Parallel and Perpendicular Lines

What you'll learn

How to recognise the gradient and y-intercept of a straight line.
How to write equations of lines that are parallel or perpendicular.
How to rearrange equations before comparing gradients.
How to use two points on a line to prove relationships or find a missing value.

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

Most straight-line questions become much easier when the equation is written in the form:

y = m x + c

Here, the coefficient of $x$ tells you the slope of the line, and the constant term tells you where the line crosses the y-axis.

Definition

Gradient and y-intercept

The gradient is the steepness of a line. In $y = m x + c$ , the gradient is $m$ .
The y-intercept is where the line crosses the y-axis. In $y = m x + c$ , the y-intercept is $c$ , so the line passes through $(0, c)$ .

Example

Reading a line equation

For the line $y = 4 x - 7$ :

Compare it with $y = m x + c$ .
The number multiplying $x$ is 4, so $m = 4$ .
The constant term is -7, so $c = - 7$ .
The line has gradient 4 and crosses the y-axis at $(0, - 7)$ .

The diagram below shows the key idea visually: parallel lines have the same gradient, while perpendicular lines meet at 90°.

Coordinate diagram showing parallel lines with the same gradient and a perpendicular line with negative reciprocal gradient

2. Parallel lines

Parallel lines go in exactly the same direction. They never meet, unless they are actually the same line.

Key Idea

Parallel gradients

Parallel straight lines have the same gradient. The y-intercept may be different, but the value of $m$ is the same.

So any line parallel to $y = 3 x + 2$ must also have gradient 3. For example, $y = 3 x - 5$ is parallel to it.

Example

Writing a parallel line through a y-axis point

Find the equation of the line parallel to $y=12x+6y=\frac{1}{2}x+6$ that passes through $(0, - 3)$ .

The original line is already in the form $y = m x + c$ .
Its gradient is $m=12m=\frac{1}{2}$ .
A parallel line has the same gradient, so start with:
$y=12x+cy=\frac{1}{2}x+c$
The point $(0, - 3)$ is on the y-axis, so the y-intercept is -3.
Therefore the equation is $y=12x−3y=\frac{1}{2}x-3$ .

Common Mistake

Changing the wrong part

For a parallel line, keep the gradient the same. You change the y-intercept only if the line crosses the y-axis somewhere else.

3. Perpendicular lines

Perpendicular lines meet at a right angle, which is 90°.

Definition

Perpendicular lines

Two lines are perpendicular if they meet at 90°. For non-horizontal lines, their gradients are negative reciprocals.

A reciprocal means “flip the fraction”. A negative reciprocal means “flip it and change the sign”.

For example:

The negative reciprocal of $12\frac{1}{2}$ is -2.
The negative reciprocal of $- 3$ is $13\frac{1}{3}$ .
The negative reciprocal of $45\frac{4}{5}$ is $−54-\frac{5}{4}$ .

Key Idea

Perpendicular gradients

If two non-horizontal lines are perpendicular, their gradients multiply to give -1:

m_1m_2=-1

Example

Writing a perpendicular line through a y-axis point

Find the equation of the line perpendicular to $y=14x−5y=\frac{1}{4}x-5$ that passes through $(0, 6)$ .

The original line has gradient $m=14m=\frac{1}{4}$ .
The negative reciprocal of $14\frac{1}{4}$ is -4, so the perpendicular gradient is $m = - 4$ .
Start with the new line:
$y = - 4 x + c$
The line passes through $(0, 6)$ , so $c = 6$ .
The equation is $y = - 4 x + 6$ .

Common Mistake

Forgetting the negative sign

For perpendicular lines, do not just flip the fraction. You must also change the sign. The perpendicular gradient to $23\frac{2}{3}$ is $−32-\frac{3}{2}$ , not $32\frac{3}{2}$ .

Common Mistake

Horizontal and vertical lines

The negative reciprocal rule assumes both lines can be written in the form $y = m x + c$ . A horizontal line such as $y = 4$ is perpendicular to a vertical line such as $x = 2$ .

4. Rearranging first

Sometimes the equation is not already written as $y = m x + c$ . Before comparing gradients, rearrange it so that $y$ is the subject.

Example

Finding a parallel line after rearranging

Find the equation of the line parallel to $2 x + 5 y = 10$ that passes through $(0, 4)$ .

