# 3.9: Perpendicular Lines in the Coordinate Plane

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

**Practice**Perpendicular Lines in the Coordinate Plane

What if you wanted to figure out if two lines in a shape met at right angles? How could you do this? After completing this Concept, you'll be able to use slope to help you determine whether or not lines are perpendicular.

### Watch This

CK-12 Foundation: Chapter3PerpendicularLinesintheCoordinatePlaneA

Watch the portion of this video that deals with Perpendicular Lines

Khan Academy: Equations of Parallel and Perpendicular Lines

### Guidance

Recall that the definition of **perpendicular** is two lines that intersect at a \begin{align*}90^\circ\end{align*}

If we take a closer look at these two lines, we see that the slope of one is -4 and the other is \begin{align*}\frac{1}{4}\end{align*}*The slopes of perpendicular lines are opposite signs and reciprocals of each other.*

#### Example A

Find the slope of the perpendicular lines to the lines below.

a) \begin{align*}y=2x+3\end{align*}

b) \begin{align*}y=-\frac{2}{3}x-5\end{align*}

c) \begin{align*}y=x+2\end{align*}

We are only concerned with the slope for each of these.

a) \begin{align*}m = 2\end{align*}

b) \begin{align*}m=-\frac{2}{3}\end{align*}

c) Because there is no number in front of \begin{align*}x\end{align*}

#### Example B

Find the equation of the line that is perpendicular to \begin{align*}y=-\frac{1}{3}x+4\end{align*}

First, the slope is the reciprocal and opposite sign of \begin{align*}-\frac{1}{3}\end{align*}*not our new line*. We need to plug in 9 for \begin{align*}x\end{align*} and -5 for \begin{align*}y\end{align*} to solve for the *new* \begin{align*}y-\end{align*}intercept \begin{align*}(b)\end{align*}.

\begin{align*}-5 & = 3(9)+b\\ -5 & = 27 + b \qquad \text{Therefore, the equation of line is} \ y=3x-32.\\ -32 & = b\end{align*}

#### Example C

Graph \begin{align*}3x-4y=8\end{align*} and \begin{align*}4x+3y=15\end{align*}. Determine if they are perpendicular.

First, we have to change each equation into slope-intercept form. In other words, we need to solve each equation for \begin{align*}y\end{align*}.

\begin{align*}3x-4y & = 8 && 4x+3y = 15\\ -4y & = -3x+8 && 3y = -4x + 15\\ y & = \frac{3}{4}x-2 && y = -\frac{4}{3}x+5\end{align*}

Now that the lines are in slope-intercept form (also called \begin{align*}y-\end{align*}intercept form), we can tell they are perpendicular because their slopes are opposite reciprocals.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter3PerpendicularLinesintheCordinatePlaneB

### Guided Practice

1. Determine which of the following pairs of lines are perpendicular.

- \begin{align*}y=-2x+3\end{align*} and \begin{align*}y=\frac{1}{2}x+3\end{align*}

- \begin{align*}y=4x-2\end{align*} and \begin{align*}y=4x+5\end{align*}

- \begin{align*}y=-x+5\end{align*} and \begin{align*}y=x+1\end{align*}

2. Find the equation of the line that is perpendicular to the line \begin{align*}y=2x+7\end{align*} and goes through the point (2, -2).

3. Give an example of a line that is perpendicular to the line \begin{align*}y=\frac{2}{3}x-4\end{align*}.

**Answers:**

1. Two lines are perpendicular if their slopes are opposite reciprocals. The only pairs of lines this is true for is the *first* pair, because \begin{align*}-2\end{align*} and \begin{align*}\frac{1}{2}\end{align*} are opposites and reciprocals.

2. The perpendicular line goes through (2, -2), but the slope is \begin{align*}-\frac{1}{2}\end{align*} because we need to take the opposite reciprocal of \begin{align*}2\end{align*}.

\begin{align*}y & = -\frac{1}{2}x+b\\ -2 & = -\frac{1}{2}(2) + b\\ -2 & = -1+b\\ -1 & =b\end{align*}

The equation is \begin{align*}y = -\frac{1}{2}x-1\end{align*}.

3. Any line perpendicular to \begin{align*}y=\frac{2}{3}x-4\end{align*} will have a slope of \begin{align*}-\frac{3}{2}\end{align*}. Any equation of the form \begin{align*}y=-\frac{3}{2}x+b\end{align*} will work.

### Interactive Practice

### Explore More

- Determine which of the following pairs of lines are perpendicular.
- \begin{align*}y=-3x+1\end{align*} and \begin{align*}y=3x-1\end{align*}
- \begin{align*}2x-3y=6\end{align*} and \begin{align*}3x+2y=6\end{align*}
- \begin{align*}5x+2y=-4\end{align*} and \begin{align*}5x+2y=8\end{align*}
- \begin{align*}x-3y=-3\end{align*} and \begin{align*}x+3y=9\end{align*}
- \begin{align*}x+y=6\end{align*} and \begin{align*}4x+4y=-16\end{align*}

Determine the equation of the line that is ** perpendicular** to the given line, through the given point.

- \begin{align*}y=x-1; \ (-6, \ 2)\end{align*}
- \begin{align*}y=3x+4; \ (9, \ -7)\end{align*}
- \begin{align*}5x-2y=6; \ (5, \ 5)\end{align*}
- \begin{align*}y = 4; \ (-1, \ 3)\end{align*}
- \begin{align*}x = -3; \ (1, \ 8)\end{align*}
- \begin{align*}x - 3y = 11; \ (0, \ 13)\end{align*}

Determine if each pair of lines is perpendicular or not.

For the line and point below, find a perpendicular line, through the given point.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 3.9.

### Image Attributions

## Description

## Learning Objectives

Here you'll learn properties of perpendicular lines in the coordinate plane, and how slope can help you to determine whether or not two lines are perpendicular.

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## Date Created:

Jul 17, 2012## Last Modified:

Feb 26, 2015## Vocabulary

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