y=mx+c Worksheets | Questions and Revision | MME

# y=mx+c Worksheets, Questions and Revision

Level 4 Level 5

## The equation of a straight line

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

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

$\textcolor{red}{m}$ is the gradient

$\textcolor{blue}{c}$ is the $y$-intercept.‏ (the point where the line touches the $y$ axis)

$\textcolor{limegreen}{x}$ and $\textcolor{limegreen}{y}$ are the coordinates the line passes through.

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

## Example 1: Finding y=mx+c

Find the equation of the straight line shown below.

So, we are looking for an equation of the form

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

First, finding $\textcolor{blue}{c}$ the $y$-intercept – looking at the graph, we can see it touches the axis at -1, therefore we have $\textcolor{blue}{c=-1}$.

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

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

Therefore, we have that the equation of the straight line is

$y=2x-1$

## Example 2

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

Like in the last example, we want an equation of the form

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

Firstly, the gradient in the same way as before,

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

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

$y=-3x+c$.

Finding $\textcolor{blue}{c}$.

We use the $x$ and $y$ valyues of one co-ordinate and plug them into the equation

$(-3,1) = (x,y)$

$x = -3, y=1$

Putting these into the equation we get

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

Now solve for $c$

$c=1-9=-8$

Now we have all the components of a line equation, we can write the resulting equation as

$y=-3x-8$.

## Example 3

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 $-\dfrac{1}{2}$ and the $y$-intercept is 7.

### Example Questions

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

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

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 the 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

$y=8x+18$.

Level 4-5

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