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Here we will learn about **parallel and perpendicular lines** including how to show when two lines are parallel or perpendicular, how to calculate the gradient of parallel and perpendicular lines and use this knowledge to calculate the gradient of tangents to circles.

There are also parallel and perpendicular worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Parallel lines **are straight lines with a constant distance between them. **Perpendicular lines** intersect one another at a

E.g.

A table has

Straight line graphs can be parallel, perpendicular or have

Two (or more) straight lines are **parallel **when they share the **same gradient**.

E.g.

Let’s look at the three lines

Each straight line has the same gradient (here,

We can therefore state that the equation of a line parallel to a line with the gradient

**Top Tip: **All vertical lines are parallel to the line

Parallel lines can be recognised by their equations by checking if their gradients are the same. This may require some rearranging.

E.g.

Show that y=-2x+4 and 4x+2y=9 are parallel.

The gradient of the first line is -2. We need to rearrange the second equation so that it is in the form y=mx+c so we can find the gradient.

4x+2y=9 2y=-4x+9 y=-2x+4.5

The gradient of the second line is also -2. The gradients of the two lines are the same, so the lines are parallel.

**Perpendicular lines** have gradients that multiply to give

In other words, the gradients of two lines that are perpendicular to each other are the negative reciprocals of each other.

E.g.

To see why let’s look at the line

The gradient of the line

Let’s now construct a line that is perpendicular to the line

The equation of a straight line that is perpendicular to the line

We can see that the gradient of each line has a different sign (one has a positive gradient whilst the other has a negative gradient).

To find out what is going on here let’s look at a second example.

E.g.

Take the line

Let’s construct a perpendicular to the line

Here the equation of the line that is perpendicular to y = 2x + 3 at the point (0, 3) is y=−\frac{1}{2}x+3

We can still state that the gradients change from being positive to negative but we now have the gradient of -\frac{1}{2} which is the negative reciprocal of 2 .

Let us look at the previous two examples together in the table below:

Gradient of line 1 | Gradient of line 2 | Product of both gradients |

1 | -1 | 1 × (-1) = -1 |

2 | -½ | 2 × -½ = -1 |

Here we can see that the **product of the gradients of each line is equal to −1.**

We can therefore state that for any two perpendicular lines

If we divide both sides of the formula by

The value of **negative reciprocal** of

The gradients of parallel lines are the same.

The gradients of perpendicular lines multiply together to make -1.

Here is a brief summary.

For any two lines to be **parallel**, the **gradient is the same** for each line.

E.g.

y = 5x + 3 and y = 5x − 9 are **parallel **as the gradient for each line is 5.

For any two lines to be **perpendicular**, the **product of their gradients must equal − 1**.

E.g.

y=4x+7 and y=-\frac{1}{4}x+9 are **perpendicular **as 4\times-\frac{1}{4}=-1

In order to find the equation of a parallel line:

**State/calculate the gradient of the original straight line.****Substitute the value for**m into the equation of a new straight line.**State/calculate the value of the \textbf{y} -intercept.**

Get your free parallel and perpendicular lines worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free parallel and perpendicular lines worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Parallel and perpendicular lines** is part of our series of lessons to support revision on **straight line graphs**. You may find it helpful to start with the main straight line graphs lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

The equation of a line A is given as 5y = x − 10 .

Write the equation of a line that is parallel to A .

**State/calculate the gradient of the original straight line**.

As 5y=x-10 .

Here, the gradient of the line is \frac{1}{5}

2**Substitute the value for m into the equation of a new straight line**.

Substituting m=\frac{1}{5} into a new equation, we get y=\frac{1}{5}x+c

3**State/calculate the value of the y-intercept**.

Here we are asked to state a parallel line so any value of

An example of an equation of the line parallel to line A is y=\frac{1}{5}x+3

The line B is drawn on a set of axes. Write the equation of a line that is parallel to B.