Rearrange the original equation into $y = m x + c$ form:
$2x+5y=105y=−2x+10y=−25x+2\begin{aligned} 2x+5y&=10\\ 5y&=-2x+10\\ y&=-\frac{2}{5}x+2 \end{aligned}$
The gradient is $m=−25m=-\frac{2}{5}$ .
A parallel line has the same gradient, so use:
$y=−25x+cy=-\frac{2}{5}x+c$
The line passes through $(0, 4)$ , so $c = 4$ .
The equation is $y=−25x+4y=-\frac{2}{5}x+4$ .

Common Mistake

Not dividing every term

When you divide to make $y$ the subject, divide every term. For example, from $5 y = - 2 x + 10$ , you get $y=−25x+2y=-\frac{2}{5}x+2$ .

5. Comparing several lines

If you are given a list of line equations, do not try to compare the whole equations at once. Just find each gradient.

Example

Finding parallel and perpendicular pairs

Here are five lines:

Line A: $y = 2 x + 1$
Line B: $3 y = x - 6$
Line C: $2 y = 4 x - 5$
Line D: $y = - 3 x + 2$
Line E: $y + x = 5$

Find one parallel pair and one perpendicular pair.

Line A is already in $y = m x + c$ form, so its gradient is 2.
Rearrange Line B:
$\Rightarrow y=\frac{1}{3}x-2$
Rearrange Line C:
$\Rightarrow y=2x-\frac{5}{2}$
Line D has gradient -3.
Rearrange Line E:
$\Rightarrow y=-x+5$
Lines A and C are parallel because they both have gradient 2.
Lines B and D are perpendicular because $13×(−3)=−1\frac{1}{3}\times(-3)=-1$ .

6. When a line is given by two points

If you are given two points instead of an equation, first calculate the gradient.

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}

This means “change in y divided by change in x”.

Example

Equation of a parallel line from two points

Line A passes through $(2, 1)$ and $(6, 13)$ . Find the equation of the line parallel to A that passes through $(3, 4)$ .

Find the gradient of Line A:
$m=13−16−2=124=3m=\frac{13-1}{6-2}=\frac{12}{4}=3$
A parallel line has the same gradient, so use:
$y = 3 x + c$
Substitute the point $(3, 4)$ into the equation:
$\Rightarrow c=-5$
The equation is $y = 3 x - 5$ .

Example

Equation of a perpendicular line from two points

Line A passes through $(1, 4)$ and $(5, 6)$ . Find the equation of the line perpendicular to A that passes through $(- 2, 3)$ .

Find the gradient of Line A:
$m=6−45−1=24=12m=\frac{6-4}{5-1}=\frac{2}{4}=\frac{1}{2}$
The perpendicular gradient is -2, because $12×(−2)=−1\frac{1}{2}\times(-2)=-1$ .
Use the form $y = - 2 x + c$ .
Substitute the point $(- 2, 3)$ :
$\Rightarrow c=-1$
The equation is $y = - 2 x - 1$ .

Tip

When the point is not on the y-axis

If the point is not of the form $(0, c)$ , you cannot read off the y-intercept. Substitute the coordinates into $y = m x + c$ to find $c$ .

7. Showing lines are parallel or perpendicular

In “show that” questions, you need to calculate both gradients and then write a clear conclusion.

Example

Showing two lines are parallel

Line A passes through $(0, 2)$ and $(4, 10)$ . Line B passes through $(- 1, - 1)$ and $(2, 5)$ . Show that the lines are parallel.

Find the gradient of Line A:
$mA=10−24−0=84=2m_A=\frac{10-2}{4-0}=\frac{8}{4}=2$
Find the gradient of Line B:
$mB=5−(−1)2−(−1)=63=2m_B=\frac{5-(-1)}{2-(-1)}=\frac{6}{3}=2$
The gradients are equal.
Therefore Line A and Line B are parallel.

8. Finding a missing value

For missing coordinate questions, write the gradient involving the unknown, then use the parallel or perpendicular rule.

Example

Finding a missing coordinate using perpendicular gradients

Line A passes through $(2, 7)$ and $(5, 1)$ . Line B passes through $(- 1, 3)$ and $(k, 5)$ . The lines are perpendicular. Find $k$ .

