular line has gradient $m = 2$ .
Passing through (0, 7) gives $c = 7$ .
The equation is $y = 2 x + 7$ .

5. Using two points to find a gradient

If a line is described using two points, calculate its gradient first.

Definition

Gradient between two points

For points $x_1,y_1)$ and $x_2,y_2)$ , the gradient is $m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}$ : change in $y$ divided by change in $x$ .

Gradient between two points is the vertical change divided by the horizontal change.

Once you know the gradient, use $y = m x + c$ . To substitute a point means to replace $x$ and $y$ with that point’s coordinates.

Example

Parallel line from two points

Line A passes through (1, 2) and (5, 14), so its gradient is $m=14−25−1=3m=\frac{14-2}{5-1}=3$ .

Line A’s two points show a rise of 12 and a run of 4, giving gradient 3; the required parallel line has the same slope through (2, 3).

A parallel line also has gradient $m = 3$ .
Start with $y = 3 x + c$ .
Use the point (2, 3): $3 = 3 (2) + c$ .
Solve to get $c = - 3$ .
The equation is $y = 3 x - 3$ .

Example

Perpendicular line from two points

Line A passes through (0, 4) and (6, 7), so its gradient is $m=7−46−0=12m=\frac{7-4}{6-0}=\frac{1}{2}$ .

The line through (1, 5) is perpendicular to Line A, so it crosses at a right angle with the negative reciprocal gradient.

A perpendicular line has gradient $m = - 2$ .
Start with $y = - 2 x + c$ .
Use the point (1, 5): $5 = - 2 (1) + c$ .
Solve to get $c = 7$ .
The equation is $y = - 2 x + 7$ .

Common Mistake

Gradient fraction upside down

Gradient is change in $y$ over change in $x$ . Keep the order of the two points consistent in the numerator and denominator.

6. Showing lines are parallel or perpendicular

When a question says “show that”, you need evidence. Calculate both gradients and then make a clear statement.

Example

Showing two lines are parallel

Line A passes through (1, 2) and (3, 8), so $mA=8−23−1=3m_A=\frac{8-2}{3-1}=3$ .
Line B passes through (-2, 1) and (0, 7), so $mB=7−10−(−2)=3m_B=\frac{7-1}{0-(-2)}=3$ .
The gradients are equal, so the two lines are parallel.

For perpendicular lines, show that the gradients multiply to -1.

7. Finding an unknown value like $k$

If a coordinate contains $k$ , write the gradient using $k$ , then set it equal to the gradient you need.

Example

Finding $k$ using perpendicular gradients

Line A passes through (-1, 2) and (3, 4), so $mA=4−23−(−1)=12m_A=\frac{4-2}{3-(-1)}=\frac{1}{2}$ .
A perpendicular line must have gradient $m = - 2$ .
Line B passes through (2, 5) and ( $k$ , -1), so its gradient is $mB=−1−5k−2=−6k−2m_B=\frac{-1-5}{k-2}=\frac{-6}{k-2}$ .

Line B uses an unknown x-coordinate k, so its horizontal change is k - 2 while its vertical change is -6.

Set this equal to -2: $−6k−2=−2\frac{-6}{k-2}=-2$ .
Solve: $- 6 = - 2 (k - 2)$ , so $- 6 = - 2 k + 4$ , giving $k = 5$ .

Exam technique

In the exam

Rearrange into $y = m x + c$ before deciding on the gradient.
For parallel lines, keep the same gradient; for perpendicular lines, flip the fraction and change the sign.
If points are given, calculate $xm=\frac{\text{change in }y}{\text{change in }x}$ carefully, then state your conclusion clearly.

Self review

Check yourself

What gradient would a line perpendicular to $y=35x−2y=\frac{3}{5}x-2$ have?
Can you rearrange $2 x + 3 y = 9$ into $y = m x + c$ and identify the gradient?
If two lines have gradients $m = 4$ and $m=−14m=-\frac{1}{4}$ , what is their relationship?

Parallel and Perpendicular Lines

What you'll learn

How to read the gradient from a straight-line equation.
How to write equations of parallel lines.
How to write equations of perpendicular lines.
How to use two points, or an unknown value like $k$ , to find a line’s gradient.

1. Start with $y = m x + c$

Most questions in this topic are about comparing gradients. So your first job is often to get the equation into the form $y = m x + c$ .

Definition

Gradient and y-intercept

The gradient tells you the steepness and direction of a straight line. In $y = m x + c$ , the gradient is $m$ .

