$y=mx+c$

In a nutshell

$y=mx+c$ is the equation for a straight line, where $m$ is the gradient of the line and $c$ is the $y$ -axis intercept. $m$ tells you how steep a line is by tracking the change in $y$ over the change in $x$ . $c$ tells you where the line intercepts the $y$ -axis. This is a handy equation because it allows you to learn a lot about a line with only a pair of coordinates. Once you understand it intuitively, you can plot out lines without graph paper.

The gradient formula

A gradient tells you how steep a line is between two points $(x_1,y_1)$ and $(x_2, y_2)$ . It's calculated as:

$\boxed{m = \dfrac{\Delta y}{\Delta x}=\dfrac{y_2-y_1}{x_2-x_1}}$

Intuitively, you can see that a greater change in $y$ or $x$ would lead to a steeper line on paper. Therefore, a higher $m$ value means a greater gradient.

Maths; Straight line graphs; KS5 Year 12; y = mx + c

Three lines with the equations $y=x+4$ , $y=5x+4$ and $y=-5x+4$ .

Once you have the coordinates of a line, you can work out the gradient easily.

Procedure

1.	Assign coordinates to $(x_1,y_1)$ and $(x_2,y_2)$ .
2.	Plug those numbers into the formula for $m$ .
3.	Solve for $m$ .

Example 1

Find the gradient for a line between points $(2,3)$ and $(6,7)$ .

Assign values for $(x_1,y_1)$ and $(x_2,y_2)$ .

$x_1 = 2$ , $y_1 = 3$ , $x_2= 6$ and $y_2=7$

Substitute into the formula for $m$ and solve.

$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\\\m&= \dfrac{7-3}{6-2}\\\\m&= \dfrac{4}{4}\\\end{aligned}$ 

$\underline{m=1}$ 

Working backwards

You can works backwards if you are given just the gradient of a line and some coordinates. This can let you find where a line terminates.

Example 2

A line has coordinates $(4,8)$ and $(-1, a)$ . The gradient of the line is $-2.5$ . What is the value of coordinate a?

Assign values to $(x_1,y_1)$ and $(x_2,y_2)$

$\begin{aligned}(4,8)&(-1,a)\\(x_1,y_1)&(x_2, y_2)\\x_1 = 4, x_2 &= -1, y_1=8, y_2=a\\m&=-2.5\\\end{aligned}$

Substitute into the formula for $m$

$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\\\-2.5 &=\dfrac{a-8}{-1-4}\\\end{aligned}$

Solve for $a$ 

$\begin{aligned}\dfrac{a-8}{-5}&=-2.5\\a&= 12.5 + 8 \\&\underline{a=20.5}\\\end{aligned}$

The y-intercept, c

$c$ is known as the $y$ -intercept. It is the point where the line has an $x$ value of zero. Visually, this is the point where the line hits the $y$ -axis, as the $y$ -axis lies where $x=0$ . To find the intercept from a complete line equation, all you need to do is make $x=0$ , i.e, remove the $m$ and $x$ from the equation.

Maths; Straight line graphs; KS5 Year 12; y = mx + c

Three lines with equations $y=x+4$ , $y=x+2$ and $y=x-4$ .

The $x$ -intercept is needed less often. It is simply where the line hits the $x$ -axis. To find it, you make $y=0$ and solve accordingly.

Example 3

What is the gradient and the y-intercept of line $3y+ 9x - 12 = 6$ ?

Rearrange the equation into the form $y=mx + c$ and solve.

$\begin{aligned}3y+9x-12&=6\\3y &= -9x +12 + 6\\3y&=-9x + 18\\y&=-3x + 6\\&\underline{m= -3, c= 6}\\\end{aligned}$