Find the gradient of Line A:
$mA=1−75−2=−63=−2m_A=\frac{1-7}{5-2}=\frac{-6}{3}=-2$
The perpendicular gradient is $12\frac{1}{2}$ , because $−2×12=−1-2\times\frac{1}{2}=-1$ .
Write the gradient of Line B using the unknown coordinate:
$mB=5−3k−(−1)=2k+1m_B=\frac{5-3}{k-(-1)}=\frac{2}{k+1}$
Set the gradient of Line B equal to $12\frac{1}{2}$ and solve:
$2k+1=12⇒4=k+1⇒k=3\frac{2}{k+1}=\frac{1}{2}\Rightarrow 4=k+1\Rightarrow k=3$

Tip

Parallel versus perpendicular missing values

For parallel lines, set the gradients equal. For perpendicular lines, set the second gradient equal to the negative reciprocal of the first.

Exam technique

In the exam

Rearrange each equation into $y = m x + c$ before comparing gradients.
For parallel lines, look for the same gradient.
For perpendicular lines, flip the gradient and change the sign, or check that $m_1m_2=-1$ .
If a line passes through $(0, c) ‘, r e a d o f f t h ey - in t er ce pt; o t h er w i ses u b s t i t u t e t h e p o in t in t o$ y=mx+c$.
In “show that” questions, always write a final sentence explaining why the lines are parallel or perpendicular.

Self review

Check yourself

What is the gradient of a line perpendicular to $y=−45x+1y=-\frac{4}{5}x+1$ ?
How would you find $c$ for a line with gradient 3 passing through $(2, - 1)$ ?
If two lines have gradients $23\frac{2}{3}$ and $32\frac{3}{2}$ , are they parallel, perpendicular, or neither?

Parallel and Perpendicular Lines

What you'll learn

How to recognise the gradient and y-intercept of a straight line.
How to write equations of lines that are parallel or perpendicular.
How to rearrange equations before comparing gradients.
How to use two points on a line to prove relationships or find a missing value.

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

Most straight-line questions become much easier when the equation is written in the form:

y = m x + c

Here, the coefficient of $x$ tells you the slope of the line, and the constant term tells you where the line crosses the y-axis.

Definition

Gradient and y-intercept

The gradient is the steepness of a line. In $y = m x + c$ , the gradient is $m$ .
The y-intercept is where the line crosses the y-axis. In $y = m x + c$ , the y-intercept is $c$ , so the line passes through $(0, c)$ .

Example

Reading a line equation

For the line $y = 4 x - 7$ :

Compare it with $y = m x + c$ .
The number multiplying $x$ is 4, so $m = 4$ .
The constant term is -7, so $c = - 7$ .
The line has gradient 4 and crosses the y-axis at $(0, - 7)$ .

The diagram below shows the key idea visually: parallel lines have the same gradient, while perpendicular lines meet at 90°.

Coordinate diagram showing parallel lines with the same gradient and a perpendicular line with negative reciprocal gradient

2. Parallel lines

Parallel lines go in exactly the same direction. They never meet, unless they are actually the same line.

Key Idea

Parallel gradients

Parallel straight lines have the same gradient. The y-intercept may be different, but the value of $m$ is the same.

So any line parallel to $y = 3 x + 2$ must also have gradient 3. For example, $y = 3 x - 5$ is parallel to it.

Example

Writing a parallel line through a y-axis point

Find the equation of the line parallel to $y=12x+6y=\frac{1}{2}x+6$ that passes through $(0, - 3)$ .

The original line is already in the form $y = m x + c$ .
Its gradient is $m=12m=\frac{1}{2}$ .
A parallel line has the same gradient, so start with:
$y=12x+cy=\frac{1}{2}x+c$
The point $(0, - 3)$ is on the y-axis, so the y-intercept is -3.
Therefore the equation is $y=12x−3y=\frac{1}{2}x-3$ .

Common Mistake

Changing the wrong part

For a parallel line, keep the gradient the same. You change the y-intercept only if the line crosses the y-axis somewhere else.

3. Perpendicular lines

Perpendicular lines meet at a right angle, which is 90°.

Definition

Perpendicular lines

Two lines are perpendicular if they meet at 90°. For non-horizontal lines, their gradients are negative reciprocals.

A reciprocal means “flip the fraction”. A negative reciprocal means “flip it and change the sign”.

For example:

The negative reciprocal of $12\frac{1}{2}$ is -2.
The negative reciprocal of $- 3$ is $13\frac{1}{3}$ .
The negative reciprocal of $45\frac{4}{5}$ is $−54-\frac{5}{4}$ .

