# 5.2: Equations of Parallel and Perpendicular Lines

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

## Learning Objectives

- Determine whether lines are parallel or perpendicular
- Write equations of perpendicular lines
- Write equations of parallel lines
- Investigate families of lines

## Introduction

In this section you will learn how **parallel lines** and **perpendicular lines** are related to each other on the coordinate plane. Let’s start by looking at a graph of two parallel lines.

We can clearly see that the two lines have different \begin{align*}y-\end{align*}

How about the slopes of the lines? The slope of line \begin{align*}A\end{align*} is \begin{align*}\frac{6-2}{0-(-2)}=\frac{4}{2}=2\end{align*}, and the slope of line \begin{align*}B\end{align*} is \begin{align*}\frac{0-(-4)}{2-0}=\frac{4}{2}=2\end{align*}. The slopes are the same.

Is that significant? Yes. By definition, parallel lines never meet. That means that when one of them slopes up by a certain amount, the other one has to slope up by the same amount so the lines will stay the same distance apart. If you look at the graph above, you can see that for any \begin{align*}x-\end{align*}value you pick, the \begin{align*}y-\end{align*}values of lines \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are the same vertical distance apart—which means that both lines go up by the same vertical distance every time they go across by the same horizontal distance. In order to stay parallel, their slopes must stay the same.

**All parallel lines** have the same slopes and different \begin{align*}y-\end{align*}intercepts.

Now let’s look at a graph of two perpendicular lines.

We can’t really say anything about the \begin{align*}y-\end{align*}intercepts. In this example, the \begin{align*}y-\end{align*}intercepts are different, but if we moved the lines four units to the right, they would both intercept the \begin{align*}y-\end{align*}axis at (0, -2). So perpendicular lines can have the same or different \begin{align*}y-\end{align*}intercepts.

What about the relationship between the slopes of the two lines?

To find the slope of line \begin{align*}A\end{align*}, we pick two points on the line and draw the blue (upper) right triangle. The legs of the triangle represent the rise and the run. We can see that the slope is \begin{align*}\frac{8}{4}\end{align*}, or 2.

To find the slope of line \begin{align*}B\end{align*}, we pick two points on the line and draw the red (lower) right triangle. Notice that the two triangles are identical, only rotated by \begin{align*}90^\circ\end{align*}. Where line \begin{align*}A\end{align*} goes 8 units up and 4 units right, line \begin{align*}B\end{align*} goes 8 units right and 4 units down. Its slope is \begin{align*}-\frac{4}{8}\end{align*}, or \begin{align*}-\frac{1}{2}\end{align*}.

This is always true for perpendicular lines; where one line goes \begin{align*}a\end{align*} units up and \begin{align*}b\end{align*} units right, the other line will go \begin{align*}a\end{align*} units right and \begin{align*}b\end{align*} units down, so the slope of one line will be \begin{align*}\frac{a}{b}\end{align*} and the slope of the other line will be \begin{align*}-\frac{b}{a}\end{align*}.

The slopes of **perpendicular lines** are always negative reciprocals of each other.

The Java applet at http://members.shaw.ca/ron.blond/perp.APPLET/index.html lets you drag around a pair of perpendicular lines to see how their slopes change. Click “Show Grid” to see the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}axes, and click “Show Constructors” to see the triangles that are being used to calculate the slopes of the lines (you can then drag the circle to make it bigger or smaller, and click on a triangle to see the slope calculations in detail.)

## Determine When Lines are Parallel or Perpendicular

You can find whether lines are parallel or perpendicular by comparing the slopes of the lines. If you are given points on the lines, you can find their slopes using the formula. If you are given the equations of the lines, re-write each equation in a form that makes it easy to read the slope, such as the slope-intercept form.

**Example 1**

*Determine whether the lines are parallel or perpendicular or neither.*

a) One line passes through the points (2, 11) and (-1, 2); another line passes through the points (0, -4) and (-2, -10).

b) One line passes through the points (-2, -7) and (1, 5); another line passes through the points (4, 1) and (-8, 4).

c) One lines passes through the points (3, 1) and (-2, -2); another line passes through the points (5, 5) and (4, -6).

**Solution**

Find the slope of each line and compare them.

a) \begin{align*}m_1=\frac{2-11}{-1-2}=\frac{-9}{-3}=3 \ \quad \ \text{and} \ \quad \ m_2=\frac{-10-(-4)}{-2-0}=\frac{-6}{-2}=3\end{align*}

The slopes are equal, so **the lines are parallel.**

b) \begin{align*}m_1=\frac{5-(-7)}{1-(-2)}=\frac{12}{3}=4 \quad \ \text{and} \ \quad m_2=\frac{4-1}{-8-4}=\frac{3}{-12}=-\frac{1}{4}\end{align*}

The slopes are negative reciprocals of each other, so **the lines are perpendicular.**

c) \begin{align*}m_1=\frac{-2-1}{-2-3}=\frac{-3}{-5}=\frac{3}{5} \ \quad \ \text{and} \ \quad \ m_2=\frac{-6-5}{4-5}=\frac{-13}{-1}=13\end{align*}

The slopes are not the same or negative reciprocals of each other, so **the lines are neither parallel nor perpendicular.**

**Example 2**

*Determine whether the lines are parallel or perpendicular or neither:*

a) \begin{align*}3x+4y=2\end{align*} and \begin{align*}8x-6y=5\end{align*}

b) \begin{align*}2x=y-10\end{align*} and \begin{align*}y=-2x+5\end{align*}

c) \begin{align*}7y+1=7x\end{align*} and \begin{align*}x+5=y\end{align*}

**Solution**

Write each equation in slope-intercept form:

a) line 1: \begin{align*}3x+4y=2 \Rightarrow 4y=-3x+2 \Rightarrow y=-\frac{3}{4}x+\frac{1}{2} \Rightarrow \ \text{slope} = - \frac{3}{4}\end{align*}

line 2: \begin{align*}8x-6y=5 \Rightarrow 8x-5=6y \Rightarrow y=\frac{8}{6} x-\frac{5}{6} \Rightarrow y=\frac{4}{3} x-\frac{5}{6} \Rightarrow \ \text{slope} = \frac{4}{3}\end{align*}

The slopes are negative reciprocals of each other, so **the lines are perpendicular.**

b) line 1: \begin{align*}2x=y-10 \Rightarrow y=2x+10 \Rightarrow \ \text{slope} = 2\end{align*}

line 2: \begin{align*}y=-2x+5 \Rightarrow \ \text{slope} = -2\end{align*}

The slopes are not the same or negative reciprocals of each other, so **the lines are neither parallel nor perpendicular.**

c) line 1: \begin{align*}7y+1=7x \Rightarrow 7y=7x-1 \Rightarrow y=x-\frac{1}{7} \Rightarrow \ \text{slope} = 1\end{align*}

line 2: \begin{align*}x+5=y \Rightarrow y=x+5 \Rightarrow \ \text{slope} = 1\end{align*}

The slopes are the same, so **the lines are parallel.**

## Write Equations of Parallel and Perpendicular Lines

We can use the properties of parallel and perpendicular lines to write an equation of a line parallel or perpendicular to a given line. You might be given a line and a point, and asked to find the line that goes through the given point and is parallel or perpendicular to the given line. Here’s how to do this:

- Find the slope of the given line from its equation. (You might need to re-write the equation in a form such as the slope-intercept form.)
- Find the slope of the parallel or perpendicular line—which is either the same as the slope you found in step 1 (if it’s parallel), or the negative reciprocal of the slope you found in step 1 (if it’s perpendicular).
- Use the slope you found in step 2, along with the point you were given, to write an equation of the new line in slope-intercept form or point-slope form.

**Example 3**

*Find an equation of the line perpendicular to the line \begin{align*}y=-3x+5\end{align*} that passes through the point (2, 6).*

**Solution**

The slope of the given line is -3, so the perpendicular line will have a slope of \begin{align*}\frac{1}{3}\end{align*}.

