**The Equation of a Straight Line**

A straight line graph will always have an equation in the form y=mx+c. You need to be able to work out the equation of a straight line from a graph, as well as manipulate the equation itself.

There are **3 key skills** you need to learn involving the equation of a straight line.

Make sure you are familiar with the following topics before continuing.

## The straight line equation

Any straight line graph can be described by the following equation:

\textcolor{red}{y}=\textcolor{limegreen}{m}\textcolor{red}{x}+\textcolor{blue}{c}

where \textcolor{red}x and \textcolor{red}y are the coordinates the line passes through, \textcolor{limegreen}m is the gradient and \textcolor{blue}c is the y-intercept (the y-coordinate where the line crosses the y axis).

## Skill 1: **Finding the Equation of a Straight Line**

We need to be able to find the equation of a straight line from the graph.

**Example: **Find the equation of the straight line graph below

**Step 1:** Find \textcolor{blue}{c}

We are looking for an equation of the form,

y=\textcolor{limegreen}{m}x+\textcolor{blue}{c}

We know \textcolor{blue}{c} = y-intercept. Looking at the graph, we can see it crosses the axis at - 1, therefore we have \textcolor{blue}{c=-1}.

**Step 2:** Find the gradient (\textcolor{limegreen}{m})

Then, to work out the gradient,

\text{Gradient } = \dfrac{\textcolor{red}{\text{change in }y}}{\textcolor{blue}{\text{change in }x}}

The triangle we have drawn has height 4 and width 2, so we get,

m=\text{gradient}=\dfrac{\textcolor{red}{4}}{\textcolor{blue}{2}}=\textcolor{limegreen}{2}

Therefore, the equation of the straight line is,

y=\textcolor{limegreen}{2}x\textcolor{blue}{-1}

## Skill 2: Fin**ding the Equation of a Line Through Two Points **

Finding the equation of a straight line between two points is an important skill.

**Example:** Find the equation of the line that passes through (-3, 1) and (2, -14).

**Step 1:** Finding the **gradient**,

m=\text{gradient}=\dfrac{(-14)-1}{2-(-3)}=\dfrac{-15}{5}=-3

Now we know that m=-3, we know that our equation must take the form,

y=-3x+c

**Step 2: ****Substitute** the x and y values of one co-ordinate, say x = -3, y=1, into the equation,

1=(-3)\times(-3)+c=9+c

**Step 3:** Rearrange to solve for c,

c=1-9=-8

**Step 4:** Now we have all the components of the **equation of a line**, we can write the resulting equation as,

y=-3x-8

## Skill 3: **Rearranging Equations into the form y=mx+c**

It is often necessary to rearrange the equation of a line to get it in the form y=mx+c. This is essential for finding the gradient and y-intercept.

**Example:** Find the gradient and y-intercept of the line x+2y=14.

We want to **rearrange this equation** to make y the subject. So, subtracting x from both sides, we get

2y=-x+14

Then, dividing both sides by 2, we get

y=-\dfrac{1}{2}x+7

Therefore, the gradient is -\dfrac{1}{2} and the y-intercept is 7.

## GCSE Maths Revision Guide

£14.99### Example Questions

**Question 1:** Determine the equation of the line shown below.

**[2 marks]**

We want an equation of the form

y=mx+c

So, we need to find the gradient, m, and y-intercept, c.

Firstly, looking at the graph we can see that the y-intercept is 2, so c=2.

Now, we will find the gradient by drawing a triangle underneath the line in question.

The triangle we have drawn has height 1 and width 3, so we get

m=\text{gradient}=\dfrac{1}{3}

Therefore, the equation of the line is

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

**Question 2: **Determine the equation of the line that passes through (-3, -6) and (2, 34).

**[2 marks]**

We want an equation of the form

y=mx+c

So, we need to find the gradient, m, and y-intercept, c.

Firstly, we will find the gradient by dividing the difference in the y coordinates by the difference in the x coordinates:

m=\text{gradient}=\dfrac{-6-34}{-3-2}=\dfrac{-40}{-5}=8

Therefore, the equation of the line is

y=8x+c

Then, to find c we will substitute one pair of coordinates that the line passes through into the equation and rearrange. Here, we’ll pick (2, 34). Subbing this in, we get

34=8\times2+c=16+c

Subtracting 16 from both sides, we get

c=34-16=18

Therefore, the equation of the line is

y=8x+18.

**Question 3:** Determine the equation of the line shown below.

**[2 marks]**

We want an equation of the form

y=mx+c

So, we need to find the gradient, m, and y-intercept, c.

Firstly, looking at the graph we can see that the y-intercept is - 1, so c=-1.

Now, we will find the gradient by drawing a triangle underneath the line in question. Hence

m=\text{gradient}=\dfrac{3}{2}

Therefore, the equation of the line is

y=\dfrac{3}{2}x-1

### Worksheets and Exam Questions

#### (NEW) y=mx+c Exam Style Questions - MME

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