Key Idea

Perpendicular gradients

If two non-horizontal lines are perpendicular, their gradients multiply to give -1:

m_1m_2=-1

Example

Writing a perpendicular line through a y-axis point

Find the equation of the line perpendicular to $y=14x−5y=\frac{1}{4}x-5$ that passes through $(0, 6)$ .

The original line has gradient $m=14m=\frac{1}{4}$ .
The negative reciprocal of $14\frac{1}{4}$ is -4, so the perpendicular gradient is $m = - 4$ .
Start with the new line:
$y = - 4 x + c$
The line passes through $(0, 6)$ , so $c = 6$ .
The equation is $y = - 4 x + 6$ .

Common Mistake

Forgetting the negative sign

For perpendicular lines, do not just flip the fraction. You must also change the sign. The perpendicular gradient to $23\frac{2}{3}$ is $−32-\frac{3}{2}$ , not $32\frac{3}{2}$ .

Common Mistake

Horizontal and vertical lines

The negative reciprocal rule assumes both lines can be written in the form $y = m x + c$ . A horizontal line such as $y = 4$ is perpendicular to a vertical line such as $x = 2$ .

4. Rearranging first

Sometimes the equation is not already written as $y = m x + c$ . Before comparing gradients, rearrange it so that $y$ is the subject.

Example

Finding a parallel line after rearranging

Find the equation of the line parallel to $2 x + 5 y = 10$ that passes through $(0, 4)$ .

Rearrange the original equation into $y = m x + c$ form:
$2x+5y=105y=−2x+10y=−25x+2\begin{aligned} 2x+5y&=10\\ 5y&=-2x+10\\ y&=-\frac{2}{5}x+2 \end{aligned}$
The gradient is $m=−25m=-\frac{2}{5}$ .
A parallel line has the same gradient, so use:
$y=−25x+cy=-\frac{2}{5}x+c$
The line passes through $(0, 4)$ , so $c = 4$ .
The equation is $y=−25x+4y=-\frac{2}{5}x+4$ .

Common Mistake

Not dividing every term

When you divide to make $y$ the subject, divide every term. For example, from $5 y = - 2 x + 10$ , you get $y=−25x+2y=-\frac{2}{5}x+2$ .

5. Comparing several lines

If you are given a list of line equations, do not try to compare the whole equations at once. Just find each gradient.

Example

Finding parallel and perpendicular pairs

Here are five lines:

Line A: $y = 2 x + 1$
Line B: $3 y = x - 6$
Line C: $2 y = 4 x - 5$
Line D: $y = - 3 x + 2$
Line E: $y + x = 5$

Find one parallel pair and one perpendicular pair.

Line A is already in $y = m x + c$ form, so its gradient is 2.
Rearrange Line B:
$\Rightarrow y=\frac{1}{3}x-2$
Rearrange Line C:
$\Rightarrow y=2x-\frac{5}{2}$
Line D has gradient -3.
Rearrange Line E:
$\Rightarrow y=-x+5$
Lines A and C are parallel because they both have gradient 2.
Lines B and D are perpendicular because $13×(−3)=−1\frac{1}{3}\times(-3)=-1$ .

6. When a line is given by two points

If you are given two points instead of an equation, first calculate the gradient.

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}

This means “change in y divided by change in x”.

Example

Equation of a parallel line from two points

Line A passes through $(2, 1)$ and $(6, 13)$ . Find the equation of the line parallel to A that passes through $(3, 4)$ .

Find the gradient of Line A:
$m=13−16−2=124=3m=\frac{13-1}{6-2}=\frac{12}{4}=3$
A parallel line has the same gradient, so use:
$y = 3 x + c$
Substitute the point $(3, 4)$ into the equation:
$\Rightarrow c=-5$
The equation is $y = 3 x - 5$ .

Example

Equation of a perpendicular line from two points

Line A passes through $(1, 4)$ and $(5, 6)$ . Find the equation of the line perpendicular to A that passes through $(- 2, 3)$ .

Find the gradient of Line A:
$m=6−45−1=24=12m=\frac{6-4}{5-1}=\frac{2}{4}=\frac{1}{2}$
The perpendicular gradient is -2, because $12×(−2)=−1\frac{1}{2}\times(-2)=-1$ .
Use the form $y = - 2 x + c$ .
Substitute the point $(- 2, 3)$ :
$\Rightarrow c=-1$
The equation is $y = - 2 x - 1$ .