Now to find the equation of a line with slope \begin{align*}\frac{1}{3}\end{align*} that passes through (2, 6):

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=\frac{1}{3}x+b\end{align*}.

Plug in the point (2, 6) to find \begin{align*}b\end{align*}: \begin{align*}6=\frac{1}{3}(2)+b \Rightarrow b=6-\frac{2}{3} \Rightarrow b=\frac{20}{3}\end{align*}.

**The equation of the line is** \begin{align*}y=\frac{1}{3}x+\frac{20}{3}\end{align*}.

**Example 4**

*Find the equation of the line perpendicular to \begin{align*}x-5y=15\end{align*} that passes through the point (-2, 5).*

**Solution**

Re-write the equation in slope-intercept form: \begin{align*}x-5y=15 \Rightarrow -5y=-x+15 \Rightarrow y=\frac{1}{5}x-3\end{align*}.

The slope of the given line is \begin{align*}\frac{1}{5}\end{align*}, so we’re looking for a line with slope -5.

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=-5x+b\end{align*}.

Plug in the point (-2, 5): \begin{align*}5=-5(-2)+b \Rightarrow b=5-10 \Rightarrow b=-5\end{align*}

**The equation of the line is** \begin{align*}y=-5x-5\end{align*}.

**Example 5**

Find the equation of the line parallel to \begin{align*}6x-5y=12\end{align*} that passes through the point (-5, -3).

**Solution**

Rewrite the equation in slope-intercept form: \begin{align*}6x-5y=12 \Rightarrow 5y=6x-12 \Rightarrow y=\frac{6}{5}x-\frac{12}{5}\end{align*}.

The slope of the given line is \begin{align*}\frac{6}{5}\end{align*}, so we are looking for a line with slope \begin{align*}\frac{6}{5}\end{align*} that passes through the point (-5, -3).

Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.

Plug in the slope: \begin{align*}y=\frac{6}{5}x+b\end{align*}.

Plug in the point (-5, -3): \begin{align*}n-3=\frac{6}{5}(-5)+b \Rightarrow -3=-6+b \Rightarrow b=3\end{align*}

**The equation of the line is** \begin{align*}y=\frac{6}{5}x+3\end{align*}.

## Investigate Families of Lines

A **family of lines** is a set of lines that have something in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the \begin{align*}y–\end{align*}intercept is the same.

**Family 1:** Keep the slope unchanged and vary the \begin{align*}y-\end{align*}intercept.

The figure below shows the family of lines with equations of the form \begin{align*}y=-2x+b\end{align*}:

All the lines have a slope of –2, but the value of \begin{align*}b\end{align*} is different for each line.

Notice that in such a family all the lines are parallel. All the lines look the same, except that they are shifted up and down the \begin{align*}y-\end{align*}axis. As \begin{align*}b\end{align*} gets larger the line rises on the \begin{align*}y-\end{align*}axis, and as \begin{align*}b\end{align*} gets smaller the line goes lower on the \begin{align*}y-\end{align*}axis. This behavior is often called a **vertical shift.**

Family 2: Keep the \begin{align*}y-\end{align*}intercept unchanged and vary the slope.

The figure below shows the family of lines with equations of the form \begin{align*}y=mx+2\end{align*}:

All the lines have a \begin{align*}y-\end{align*}intercept of two, but the slope is different for each line. The steeper lines have higher values of \begin{align*}m\end{align*}.

**Example 6**

*Write the equation of the family of lines satisfying the given condition.*

a) parallel to the \begin{align*}x-\end{align*}axis

b) through the point (0, -1)

c) perpendicular to \begin{align*}2x+7y-9=0\end{align*}

d) parallel to \begin{align*}x+4y-12=0\end{align*}

**Solution**

a) All lines parallel to the \begin{align*}x-\end{align*}axis have a slope of zero; the \begin{align*}y-\end{align*}intercept can be anything. So the family of lines is \begin{align*}y=0x+b\end{align*} or just \begin{align*}y=b\end{align*}.

b) All lines passing through the point (0, -1) have the same \begin{align*}y-\end{align*}intercept, \begin{align*}b = -1\end{align*}. The family of lines is: \begin{align*}y=mx-1\end{align*}.

c) First we need to find the slope of the given line. Rewriting \begin{align*}2x+7y-9=0\end{align*} in slope-intercept form, we get \begin{align*}y=-\frac{2}{7}x+\frac{9}{7}\end{align*}. The slope of the line is \begin{align*}-\frac{2}{7}\end{align*}, so we’re looking for the family of lines with slope \begin{align*}\frac{7}{2}\end{align*}.