**State/calculate the gradient of the original straight line**.

Here,

\[\\m=\frac{-3}{3}=-1\\\]

**Substitute the value for m into the equation of a new straight line**.

We now have

**State/calculate the value of the y-intercept**.

Here we are asked to state a parallel line so any value of

A possible solution would be

Two lines, D and E are written in the form

The gradient of the line D is equal to

The line E is parallel to D and goes through the point (

Work out the equation of line E.

**State/calculate the gradient of the original straight line**.

We know from the question that the gradient of line D is

**Substitute the value for m into the equation of a new straight line**.

We now have

**State/calculate the value of the y-intercept**.

We need to calculate the value of

\begin{aligned}
\\12&=0.4\times{10}+c\\
\\12&=4+c\\
\\c&=8
\end{aligned}

The equation of line E is

In order to find the equation of a perpendicular line:

**State/calculate the gradient of the original straight line.****Calculate the negative reciprocal of the gradient**.**Substitute the negative reciprocal into the equation of a new straight line**.**State/calculate the value of the \textbf{y} -intercept**.

The equation of a line A is given as

**State/calculate the gradient of the original straight line**.

First we need to rearrange the equation to make

Now we can see that

**Calculate the negative reciprocal of the gradient**.

The gradient of the line perpendicular to A is equal to

\[\\n=\frac{-1}{m}=\frac{-1}{4}\\\]

**Substitute the negative reciprocal into the equation of a new straight line**.

We now have the equation y=-\frac{1}{4}x+c for the line perpendicular to A.

**State/calculate the value of the y-intercept**.

Here, we are asked to state a perpendicular line so any value of

An example of an equation of the line perpendicular to line A is y=-\frac{1}{4}x+1

The line B is drawn on a set of axes. Write the equation of a line that is perpendicular to B at the point (

**State/calculate the gradient of the original straight line**.

Here,

\[\\m=\frac{-8}{4}=-2\\\]

**Calculate the negative reciprocal of the gradient**.

The gradient of the line perpendicular to B is equal to

\[\\n=\frac{-1}{m}=\frac{-1}{-2}=\frac{1}{2}\\\]

**Substitute the negative reciprocal into the equation of a new straight line**.

We now have the equation y=\frac{1}{2}x+c for the line perpendicular to B.

**State/calculate the value of the y-intercept**.

Here we have to substitute the coordinate (

\[\\4=\frac{1}{2}\times{1}+c\\\]

\[\\4=\frac{1}{2}+c\\\]

\[\\c=3\frac{1}{2}\\\]

The equation of the line that is perpendicular to the line B is y=\frac{1}{2}x+3\frac{1}{2} or 2y=x+7 .

Two lines, D and E are written in the form

The gradient of the line D is equal to -\frac{2}{5} and intersects the y-axis at the origin. The line E is perpendicular to D. Work out the equation of line E.

**State/calculate the gradient of the original straight line**.

The gradient of line D is -\frac{2}{5} as stated in the question.

**Calculate the negative reciprocal of the gradient**.

Here, we have

\[\\n=-1\div{-\frac{2}{5}}\\
\\n=-1\times{-\frac{5}{2}}\\
\\n=\frac{5}{2}\\\]

**Substitute the negative reciprocal into the equation of a new straight line**.

Substituting m=\frac{5}{2} into y=mx+c we get y=\frac{5}{2}x+c

**State/calculate the value of the y-intercept**.

Here we have to substitute the coordinate (

\[\\0=0\times{\frac{5}{2}}+c\\\]

\[c=0\]

The equation of line E is therefore y=\frac{5}{2}x

**Using the y-intercept instead of the coefficient of**x (parallel lines)

E.g.

Take example *“The equation of a line A is given as 5y = x − 10. Write the equation of a line that is parallel to A.” *

After making

**Not/incorrectly rearranging to the form**y = mx + c

The gradient is the coefficient of

If the equation of a straight line is not written with

E.g.

Take example *“The equation of a line A is given as 5y = x − 10. Write the equation of a line that is parallel to A.” *

Here, the value of the gradient could be incorrectly stated as

**The reciprocal vs the negative reciprocal**

A common mistake is to state the gradient of the perpendicular to a line to be the reciprocal of the original gradient, and not the negative reciprocal.