A line in the form y = mx + c, showing the gradient as rise divided by run and the y-intercept c.

The y-intercept is where the line crosses the y-axis. In $y = m x + c$ , the y-intercept is $c$ .

“Make $y$ the subject” means rearrange the equation so that $y$ is on its own.

Example

Finding the gradient

Start with the equation $6 y = 3 x - 12$ .
Divide every term by 6 to make $y$ the subject: $y=12x−2y=\frac{1}{2}x-2$ .
Compare it with $y = m x + c$ .
The gradient is $m=12m=\frac{1}{2}$ and the y-intercept is $c = - 2$ .

Common Mistake

Reading the wrong coefficient

In $2 x + 5 y = 10$ , the coefficient of $x$ is not the gradient yet. Rearrange first until $y$ is on its own.

2. Parallel lines

Definition

Parallel lines

Parallel lines are straight lines that never meet. In $y = m x + c$ form, parallel lines have the same gradient.

Parallel lines have the same steepness but different y-intercepts.

The y-intercept can change, but the gradient must stay the same.

Example

Parallel line through the y-axis

The line $y = - 3 x + 8$ has gradient $m = - 3$ .
A parallel line has the same gradient, so start with $y = - 3 x + c$ .
Passing through (0, 5) means the y-intercept is 5, so $c = 5$ .
The equation is $y = - 3 x + 5$ .

Tip

When it says “a line”

If no point is given, there are many correct answers. For example, a line parallel to $y = 3 x + 1$ could be $y = 3 x - 4$ .

3. Perpendicular lines

Definition

Perpendicular lines

Perpendicular lines meet at 90°, and their gradients are negative reciprocals.

For example:

$m = 4$ becomes $m=−14m=-\frac{1}{4}$ .
$m=−23m=-\frac{2}{3}$ becomes $m=32m=\frac{3}{2}$ .

Example

Perpendicular line through the y-axis

The line $y=14x−2y=\frac{1}{4}x-2$ has gradient $m=14m=\frac{1}{4}$ .
A perpendicular line has gradient $m = - 4$ .
Passing through (0, -6) means the y-intercept is -6, so $c = - 6$ .
The equation is $y = - 4 x - 6$ .

Common Mistake

Only doing half the reciprocal rule

For perpendicular gradients, you must do both parts: flip the fraction and change the sign.

Common Mistake

Horizontal and vertical lines

The negative-reciprocal rule assumes both lines can be written as $y = m x + c$ . A horizontal line like $y = 4$ is perpendicular to a vertical line like $x = - 1$ .

A horizontal line and a vertical line are perpendicular, even though the vertical line is not in y = mx + c form.

4. If the equation is not already in $y = m x + c$

Sometimes the given line is written in a different form, such as $3 x + 6 y = 12$ . Always rearrange first, then use the parallel or perpendicular rule.

Example

Rearrange first, then use the rule

Rearrange $3 x + 6 y = 12$ by subtracting $3 x$ : $6 y = - 3 x + 12$ .
Divide by 6: $y=−12x+2y=-\frac{1}{2}x+2$ .
The given line has gradient $m=−12m=-\frac{1}{2}$ .
A perpendicular line has gradient $m = 2$ .
Passing through (0, 7) gives $c = 7$ .
The equation is $y = 2 x + 7$ .

5. Using two points to find a gradient

If a line is described using two points, calculate its gradient first.

Definition

Gradient between two points

For points $x_1,y_1)$ and $x_2,y_2)$ , the gradient is $m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}$ : change in $y$ divided by change in $x$ .

Gradient between two points is the vertical change divided by the horizontal change.

Once you know the gradient, use $y = m x + c$ . To substitute a point means to replace $x$ and $y$ with that point’s coordinates.

Example

Parallel line from two points

Line A passes through (1, 2) and (5, 14), so its gradient is $m=14−25−1=3m=\frac{14-2}{5-1}=3$ .

Line A’s two points show a rise of 12 and a run of 4, giving gradient 3; the required parallel line has the same slope through (2, 3).

A parallel line also has gradient $m = 3$ .
Start with $y = 3 x + c$ .
Use the point (2, 3): $3 = 3 (2) + c$ .
Solve to get $c = - 3$ .
The equation is $y = 3 x - 3$ .

Example

Perpendicular line from two points

Line A passes through (0, 4) and (6, 7), so its gradient is $m=7−46−0=12m=\frac{7-4}{6-0}=\frac{1}{2}$ .