**The family of lines is** \begin{align*}y=\frac{7}{2}x+b\end{align*}.

d) Rewrite \begin{align*}x+4y-12=0\end{align*} in slope-intercept form: \begin{align*}y=-\frac{1}{4}x+3\end{align*}. The slope is \begin{align*}-\frac{1}{4}\end{align*}, so that’s also the slope of the family of lines we are looking for.

**The family of lines is** \begin{align*}y=-\frac{1}{4}x+b\end{align*}.

## Review Questions

For questions 1-10, determine whether the lines are parallel, perpendicular or neither.

- One line passes through the points (-1, 4) and (2, 6); another line passes through the points (2, -3) and (8, 1).
- One line passes through the points (4, -3) and (-8, 0); another line passes through the points (-1, -1) and (-2, 6).
- One line passes through the points (-3, 14) and (1, -2); another line passes through the points (0, -3) and (-2, 5).
- One line passes through the points (3, 3) and (-6, -3); another line passes through the points (2, -8) and (-6, 4).
- One line passes through the points (2, 8) and (6, 0); another line has the equation \begin{align*}x-2y=5\end{align*}.
- One line passes through the points (-5, 3) and (2, -1); another line has the equation \begin{align*}2x+3y=6\end{align*}.
- Both lines pass through the point (2, 8); one line also passes through (3, 5), and the other line has slope 3.
- 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*}
- Lines \begin{align*}A, B, C, D\end{align*}, and \begin{align*}E\end{align*} all pass through the point (3, 6). Line \begin{align*}A\end{align*} also passes through (7, 12); line \begin{align*}B\end{align*} passes through (8, 4); line \begin{align*}C\end{align*} passes through (-1, -3); line \begin{align*}D\end{align*} passes through (1, 1); and line \begin{align*}E\end{align*} passes through (6, 12).
- Are any of these lines perpendicular? If so, which ones? If not, why not?
- Are any of these lines parallel? If so, which ones? If not, why not?

- Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through point (3, -2).
- Find the equation of the line perpendicular to \begin{align*}y=-\frac{2}{5}x-3\end{align*} that passes through point (2, 8).
- Find the equation of the line parallel to \begin{align*}7y+2x-10=0\end{align*} that passes through the point (2, 2).
- Find the equation of the line perpendicular to \begin{align*}y+5=3(x-2)\end{align*} that passes through the point (6, 2).
- Line \begin{align*}S\end{align*} passes through the points (2, 3) and (4, 7). Line \begin{align*}T\end{align*} passes through the point (2, 5). If Lines \begin{align*}S\end{align*} and \begin{align*}T\end{align*} are parallel, name one more point on line \begin{align*}T\end{align*}. (
**Hint:**you don’t need to find the slope of either line.) - Lines \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} both pass through (-1, 5). Line \begin{align*}P\end{align*} also passes through (-3, -1). If \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} are perpendicular, name one more point on line \begin{align*}Q\end{align*}. (This time you will have to find the slopes of both lines.)
- Write the equation of the family of lines satisfying the given condition.
- All lines that pass through point (0, 4).
- All lines that are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
- All lines that are parallel to \begin{align*}y-3=4x+2\end{align*}.
- All lines that pass through the point (0, -1).

- Name two lines that pass through the point (3, -1) and are perpendicular to each other.
- Name two lines that are each perpendicular to \begin{align*}y=-4x-2\end{align*}. What is the relationship of those two lines to each other?
- Name two perpendicular lines that both pass through the point (3, -2). Then name a line parallel to one of them that passes through the point (-2, 5).