y=mx+c

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y=mx+c Revision

The Equation of a Straight Line ( $y=mx+c$ )

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

y=mx+c finding the equation of a straight line

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 $\textcolor{blue}{c=-1}$ .

y=mx+c finding the gradient of a straight line

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: Finding 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 $7$ .

y=mx+c Example Questions

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

[2 marks]

y=mx+c example 1

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

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

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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

$y=8x+18$ .

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