Linear Equations in Point-Slope Form

If you are given an equation in point-slope form, it is not necessary to re-write it in slope-intercept form in order to graph it. The point-slope form of the equation gives you enough information so you can graph the line.

Let's make a graph of the line given by the equation \(\begin{align*}y-2=\frac{2}{3}(x+2)\end{align*}\):

Begin by rewriting the equation to make it point-slope form: \(\begin{align*}y-2= \frac{2}{3}(x-(-2))\end{align*}\) Now we see that point (–2, 2) is on the line and that the \(\begin{align*}\text{slope}=\frac{2}{3}\end{align*}\). First plot point (–2, 2) on the graph.

[Figure1]

A slope of \(\begin{align*}\frac{2}{3}\end{align*}\) tells you that from your point you should move 2 units up and 3 units to the right and draw another point.

[Figure2]

Now draw a line through the two points and extend the line in both directions.

[Figure3]

Writing a Linear Function in Point-Slope Form

Remember from the previous Concept that \(\begin{align*}f(x)\end{align*}\) and \(\begin{align*}y\end{align*}\) are used interchangeably. Therefore, to write a function in point-slope form, you replace \(\begin{align*}y-y_1\end{align*}\) with \(\begin{align*}f(x)-y_1\end{align*}\).

Let's write the equation of the linear function with \(\begin{align*}m=9.8\end{align*}\) and \(\begin{align*}f(5.5)=12.5\end{align*}\) in point-slope form:

This function has a slope of 9.8 and contains the ordered pair (5.5, 12.5). Substituting the appropriate values into point-slope form, we get the following:

\(\begin{align*}y-12.5=9.8(x-5.5)\end{align*}\)

Replacing \(\begin{align*}y-y_1\end{align*}\) with \(\begin{align*}f(x)-y_1\end{align*}\), the equation in point-slope form is:

\(\begin{align*}f(x)-12.5& =9.8(x-5.5)\\ f(x)-12.5 =9.8x-53.9\\ f(x) =9.8x - 41.4 \end{align*}\)

where the last equation is in slope-intercept form.