Suppose a coordinate grid is transposed over the blueprint for a house under construction, and two lines on the blueprint are perpendicular to each other. If one of the lines has the equation

### Equations of Perpendicular Lines

Lines can be parallel, **coincident** (overlap each other), or intersecting (crossing). Lines that intersect at

**Perpendicular lines** form a right angle. The product of their slopes is –1.

#### Let's verify that the following lines are perpendicular:

Line

Find the slopes of each line.

To verify that the lines are perpendicular, the product of their slopes must equal –1.

Because the product of their slopes is

#### Now, let's determine whether the following two lines are parallel, perpendicular, or neither:

Line 1:

Begin by finding the slopes of lines 1 and 2.

The slope of the first line is 2.

The slope of the second line is –2.

These slopes are not identical, so these lines are not parallel.

To check if the lines are perpendicular, find the product of the slopes.

Lines 1 and 2 are neither parallel nor perpendicular.

#### Writing Equations of Perpendicular Lines

Writing equations of perpendicular lines is slightly more difficult than writing parallel line equations. The reason is because you must find the slope of the perpendicular line before you can proceed with writing an equation.

#### Let's find the equation of the following line:

Find the equation of the line perpendicular to the line

Begin by finding the slopes of the perpendicular line. Using the perpendicular line definition,

Solve for

The slope of the line perpendicular to

You now have the slope and a point. Use point-slope form to write its equation.

You can rewrite this in slope-intercept form:

### Examples

#### Example 1

Earlier, you were asked to suppose that when a coordinate grid is transposed over the blueprint of a house under construction, two lines on the blueprint are perpendicular. If one of the lines had the equation

First, we need to determine what the slope of the other line is. Since the two lines are perpendicular, we can substitute the slope of the first line,

Solve for

The slope of the line perpendicular to

You now have the slope and a point. Use point-slope form to write its equation.

You can rewrite this in slope-intercept form:

The equation of the second line is

#### Example 2

Find the equation of the line perpendicular to the line

The line

Lines that make a

Vertical lines are in the form

Since the vertical line must go through (5, 4), the equation is

### Review

- Define perpendicular lines.
- What is true about the slopes of perpendicular lines?

In 3-7, determine the slope of a line perpendicular to the line given.

y=−5x+7 2x+8y=9 x=8 y=−4.75 y−2=15(x+3)

In 8–14, determine whether the lines are parallel, perpendicular, or neither.

- Line
a: passing through points (–1, 4) and (2, 6); Lineb: passing through points (2, –3) and (8, 1). - Line
a: passing through points (4, –3) and (–8, 0); Line \begin{align*}b:\end{align*} passing through points (–1, –1) and (–2, 6). - Line \begin{align*}a:\end{align*} passing through points (–3, 14) and (1, –2); Line \begin{align*}b:\end{align*} passing through points (0, –3) and (–2, 5).
- Line \begin{align*}a:\end{align*} passing through points (3, 3) and (–6, –3); Line \begin{align*}b:\end{align*} passing through points (2, –8) and (–6, 4).
- Line 1: \begin{align*}4y+x=8\end{align*}; Line 2: \begin{align*}12y+3x=1\end{align*}
- Line 1: \begin{align*}5y+3x+1\end{align*}; Line 2: \begin{align*}6y+10x=-3\end{align*}
- Line 1: \begin{align*}2y-3x+5=0\end{align*}; Line 2: \begin{align*}y+6x=-3\end{align*}

In 15-21, find the line perpendicular to it through the given point.

- \begin{align*}x+4y=12; (-3,-2)\end{align*}
- \begin{align*}y=\frac{1}{3}x+2; (-3,-1)\end{align*}
- \begin{align*}y=\frac{3}{5}x-4; (6,-2)\end{align*}
- \begin{align*}2x+y=5; (2,-2)\end{align*}
- \begin{align*}y=x-6; (-2,0)\end{align*}
- \begin{align*}5x-7=3y; (8,-2)\end{align*}
- \begin{align*}y=\frac{2}{3}x-1; (4,7)\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.8.