### Perpendicular Lines in the Coordinate Plane

**Perpendicular** lines are 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, 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 reciprocals of each other.

What if you were given two perpendicular lines in the coordinate plane? What could you say about their slopes?

### Examples

#### Example 1

Find the slope of the line that is perpendicular to this line: \begin{align*}y = -\frac{2}{3}x-5\end{align*}.

\begin{align*}m = -\frac{2}{3}\end{align*}, take the reciprocal and change the sign, \begin{align*}m_\perp=\frac{3}{2}\end{align*}.

#### Example 2

Find the slope of the line that is perpendicular to this line: \begin{align*}y=x+2\end{align*}.

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 3

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 opposite reciprocal of \begin{align*}-\frac{1}{3}\end{align*}. So, \begin{align*}m = 3\end{align*}. 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\\ -32 & = b\end{align*}

Therefore, the equation of the perpendicular line is \begin{align*}y=3x-32\end{align*}.

#### Example 4

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.

#### Example 5

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).

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*}.

### Review

Determine if each pair of lines are perpendicular. Then, graph each pair on the same set of axes.

- \begin{align*}y=-2x+3\end{align*} and \begin{align*}y = \frac{1}{2}x+3\end{align*}
- \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*}x-3y=-3\end{align*} and \begin{align*}x+3y=9\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*}

Find the equations of the two lines in each graph below. Then, determine if the two lines are perpendicular.

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

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.9.