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3.9: Perpendicular Lines in the Coordinate Plane

Difficulty Level: At Grade Created by: CK-12
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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


Recall that the definition of perpendicular is two lines that intersect at a \begin{align*}90^\circ\end{align*}, or right, angle. In the coordinate plane, that would look like this:

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*}. This can be generalized to any pair of perpendicular lines in the coordinate plane. 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*}, so \begin{align*}m_\perp\end{align*} is the reciprocal and negative, \begin{align*}m_\perp=-\frac{1}{2}\end{align*}.

b) \begin{align*}m=-\frac{2}{3}\end{align*}, take the reciprocal and make the slope positive, \begin{align*}m_\perp=\frac{3}{2}\end{align*}.

c) Because there is no number in front of \begin{align*}x\end{align*}, the slope is 1. The reciprocal of 1 is 1, so the only thing to do is make it negative, \begin{align*}m_\perp=-1\end{align*}.

Example B

Find the equation of the line that is perpendicular to \begin{align*}y=-\frac{1}{3}x+4\end{align*} and passes through (9, -5).

First, the slope is the reciprocal and opposite sign of \begin{align*}-\frac{1}{3}\end{align*}. So, \begin{align*}m = 3\end{align*}. Now, we need to find the \begin{align*}y-\end{align*}intercept. 4 is the \begin{align*}y-\end{align*}intercept of the given line, 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*}.


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

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

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

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Difficulty Level:
At Grade
Date Created:
Jul 17, 2012
Last Modified:
Feb 26, 2015
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