Calculating the Equation of a Straight Line

What determines a straight line?

A non-vertical straight line can be written as

$y=mx+c$

Here, $m$ is the gradient and $c$ is the $y$-coordinate of the point where the line crosses the $y$-axis. Because every point on the $y$-axis has $x=0$, this intercept is the point $(0,c)$.

To determine one particular line, you need enough information to find both $m$ and $c$. One point alone is not enough: infinitely many lines pass through it. A point together with a gradient does determine a line. Two distinct points usually determine both the gradient and the intercept.

A line from one point and its gradient

Suppose a line has gradient and passes through . Begin with and substitute the known gradient and the coordinates of the point:

$1=(-1)(3)+c$

Solving gives $c=4$, so the equation is

$y=-x+4$

This works because the coordinates of every point on a line satisfy its equation. Substitution uses the known point to reveal the missing intercept.

An equivalent shortcut is the point–gradient form

$y-y_1=m(x-x_1),$

where $(x_1,y_1)$ is a known point. For the same information,

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

which expands to $y=-x+4$. This form is especially useful when the intercept is not immediately visible, although you may need to rearrange the answer into $y=mx+c$.

A line through two points

For points $(x_1,y_1)$ and $(x_2,y_2)$, first calculate the change in divided by the corresponding change in :

$m=\frac{y_2-y_1}{x_2-x_1}.$

The subtraction order must match. If you calculate $y_2-y_1$ on top, calculate $x_2-x_1$ underneath.

For example, take the points $(-1,-4)$ and $(2,11)$:

$m=\frac{11-(-4)}{2-(-1)}=\frac{15}{3}=5.$

Now use either point in $y=mx+c$. Using $(-1,-4)$ gives

$-4=5(-1)+c,$

so $c=1$. Therefore,

$y=5x+1.$

The gradient means that increases by whenever increases by . The constant means that the line crosses the -axis at .

Data for Finding a line from two points

Checking the equation

Substitute the other given point into the completed equation. For $(2,11)$:

$5(2)+1=11,$

so the point satisfies $y=5x+1$. You can also check that the gradient and in the answer are the values you found. Merely reusing the same point that found is a weaker check, because it may repeat an earlier error.

Horizontal and vertical lines

If two points have the same $y$-coordinate, the change in $y$ is zero, so $m=0$. The equation is horizontal, such as $y=7$.

If two distinct points have the same $x$-coordinate, then $x_2-x_1=0$, so the gradient calculation would require division by zero. The line is vertical and cannot be written as . If both points have , for example, its equation is simply .

Recall the structure

For a non-vertical line:

$y=mx+c$

• $m$: gradient