Tip

When the point is not on the y-axis

If the point is not of the form $(0, c)$ , you cannot read off the y-intercept. Substitute the coordinates into $y = m x + c$ to find $c$ .

7. Showing lines are parallel or perpendicular

In “show that” questions, you need to calculate both gradients and then write a clear conclusion.

Example

Showing two lines are parallel

Line A passes through $(0, 2)$ and $(4, 10)$ . Line B passes through $(- 1, - 1)$ and $(2, 5)$ . Show that the lines are parallel.

Find the gradient of Line A:
$mA=10−24−0=84=2m_A=\frac{10-2}{4-0}=\frac{8}{4}=2$
Find the gradient of Line B:
$mB=5−(−1)2−(−1)=63=2m_B=\frac{5-(-1)}{2-(-1)}=\frac{6}{3}=2$
The gradients are equal.
Therefore Line A and Line B are parallel.

8. Finding a missing value

For missing coordinate questions, write the gradient involving the unknown, then use the parallel or perpendicular rule.

Example

Finding a missing coordinate using perpendicular gradients

Line A passes through $(2, 7)$ and $(5, 1)$ . Line B passes through $(- 1, 3)$ and $(k, 5)$ . The lines are perpendicular. Find $k$ .

Find the gradient of Line A:
$mA=1−75−2=−63=−2m_A=\frac{1-7}{5-2}=\frac{-6}{3}=-2$
The perpendicular gradient is $12\frac{1}{2}$ , because $−2×12=−1-2\times\frac{1}{2}=-1$ .
Write the gradient of Line B using the unknown coordinate:
$mB=5−3k−(−1)=2k+1m_B=\frac{5-3}{k-(-1)}=\frac{2}{k+1}$
Set the gradient of Line B equal to $12\frac{1}{2}$ and solve:
$2k+1=12⇒4=k+1⇒k=3\frac{2}{k+1}=\frac{1}{2}\Rightarrow 4=k+1\Rightarrow k=3$

Tip

Parallel versus perpendicular missing values

For parallel lines, set the gradients equal. For perpendicular lines, set the second gradient equal to the negative reciprocal of the first.

Exam technique

In the exam

Rearrange each equation into $y = m x + c$ before comparing gradients.
For parallel lines, look for the same gradient.
For perpendicular lines, flip the gradient and change the sign, or check that $m_1m_2=-1$ .
If a line passes through $(0, c) ‘, r e a d o f f t h ey - in t er ce pt; o t h er w i ses u b s t i t u t e t h e p o in t in t o$ y=mx+c$.
In “show that” questions, always write a final sentence explaining why the lines are parallel or perpendicular.

Self review

Check yourself

What is the gradient of a line perpendicular to $y=−45x+1y=-\frac{4}{5}x+1$ ?
How would you find $c$ for a line with gradient 3 passing through $(2, - 1)$ ?
If two lines have gradients $23\frac{2}{3}$ and $32\frac{3}{2}$ , are they parallel, perpendicular, or neither?

Parallel and Perpendicular Lines

What you'll learn

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

2. Parallel lines

3. Perpendicular lines

4. Rearranging first

5. Comparing several lines

6. When a line is given by two points

7. Showing lines are parallel or perpendicular

8. Finding a missing value

Recap questions

You've reached the end

Parallel and Perpendicular Lines

What you'll learn

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

2. Parallel lines

3. Perpendicular lines

4. Rearranging first

5. Comparing several lines

6. When a line is given by two points

7. Showing lines are parallel or perpendicular

8. Finding a missing value

Recap questions

You've reached the end

Parallel and Perpendicular Lines

What you'll learn

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

2. Parallel lines

3. Perpendicular lines

4. Rearranging first

5. Comparing several lines

6. When a line is given by two points

7. Showing lines are parallel or perpendicular

8. Finding a missing value

Recap questions

You've reached the end

Parallel and Perpendicular Lines Revision Guide

Parallel and Perpendicular Lines

What you'll learn

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

2. Parallel lines

3. Perpendicular lines

4. Rearranging first

5. Comparing several lines

6. When a line is given by two points

7. Showing lines are parallel or perpendicular

8. Finding a missing value

Recap questions

You've reached the end

Parallel and Perpendicular Lines Revision Guide

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

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