How to write the equation of a line in point-slope form

If you know a point on a line and how steep the line is, you already have what you need to write its equation. We’ll plug those values into point-slope form, then see how to find the slope when we’re given two points.

Start with a point and the slope

If a line has slope $m$ and passes through $(x_{1}, y_{1})$ , its point-slope form is

y - y_{1} = m (x - x_{1})

The slope describes how much $y$ changes for each unit of change in $x$ . So the change in $y$ , which is $y - y_{1}$ , equals the slope times the change in $x$ , which is $x - x_{1}$ .

Example: Use the given point and slope

The slope is $- \frac{1}{4}$ and the point is $(- 6, 1)$ . Use $x_{1} = - 6$ and $y_{1} = 1$ . Subtracting the negative x-coordinate gives $x + 6$ .

\begin{array}{l} y - 1 = - \frac{1}{4} (x - (- 6)) \\ y - 1 = - \frac{1}{4} (x + 6) \end{array}

Example: Find the slope from two points

Use $(- 2, - 4)$ and $(3, 5)$ . Subtract the y-coordinates and the x-coordinates in the same order. The change in $y$ is $9$ and the change in $x$ is $5$ .

From (−2, −4) to (3, 5), x increases by 5 and y increases by 9.

\begin{array}{l} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m = \frac{5 - (- 4)}{3 - (- 2)} = \frac{9}{5} \\ y + 4 = \frac{9}{5} (x + 2) \end{array}

We used $(- 2, - 4)$ as the known point. Using $(3, 5)$ gives $y - 5 = \frac{9}{5} (x - 3)$ , which describes the same line.

Example: Keep the point visible in the equation

For the points $(4, 2)$ and $(6, 3)$ , the slope is $\frac{3 - 2}{6 - 4} = \frac{1}{2}$ . Using $(4, 2)$ gives

y - 2 = \frac{1}{2} (x - 4)

Leaving the equation in this form keeps the point and slope visible. If you want the y-intercept, you can rewrite it in slope-intercept form.

Write a vertical line as x equals a constant

If two distinct points have the same x-coordinate $a$ , then $x_{2} - x_{1} = 0$ and the slope is undefined. The equation of that vertical line is $x = a$ . A horizontal line has slope zero, so point-slope form simplifies to $y = y_{1}$ .

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