# Equation of a Line

Here we will learn about the equation of a line, including recognising the gradient and y-intercept of a straight line, and finding the equation of a line from a graph.

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

## What is the equation of a line?

The equation of a line is the algebraic representation of a line using cartesian coordinates.

The general form of the equation of a straight line is written as y = mx + c .
Where m is the gradient of the straight line
and c is the y -intercept of the straight line.

The coordinate ( x,y) lies on the line y = mx + c giving us a linear relationship between x and y . The term linear equation is given to any straight line.

E.g.
Let’s look at the line y = 3x − 4 .

Here, we can find any y -coordinate given the value for x by substituting into the equation y = 3x − 4 .

E.g.
When x = 1 ,

\begin{aligned} &y=3 \times 1-4 \\\\ &y=3-4 \\\\ &y=-1 \end{aligned}

So when the x value is 1 and the y value is -1 , this gives us the coordinate ( 1, -1 ) which lies on the line as shown below.

Not all straight lines will appear to be exactly in the form y = mx + c so we need to understand how we determine the gradient (or steepness of the line) m , and the y -intercept (the point where the line intersects the y -axis) c from equations that are not in the form y = mx + c .

Here are some examples of linear equations not in the form y = mx + c

• y + 17 = 6x
• 2y = 10x + 3
• x = 6y −1
• y = −3 (a horizontal line)
• x = y
• x = 0 (a vertical line)

In order to easily determine m and c we need to rearrange the equation to make y the subject.

E.g.
Take the equation above of y + 17 = 6x and make y the subject.

By rearranging the equation into the form y = mx + c we can clearly state that the gradient m = 6 and the y -intercept c = −17 .

From this, we can draw the straight line onto a set of axes.

Check out our lessons on y = mx + c and gradient of a line to learn more about this.

Step-by-step guide: y = mx + c

Step-by-step guide: Gradient of a line

## How to find the equation of a line

In order to find the equation of a straight line:

1. Calculate the gradient of the line
2. State the y -intercept of the straight line
3. Write the equation of the line in the form y = mx + c

## Equation of a line examples

### Example 1: Positive gradient, positive y-intercept

Work out the equation of the straight line given in the diagram below.

1. Calculate the gradient of the line.

Two points that lie on the line are: ( 0, 4 ) and ( 2, 8 ).The line passes through these two given points.

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-4}{2-0}=\frac{4}{2}=2\\\\ &m=2 \end{aligned}

2State the y-intercept of the straight line.

The y -intercept occurs when x = 0 .

Here, c = 4

3Write the equation of the line in the form y = mx + c .

As the slope m = 2 and y intercept c = 4 ,

y=2x+4

### Example 2: Positive gradient, negative y-intercept

Work out the equation of the straight line given in the diagram below.

Two points that lie on the line are: ( 1, −4 ) and ( 3, 2 ).

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2--4}{3-1}=\frac{6}{2}=3\\\\ &m=3 \end{aligned}

The y -intercept occurs when x = 0 .

Here, c = −7

As m = 3 and c = −7, y = 3x − 7

### Example 3: Negative gradient, positive y-intercept

Work out the equation of the straight line given in the diagram below.

Two points that lie on the line are: ( 1, −2 ) and ( −1, 8 ).

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8--2}{-1-1}=\frac{10}{-2}=-5\\\\ &m=-5 \end{aligned}

The y -intercept occurs when x = 0

Here, c = 3

As m = −5 and c = 3, y = −5x + 3

### Example 4: Negative gradient, negative y-intercept

Work out the equation of the straight line given in the diagram below.

Two points that lie on the line are: ( −3, −2 ) and ( 2, −7 ).

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7--2}{2--3}=\frac{-5}{5}=-1\\\\ &m=-1 \end{aligned}

The y -intercept occurs when x = 0

Here, c = −5

As m = −1 and c = −5, y = −x − 5

### Example 5: Positive fractional gradient, positive y-intercept

Work out the equation of the straight line given in the diagram below.

Two points that lie on the line are: ( 4, 10 ) and ( −2, 7 ).

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-10}{-2-4}=\frac{-3}{-6}=\frac{1}{2}\\\\ &m=\frac{1}{2} \end{aligned}

The y -intercept occurs when x = 0

Here, c = 8

As m = \frac{1}{2} and c = 8, \;y = \frac{1}{2}x + 8

### Example 6: Negative fractional gradient, negative y-intercept

Work out the equation of the straight line given in the diagram below.

Two points that lie on the line are: ( −2, −4 ) and ( 2, −5 ).

\begin{aligned} &m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-5--4}{2--2}=\frac{-1}{4}\\\\ &m=\frac{-1}{4} \end{aligned} .

