Linear Equations in Point-Slope Form

Equations can be written in many forms. The previous Concepts taught you how to write equations of lines in slope-intercept form. This Concept will provide a second way to write an equation of a line: point-slope form.

The equation of the line between any two points \(\begin{align*}(x_1,y_1)\end{align*}\) and \(\begin{align*}(x_2,y_2)\end{align*}\) can be written in the following form: \(\begin{align*}y-y_1=m(x-x_1)\end{align*}\).

To write an equation in point-slope form, you need two things:

1. The slope of the line
2. A point on the line

Let's write the equation for the a line containing (9, 3) and (4, 5) in point-slope form:

Begin by finding the slope.

\(\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{4-9}=-\frac{2}{5}\end{align*}\)

Instead of trying to find \(\begin{align*}b\end{align*}\) (the \(\begin{align*}y-\end{align*}\)intercept), you will use the point-slope formula.

\(\begin{align*}y-y_1& =m(x-x_1)\\ y-3& = \frac{-2}{5}(x-9)\end{align*}\)

It doesn't matter which point you use.

You could also use the other ordered pair to write the equation:

\(\begin{align*}y-5= \frac{-2}{5}(x-4)\end{align*}\)

These equations may look completely different, but by solving each one for \(\begin{align*}y\end{align*}\), you can compare the slope-intercept form to check your answer.

\(\begin{align*}y-3& = \frac{-2}{5} (x-9) \Rightarrow y=\frac{-2}{5} x+\frac{18}{5}+3\\ y& =\frac{-2}{5} x+\frac{33}{5}\\ y-5& =\frac{-2}{5} (x-4) \\ y& =\frac{-2}{5} x+\frac{8}{5}+5\\ y& =\frac{-2}{5} x+\frac{33}{5}\end{align*}\)

This process is called rewriting in slope-intercept 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.

Example 1

Earlier, you were told that the cost of a wedding was a function of the number of guests attending. If you knew the slope of the function and you also knew how much the wedding would cost if 150 guests attended, could you write a linear equation representing this situation? If so, what form of the equation would be easiest to use?

Yes, you could write a linear equation if you knew the slope of the function and how much the wedding would cost if 150 guests attended. The price of the wedding if 150 guests attended would be one point on the line. Then, you could use point-slope form to write an equation. Point-slope form would be the best because those are the pieces of information that you have. If absolutely necessary, you could transform the equation into a function in slope-intercept form. 

Example 2

Rewrite \(\begin{align*}y-5=3(x-2)\end{align*}\) in slope-intercept form.

Use the Distributive Property to simplify the right side of the equation:

\(\begin{align*}y-5=3x-6\end{align*}\)

Solve for \(\begin{align*}y\end{align*}\):

\(\begin{align*}y-5+5& =3x-6+5\\ y& =3x-1\end{align*}\)

Write the equation for the line in point-slope form.

3. The slope is \(\begin{align*}\frac{1}{3}\end{align*}\); the \(\begin{align*}y-\end{align*}\)intercept is –4.

4. The slope is \(\begin{align*}-\frac{1}{10}\end{align*}\) and contains the point (10, 2).

7. The line contains the points (–2, 3) and (–1, –2).

Write each equation in slope-intercept form.

14. \(\begin{align*}y-2=3(x-1)\end{align*}\)

15. \(\begin{align*}y+4=\frac{-2}{3}(x+6)\end{align*}\)

Part 1

3. \(y+4=\frac{1}{3}(x-0) \quad\) or \(\quad y+4=\frac{1}{3} x\)

4. \(y-2=-\frac{1}{10}(x-10)\)

7. \(y-3=-5(x+2) \quad\) or \(\quad y+2=-5(x+1)\)

8. \(y-0=2(x-0)\) or \(y-2=2(x-1)\)

Part 2

14. \(y=3 x-1\)

15. \(y=-\frac{2}{3} x-8\)