# 5.2: Linear Equations in Point-Slope Form

**At Grade**Created by: CK-12

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

The line between any two points \begin{align*}(x_1,y_1)\end{align*}

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

- The slope of the line
- A point on the line

**Example 1:** *Write an equation for a line containing* (9, 3) and (4, 5).

**Solution:** 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*}

\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*}

\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.**

**Example 2:** Rewrite \begin{align*}y-5=3(x-2)\end{align*}

**Solution:** 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*}

## Graphing Equations Using 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.

**Example 3:** *Make a graph of the line given by the equation \begin{align*}y-2=\frac{2}{3}(x+2)\end{align*} y−2=23(x+2)*

**Solution:** Begin by rewriting the equation to make point-slope form: \begin{align*}y-2= \frac{2}{3}(x-(-2))\end{align*}

A slope of \begin{align*}\frac{2}{3}\end{align*}

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

## Writing a Linear Function in Point-Slope Form

Remember from the previous lesson that \begin{align*}f(x)\end{align*}

**Example 4:** *Write the equation of the linear function in point-slope form.*

\begin{align*}m=9.8\end{align*}

**Solution:** 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,

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

Replacing \begin{align*}y-y_1\end{align*}

\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*}

## Solving Situations Involving Point-Slope Form

Let’s solve some word problems where point-slope form is needed.

**Example 5:** *Marciel rented a moving truck for the day. Marciel remembers only that the rental truck company charges $40 per day and some amount of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?*

**Solution:** Define the variables: \begin{align*}x=\end{align*} distance in miles; \begin{align*}y=\end{align*} cost of the rental truck in dollars. There are two ordered pairs: (0, 40) and (46, 63).

**Step 1:** Begin by finding the slope: \begin{align*}\frac{63-40}{46-0}=\frac{23}{46}=\frac{1}{2}\end{align*}.

**Step 2:** Substitute the slope for \begin{align*}m\end{align*} and one of the coordinates for \begin{align*}(x_1,y_1)\end{align*}.

\begin{align*}y-40= \frac{1}{2} (x-0)\end{align*}

To find out how much will it cost to rent the truck for 220 miles, substitute 220 for the variable \begin{align*}x\end{align*}.

\begin{align*}y-40 & = \frac{1}{2} (220-0)\\ y-40 & =0.5(220) \Rightarrow y=\$150\end{align*}

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Linear Equations in Point-Slope Form (9:38)

- What is the equation for a line containing the points \begin{align*}(x_1,y_1)\end{align*} and \begin{align*}(x_2,y_2)\end{align*} in point-slope form?
- In what ways is it easier to use point-slope form rather than slope-intercept form?

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

- Slope is \begin{align*}\frac{1}{3}\end{align*}; \begin{align*}y-\end{align*}intercept –4.
- Slope is \begin{align*}-\frac{1}{10}\end{align*} and contains (10, 2).
- Slope is –75 and contains point (0, 125).
- Slope is 10 and contains point (8, –2).
- The line contains points (–2, 3) and (–1, –2).
- The line contains the points (0, 0) and (1, 2).
- The line contains points (10, 12) and (5, 25).
- The line contains points (2, 3) and (0, 3).
- The line has slope \begin{align*}\frac{3}{5}\end{align*} and \begin{align*}y-\end{align*}intercept –3.
- The line has slope –6 and \begin{align*}y-\end{align*}intercept 0.5.
- The line contains the points (–4, –2) and (8, 12).

Write each equation in slope-intercept form.

- \begin{align*}y-2=3(x-1)\end{align*}
- \begin{align*}y+4=\frac{-2}{3}(x+6)\end{align*}
- \begin{align*}0=x+5\end{align*}
- \begin{align*}y=\frac{1}{4}(x-24)\end{align*}

In 18 – 25, write the equation of the linear function in point-slope form.

- \begin{align*}m=-\frac{1}{5}\end{align*} and \begin{align*}f(0)=7\end{align*}
- \begin{align*}m=-12\end{align*} and \begin{align*}f(-2)=5\end{align*}
- \begin{align*}f(-7)=5\end{align*} and \begin{align*}f(3)=-4\end{align*}
- \begin{align*}f(6)=0\end{align*} and \begin{align*}f(0)=6\end{align*}
- \begin{align*}m=3\end{align*} and \begin{align*}f(2)=-9\end{align*}
- \begin{align*}m=-\frac{9}{5}\end{align*} and \begin{align*}f(0)=32\end{align*}
- \begin{align*}m=25\end{align*} and \begin{align*}f(0)=250\end{align*}
- \begin{align*}f(32)=0\end{align*} and \begin{align*}f(77)=25\end{align*}
- Nadia is placing different weights on a spring and measuring the length of the stretched spring. She finds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm. Write an equation in point-slope form that describes this situation. What is the unstretched length of the spring?
- Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet. Write an equation in point-slope form that describes this situation. What was the submarine’s depth five minutes after it started surfacing?
- Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month, she adds up her sales and she figures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary?

**Mixed Review**

- Translate into a sentence: \begin{align*}4(j+2)=400\end{align*}.
- Evaluate \begin{align*}0.45 \cdot 0.25-24 \div \frac{1}{4}\end{align*}.
- The formula to convert Fahrenheit to Celsius is \begin{align*}C(F)=\frac{F-32}{1.8}\end{align*}. What is the Celsius equivalent to \begin{align*}35^\circ F\end{align*}?
- Find the rate of change: The diver dove 120 meters in 3 minutes.
- What percent of 87.4 is 106?
- Find the percent of change: The original price was $25.00. The new price is $40.63.
- Solve for \begin{align*}w: \ 606=0.045(w-4000)+0.07w\end{align*}.

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## Date Created:

Feb 22, 2012## Last Modified:

Dec 11, 2014**Save or share your relevant files like activites, homework and worksheet.**

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