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Rearranging equations Plotting graphs Gradient of a lineThis topic is relevant for:

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 youβre still stuck.

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.

Letβs look at the line y = 3x β 4 .

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

E.g.

When x = 1 ,

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 β1
- y = β3 (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 = β17 .

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

In order to find the equation of a straight line:

**Calculate the gradient of the line****State the y -intercept of the straight line****Write the equation of the line in the form y = mx + c**

Get your free equation of a line worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free equation of a line worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Equation of a line** 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:

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

**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.

So the gradient

\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}2**State the y-intercept of the straight line**.

The y -intercept occurs when x = 0 .

Here, c = 4

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

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

y=2x+4Work out the equation of the straight line given in the diagram below.

**Calculate the gradient of the line**.

Two points that lie on the line are: ( 1, β4 ) and ( 3, 2 ).

So the gradient

\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}

**State the ** y **-intercept of the straight line**.

The y -intercept occurs when x = 0 .

Here, c = β7

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

As m = 3 and c = β7, y = 3x β 7

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

**Calculate the gradient of the line**.

Two points that lie on the line are: ( 1, β2 ) and ( β1, 8 ).

So the gradient

\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}

**State the ** y **-intercept of the straight line**.

The y -intercept occurs when x = 0 .

Here, c = 3

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

As m = β5 and c = 3, y = β5x + 3

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

**Calculate the gradient of the line**.

Two points that lie on the line are: ( β3, β2 ) and ( 2, β7 ).

So the gradient

\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}

**State the ** y **-intercept of the straight line**.

The y -intercept occurs when x = 0 .

Here, c = β5

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

As m = β1 and c = β5, y = βx β 5

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

**Calculate the gradient of the line**.

Two points that lie on the line are: ( 4, 10 ) and ( β2, 7 ).

So the gradient

\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}

**State the ** y **-intercept of the straight line**.

The y -intercept occurs when x = 0 .

Here, c = 8

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

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

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

**Calculate the gradient of the line**.

Two points that lie on the line are: ( β2, β4 ) and ( 2, β5 ).

So the gradient

\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} .

**State the ** y **-intercept of the straight line**.

The y -intercept occurs when x = 0 .

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

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

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

y**Mixing up the gradient and the****-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, β2 ) and ( β1,8 ).

The change in y could be correctly calculated as 8 \; β \; β2 but the change in x is then incorrectly calculated as 1 \; β \; β1 . This would result in the gradient of the line being 5 and not β5 .

Correct answer: m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8--2}{-1-1}=\frac{10}{-2}=-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 .

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-Β -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-Β -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-Β -8}{-3-3}=\frac{3}{-6}=-\frac{1}{2}

c=-\frac{13}{2}

1. Here is the graph of a straight line.

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

**(3 marks)**

Show answer

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

**(1)**

c=6

**(1)**

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

**(1)**

2. Β (a)Β 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)**

Show answer

(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)**

Show answer

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)**

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

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