The line through (1, 5) is perpendicular to Line A, so it crosses at a right angle with the negative reciprocal gradient.

A perpendicular line has gradient $m = - 2$ .
Start with $y = - 2 x + c$ .
Use the point (1, 5): $5 = - 2 (1) + c$ .
Solve to get $c = 7$ .
The equation is $y = - 2 x + 7$ .

Common Mistake

Gradient fraction upside down

Gradient is change in $y$ over change in $x$ . Keep the order of the two points consistent in the numerator and denominator.

6. Showing lines are parallel or perpendicular

When a question says “show that”, you need evidence. Calculate both gradients and then make a clear statement.

Example

Showing two lines are parallel

Line A passes through (1, 2) and (3, 8), so $mA=8−23−1=3m_A=\frac{8-2}{3-1}=3$ .
Line B passes through (-2, 1) and (0, 7), so $mB=7−10−(−2)=3m_B=\frac{7-1}{0-(-2)}=3$ .
The gradients are equal, so the two lines are parallel.

For perpendicular lines, show that the gradients multiply to -1.

7. Finding an unknown value like $k$

If a coordinate contains $k$ , write the gradient using $k$ , then set it equal to the gradient you need.

Example

Finding $k$ using perpendicular gradients

Line A passes through (-1, 2) and (3, 4), so $mA=4−23−(−1)=12m_A=\frac{4-2}{3-(-1)}=\frac{1}{2}$ .
A perpendicular line must have gradient $m = - 2$ .
Line B passes through (2, 5) and ( $k$ , -1), so its gradient is $mB=−1−5k−2=−6k−2m_B=\frac{-1-5}{k-2}=\frac{-6}{k-2}$ .

Line B uses an unknown x-coordinate k, so its horizontal change is k - 2 while its vertical change is -6.

Set this equal to -2: $−6k−2=−2\frac{-6}{k-2}=-2$ .
Solve: $- 6 = - 2 (k - 2)$ , so $- 6 = - 2 k + 4$ , giving $k = 5$ .

Exam technique

In the exam

Rearrange into $y = m x + c$ before deciding on the gradient.
For parallel lines, keep the same gradient; for perpendicular lines, flip the fraction and change the sign.
If points are given, calculate $xm=\frac{\text{change in }y}{\text{change in }x}$ carefully, then state your conclusion clearly.

Self review

Check yourself

What gradient would a line perpendicular to $y=35x−2y=\frac{3}{5}x-2$ have?
Can you rearrange $2 x + 3 y = 9$ into $y = m x + c$ and identify the gradient?
If two lines have gradients $m = 4$ and $m=−14m=-\frac{1}{4}$ , what is their relationship?

Parallel and Perpendicular Lines

What you'll learn

1. Start with $y = m x + c$

2. Parallel lines

3. Perpendicular lines

4. If the equation is not already in $y = m x + c$

5. Using two points to find a gradient

6. Showing lines are parallel or perpendicular

7. Finding an unknown value like $k$

Recap questions

You've reached the end

Parallel and Perpendicular Lines

What you'll learn

1. Start with $y = m x + c$

2. Parallel lines

3. Perpendicular lines

4. If the equation is not already in $y = m x + c$

5. Using two points to find a gradient

6. Showing lines are parallel or perpendicular

7. Finding an unknown value like $k$

Recap questions

You've reached the end

Parallel and Perpendicular Lines

What you'll learn

1. Start with y=mx+cy=mx+cy=mx+c

2. Parallel lines

3. Perpendicular lines

4. If the equation is not already in y=mx+cy=mx+cy=mx+c

5. Using two points to find a gradient

6. Showing lines are parallel or perpendicular

7. Finding an unknown value like kkk

Recap questions

You've reached the end

Parallel and Perpendicular Lines Revision Guide

Parallel and Perpendicular Lines

What you'll learn

1. Start with y=mx+cy=mx+cy=mx+c

2. Parallel lines

3. Perpendicular lines

4. If the equation is not already in y=mx+cy=mx+cy=mx+c

5. Using two points to find a gradient

6. Showing lines are parallel or perpendicular

7. Finding an unknown value like kkk

Recap questions

You've reached the end

Parallel and Perpendicular Lines Revision Guide

1. Start with $y = m x + c$

4. If the equation is not already in $y = m x + c$

7. Finding an unknown value like $k$

1. Start with $y = m x + c$

4. If the equation is not already in $y = m x + c$

7. Finding an unknown value like $k$