The y -intercept occurs when x = 0

Here, c = -4.5 or c=-\frac{9}{2} .

As m=-\frac{1}{4} and c=-\frac{9}{2}, \;y=-\frac{1}{4}x-\frac{9}{2} .

### Common misconceptions

• Mixing up the gradient and the y -intercept

The coefficient of x is the gradient m and the constant term is the y -intercept c . If the coefficient of x is 1 , remember this is written as x only as 1x = x .

• The gradient is positive or negative

Ignoring the negative values in a coordinate means that the gradient will be calculated incorrectly. When picking two coordinates, make sure that the coordinates go through the corner of a grid square. If there are negative values, make sure you use them with the negative symbol.

• Mixing up the coordinates

When calculating the gradient of a straight line, be careful to not mix up the coordinates.
E.g.
Looking back at example 5 , we have the two points that lie on the line being ( 1, −2 ) and ( −1,8 ).

The change in y could be correctly calculated as 8 \; − \; −2 but the change in x is then incorrectly calculated as 1 \; − \; −1 . This would result in the gradient of the line being 5 and not −5

• Incorrect change in x or y

Avoid counting squares to work out the change in x or y . Use the axes scales or label the coordinates, then find the difference between them.

• Not calculating the gradient

A common error is to not calculate the gradient of the line.
Let’s look at example 1 ,

A error could be made by identifying the y -intercept and determining that the equation of the line is y = x + 4 .
This will only work when x = 0 but not for any other value for x .

### Practice equation of a line questions

1. Work out the equation of the straight line

y=x+5

y=5x+1

x+1

y=2.5x+1

Two coordinates: (0,1) and (2,11)

m=\frac{11-1}{2-0}=\frac{10}{2}=5

c=1

2. Work out the equation of the straight line

y=-8x+6

x-8

y=6x-8

y=3x-8

Two coordinates: (0,-8) and (1,-2)

m=\frac{-2-  -8}{1-0}=\frac{6}{1}=6

c=-8

3. Work out the equation of the straight line

y=-5x+10

y=10x-5

x+10

y=-2.5x+10

Two coordinates: (2,0) and (1,5)

m=\frac{5-0}{1-2}=\frac{5}{-1}=-5

c=10

4. Work out the equation of the straight line

y=x-1

x-1

y=-0.5x

y=-x

Two coordinates: (3,-3) and (4,-4)

m=\frac{-4-  -3}{4-3}=\frac{-1}{1}=-1

c=0

5. Work out the equation of the straight line

y=\frac{3}{2}x+2

x+2

y=\frac{2}{3}x+2

y=\frac{1}{3}x+2

Two coordinates: (0,2) and (3,4)

m=\frac{4-2}{3-0}=\frac{2}{3}

c=2

6 Work out the equation of the straight line

y=-\frac{1}{2}x-\frac{13}{2}

y=-\frac{13}{2}x-\frac{1}{2}

x-6.5

y=-\frac{1}{4}x-6.5

Two coordinates: (3,-8) and (-3,-5)

m=\frac{-5-  -8}{-3-3}=\frac{3}{-6}=-\frac{1}{2}

c=-\frac{13}{2}

### Equation of a line GCSE questions

1. Here is the graph of a straight line.

Work out the equation of the line in the form y=mx+c.

(3 marks)

Two coordinates: (12,-2) and (0,6) and Gradient m=\frac{6-  -2}{0-12}=\frac{8}{-12}=-\frac{3}{4}

(1)

c=6

(1)

y=-\frac{3}{4}x+6

(1)

2.  (a)  Calculate the slope of the line in the diagram below.

(b) Show that the equation of the line is the same as \frac{1}{2}y=x+2.

(4 marks)

(a)

Two coordinates: (0,4) and (2,8) and Gradient m=\frac{8-4}{2-0}=\frac{4}{2}=2

(1)

(b)

c=4

(1)

y=2x+4

(1)

(1)

3. The two straight lines A and B are parallel. The equation of line A is y=\frac{1}{2}x+7.

Given that equation B passes through the coordinate (4,3) , work out the equation of the straight line, B .

(4 marks)

Parallel line so m=\frac{1}{2}

(1)

At (4,3), \; 3=4\times\frac{1}{2}+c

(1)

c=1

(1)

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

(1)

## Learning checklist

You have now learned how to:

• Reduce a given linear equation in 2 variables to the standard form y = mx + c ; calculate and interpret gradients and intercepts of graphs of such linear equations numerically, graphically and algebraically

## Still stuck?

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#### GCSE Maths Papers - November 2022 Topics

Practice paper packs based on the November advanced information for Edexcel 2022 Foundation and Higher exams.

Designed to help your GCSE students revise some of the topics that are likely to come up in November exams.