<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 5.4: Equations of Parallel and Perpendicular Lines

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

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

## Vocabulary

Terms introduced in this lesson:

parallel lines
perpendicular lines
family of lines
vertical shift

## Teaching Strategies and Tips

Use the introduction to make observations such as:

• Parallel lines have different \begin{align*}y-\end{align*}intercepts.
• Parallel lines have the same slope. Allow students to come to this conclusion using a rise-over-run argument: If one line runs (or rises) more than another line, then the lines cannot be parallel; they will eventually meet.
• Perpendicular lines have opposite reciprocal slopes. Encourage students to draw two perpendicular lines. Since one line will be increasing and the other decreasing, this shows that the slopes have opposite signs. Have students also construct the rise-over-run triangles for each line. This will demonstrate that the rise of one line is the run of the other and vice-versa; the slopes are also reciprocal.

Have students recognize in Example 5 that \begin{align*}y=-2\end{align*} is the equation of a horizontal line. The problem can then be recast as finding the equation of a vertical line through \begin{align*}(4,-2)\end{align*}.

What is the equation of a line parallel to the \begin{align*}x-\end{align*}axis and passing through \begin{align*}(2, -1)\end{align*}?

Solution: A line parallel to the \begin{align*}x-\end{align*}axis is horizontal and therefore has equation, \begin{align*}y = k\end{align*}. Since it passes through \begin{align*}(2, -1), y = -1\end{align*}.

Find the equation of a line perpendicular to \begin{align*}x = 3\end{align*}.

Solution: Since \begin{align*}x = 3\end{align*} is a vertical line, a perpendicular line will be horizontal. Therefore any equation of the form \begin{align*}y = k\end{align*} is perpendicular to \begin{align*}x = 3\end{align*}.

Find the equation of a line perpendicular to \begin{align*}x = 3\end{align*} that passes through the point \begin{align*}(14, 15)\end{align*}.

Solution: \begin{align*}y = 15\end{align*}.

Use Example 9 and a graphing utility to investigate families of lines.

• Fixing the slope \begin{align*}m\end{align*} and varying the \begin{align*}y-\end{align*}intercept b in \begin{align*}y = mx + b\end{align*} results in a set of parallel lines; one through each point on the \begin{align*}y-\end{align*}axis.
• Fixing the \begin{align*}y-\end{align*}intercept and varying the slope results in all possible lines through that \begin{align*}y-\end{align*}intercept; one for every real number angle.

## Error Troubleshooting

NONE

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: