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# 3.9: Perpendicular Lines in the Coordinate Plane

Difficulty Level: At Grade Created by: CK-12
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Practice Perpendicular Lines in the Coordinate Plane

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### Perpendicular Lines in the Coordinate Plane

Recall that the definition of perpendicular is two lines that intersect at a 90\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 14\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.

#### Calculating the Slope of Perpendicular Lines

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

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

m=2\begin{align*}m = 2\end{align*}, so m\begin{align*}m_\perp\end{align*} is the reciprocal and negative, m=12\begin{align*}m_\perp=-\frac{1}{2}\end{align*}.

b) y=23x5\begin{align*}y=-\frac{2}{3}x-5\end{align*}

m=23\begin{align*}m=-\frac{2}{3}\end{align*}, take the reciprocal and make the slope positive, m=32\begin{align*}m_\perp=\frac{3}{2}\end{align*}

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

Because there is no number in front of x\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, m=1\begin{align*}m_\perp=-1\end{align*}.

#### Finding the Equation of a Perpendicular Line

Find the equation of the line that is perpendicular to y=13x+4\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 13\begin{align*}-\frac{1}{3}\end{align*}. So, m=3\begin{align*}m = 3\end{align*}. Now, we need to find the y\begin{align*}y-\end{align*}intercept. 4 is the y\begin{align*}y-\end{align*}intercept of the given line, not our new line. We need to plug in 9 for x\begin{align*}x\end{align*} and -5 for y\begin{align*}y\end{align*} to solve for the new y\begin{align*}y-\end{align*}intercept (b)\begin{align*}(b)\end{align*}.

5532=3(9)+b=27+bTherefore, the equation of line is y=3x32.=b\begin{align*}-5 & = 3(9)+b\\ -5 & = 27 + b \qquad \text{Therefore, the equation of line is} \ y=3x-32.\\ -32 & = b\end{align*}

#### Graphing the Equation of a Line

Graph 3x4y=8\begin{align*}3x-4y=8\end{align*} and 4x+3y=15\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 y\begin{align*}y\end{align*}.

3x4y4yy=8=3x+8=34x24x+3y=153y=4x+15y=43x+5\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 y\begin{align*}y-\end{align*}intercept form), we can tell they are perpendicular because their slopes are opposite reciprocals.

### Examples

#### Example 1

Determine which of the following pairs of lines are perpendicular.

• y=2x+3\begin{align*}y=-2x+3\end{align*} and y=12x+3\begin{align*}y=\frac{1}{2}x+3\end{align*}
• y=4x2\begin{align*}y=4x-2\end{align*} and y=4x+5\begin{align*}y=4x+5\end{align*}
• y=x+5\begin{align*}y=-x+5\end{align*} and y=x+1\begin{align*}y=x+1\end{align*}

Two lines are perpendicular if their slopes are opposite reciprocals. The only pairs of lines this is true for is the first pair, because 2\begin{align*}-2\end{align*} and 12\begin{align*}\frac{1}{2}\end{align*} are opposites and reciprocals.

#### Example 2

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

The perpendicular line goes through (2, -2), but the slope is 12\begin{align*}-\frac{1}{2}\end{align*} because we need to take the opposite reciprocal of 2\begin{align*}2\end{align*}.

y221=12x+b=12(2)+b=1+b=b\begin{align*}y & = -\frac{1}{2}x+b\\ -2 & = -\frac{1}{2}(2) + b\\ -2 & = -1+b\\ -1 & =b\end{align*}

The equation is y=12x1\begin{align*}y = -\frac{1}{2}x-1\end{align*}.

#### Example 3

Give an example of a line that is perpendicular to the line y=23x4\begin{align*}y=\frac{2}{3}x-4\end{align*}.

3. Any line perpendicular to y=23x4\begin{align*}y=\frac{2}{3}x-4\end{align*} will have a slope of 32\begin{align*}-\frac{3}{2}\end{align*}. Any equation of the form y=32x+b\begin{align*}y=-\frac{3}{2}x+b\end{align*} will work.

### Review

1. Determine which of the following pairs of lines are perpendicular.
1. y=3x+1\begin{align*}y=-3x+1\end{align*} and y=3x1\begin{align*}y=3x-1\end{align*}
2. 2x3y=6\begin{align*}2x-3y=6\end{align*} and 3x+2y=6\begin{align*}3x+2y=6\end{align*}
3. 5x+2y=4\begin{align*}5x+2y=-4\end{align*} and 5x+2y=8\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.

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Date Created:
Jul 17, 2012
Oct 05, 2016
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