If the negative reciprocal was not used his would mean that both lines would have a positive gradient, or they would both have a negative gradient. This would mean that although the two lines may intersect they will not intersect at

1. Which equation is parallel to the line y=4x+3 ?

y=7+4x

y=2x+5

y=-\frac{1}{4}x+3

y=-4x+3

Any equation of the form y=4x+c for any value of c .

2. Which of the following equations is parallel to the line 2y-5=3x ?

4y-10=3x

y=3x+5

2y=-4x+5

y=1.5x+4

Both have the gradient of 1.5 (or m=\frac{3}{2} )

3. Calculate the gradient of the line that is perpendicular to the line y=-0.4x+3 .

m=0.4

m=4

m=2.5

m=-0.4

The product of the two gradients must equal -1 .

Here, -1\div-0.4 = 1\div0.4 = 2.5

4. Calculate the gradient of the line that is perpendicular to the line y=-\frac{7}{10}x+4

m=\frac{7}{10}

m=-1

m=-\frac{10}{7}

m=1\frac{3}{7}

The product of the two gradients must equal -1 .

Here, \frac{-7}{10}\times\frac{10}{7}=\frac{-70}{70}=-1

5. State the equation of the line that is parallel to the line y=2x+7 at the point ( 3,8 )

y=2x+8

y=2x+2

y=2x+8

y=-\frac{1}{2}x+2

The line must have a gradient of 2 so y=2x+c .

At the point ( 3,8 ), 8=2\times3+c , so c=2

Hence, y=2x+2

6. Calculate the equation of the line that is perpendicular to the line y=-\frac{1}{3}x+1 that intersects at the point ( 6,-1 )

y=\frac{1}{3}x+6\

y=3x-19

y=-3x+1

y=3x+1

New gradient = -1\div{-\frac{1}{3}}=-1\times{-3}=3

y=3x+c at the point ( 6,-1 ) means -1=(3\times6)+c

c=-19

1. Two straight lines A and B are shown below. Prove that the lines are parallel.

**(3 marks)**

Show answer

Gradient of line A = \frac{2−1}{2−0}=\frac{1}{2}

**(1)**

Gradient of line B = \frac{0−−2}{4−0}=\frac{2}{4}=\frac{1}{2}

**(1)**

Gradients are the same, therefore parallel

**(1)**

2. (a) Circle the equation of the line that is parallel to the y-axis.

x=3 | x+y=0 | y=−1 | xy=4 |

(b) Use the straight line graph L below to write the equation of a parallel line passing through the coordinate ( 3,4 )

**(4 marks)**

Show answer

a) x = 3

**(1)**

b) Gradient of line L is equal to − 1

**(1)**

**(1)**

4 = −3 + c so c = 7 or y = −x + 7

**(1)**

3. (a) The line P has the equation y=6−\frac{3}{5}x . Circle the gradient of a line that is perpendicular to P.

−\frac{5}{3} | \frac{3}{5} | \frac{5}{3} | 5 |

(b) Point C (3,6) lies on a circle, centre O. The tangent at C intersects the axes at points A and B. Show that the gradient of the tangent is -\frac{1}{2}

**(3 marks)**

Show answer

a) \frac{5}{3}

**(1)**

b) Gradient of OC= \frac{6}{3}=2

**(1)**

Gradient of perpendicular (tangent):

AB=−1\div2

AB=−\frac{1}{2}

**(1)**

4. Use the graph below to show that the two lines are parallel. Clearly state any conclusions.

**(4 marks)**

Show answer

**(2)**

m=-\frac{6}{5} for both lines

**(1)**

The gradients are the same so the lines are parallel

**(1)**

You have now learned how to:

- Identify horizontal and vertical lines and pairs of perpendicular and parallel lines.
- Use the form y = mx + c to identify parallel {and perpendicular} lines; find the equation of the line through 2 given points, or through 1 point with a given gradient.

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