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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 ,
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.
What is the equation of a line?
How to find the equation of a line
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
Explain how to find the equation of a line
Equation of a line worksheet
Get your free equation of a line worksheet of 20+ questions and answers. Includes reasoning and applied questions.
DOWNLOAD FREE Equation of a line worksheet
Get your free equation of a line worksheet of 20+ questions and answers. Includes reasoning and applied questions.
DOWNLOAD FREERelated lessons on straight line graphs
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:
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.
- 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}2State the
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+4Example 2: Positive gradient, negative y -intercept
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, −4 ) 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 = −7
Write the equation of the line in the form y=mx+c .
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.
Calculate the gradient of the line.
Two points that lie on the line are: ( 1, −2 ) and ( −1, 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 = −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.
Calculate the gradient of the line.
Two points that lie on the line are: ( −3, −2 ) and ( 2, −7 ).
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 = −5
Write the equation of the line in the form y=mx+c .
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.
Calculate the gradient of the line.
Two points that lie on the line are: ( 4, 10 ) and ( −2, 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
Example 6: Negative fractional gradient, negative y -intercept
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: ( −2, −4 ) and ( 2, −5 ).
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} .
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 .
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 .
Practice equation of a line questions
1. Work out the equation of the straight line
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
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
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
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
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
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
The next lessons are
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