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

# 4.6: Parallel and Perpendicular Lines

Difficulty Level: At Grade Created by: CK-12

## Parallel and Perpendicular Lines

Introduction

In this lesson you will learn that parallel lines are lines in the same plane that never intersect. Parallel lines maintain the same slope, or no slope (vertical lines), and the same distance from each other. You will also learn that two lines in the same plane that intersect or cross each other at right angles are perpendicular lines. Perpendicular lines have slopes that are negative reciprocals.

The fact that parallel lines have the same slope and perpendicular lines have slopes that are negative reciprocals will be used to write the equation of a line.

Objectives

The lesson objectives for Parallel and Perpendicular Lines are:

• Understanding the slope of parallel lines.
• Understanding the slope of perpendicular lines.
• Using the slopes of parallel and perpendicular lines to write the equations of related lines.

Introduction

The following graph shows two lines with the same slope. The slope of each line is 2. Notice that the lines are the same distance apart for the entire length of the lines. The lines will never intersect. The following lines are parallel.

The following graph shows two lines with slopes that are negative reciprocals. The slope of one line is \begin{align*}\frac{3}{4}\end{align*} and the slope of the other line is \begin{align*}-\frac{4}{3}\end{align*}. The product of the slopes is negative one. \begin{align*}\left(\frac{3}{4}\right)\left(\frac{-4}{3}\right)=\frac{-12}{12}=-1\end{align*}. Notice that the lines intersect at right angles. The lines are perpendicular lines.

Watch This

{fileref:h://authottprs.ck12.org/wiki/index.php/Concept_95:_Equations_ of_Parallel_ and_Perpendicular_Lines}

Guidance

The relationship between the slopes of parallel lines (slopes are the same) and the relationship between the slopes of perpendicular lines (slopes are negative reciprocals) can be used to determine the equations of lines.

Example A

Given the slopes of two lines, tell whether the lines are parallel, perpendicular or neither parallel nor perpendicular.

i) \begin{align*}m_1=4, m_2=\frac{1}{4}\end{align*}

ii) \begin{align*}m_1=-3, m_2=\frac{1}{3}\end{align*}

iii) \begin{align*}m_1=\frac{3}{12}, m_2=\frac{1}{4}\end{align*}

iv) \begin{align*}m_1=-1, m_2=1\end{align*}

v) \begin{align*}m_1=-\frac{1}{3}, m_2=\frac{1}{3}\end{align*}

Solutions

i) \begin{align*}m_1=4, m_2=\frac{1}{4}\end{align*} The slopes are reciprocals but not negative reciprocals. The lines are neither parallel nor perpendicular.

ii) \begin{align*}m_1=-3, m_2=\frac{1}{3}\end{align*} The slopes are reciprocals and are also negative reciprocals. The lines are perpendicular.

iii) \begin{align*}m_1=\frac{3}{12}, m_2=\frac{1}{4}\end{align*} The slopes are the same. The fractions are equivalent. The lines are parallel.

iv) \begin{align*}m_1=-1, m_2=1\end{align*} The slopes are reciprocals and are also negative reciprocals. The lines are perpendicular.

v) \begin{align*}m_1=-\frac{1}{3}, m_2=\frac{1}{3}\end{align*} The slopes are not the same. The lines are neither parallel nor perpendicular.

Example B

Determine the equation of the line passing through the point (–4, 6) and parallel to the graph of \begin{align*}3x+2y-7=0\end{align*}. Write the equation in standard form.

Begin by expressing \begin{align*}3x+2y-7=0\end{align*} in slope-intercept form.

\begin{align*}3x+2y-7&=0\\ 3x{\color{red}-3x}+2y-7&=0{\color{red}-3x}\\ 2y-7&=-3x\\ 2y-7{\color{red}+7}&=-3x{\color{red}+7}\\ 2y&=-3x+7\\ \frac{\cancel{2}y}{\cancel{2}}&=-\frac{3x}{2}+\frac{7}{2}\\ y&=-\frac{3}{2}x+\frac{7}{2}\\ & \qquad {\color{red}\updownarrow}\\ y&= \ mx+b && \text{The slope of the line is} \ -\frac{3}{2}. \ \text{The line passes through the point} \ (-4, 6).\\ y-y_1&=m(x-x_1) && \text{Substitute the values into this equation.}\\ y-{\color{red}6}&={\color{red}\frac{-3}{2}}(x-{\color{red}-4})\\ y-6&=\frac{-3}{2}(x{\color{red}+}4)\\ y-6&=\frac{-3x}{2}-\frac{12}{2}\\ 2(y)-2(6)&=\cancel{2}\left(\frac{-3x}{\cancel{2}}\right)-\cancel{2}\left(\frac{12}{\cancel{2}}\right)\\ 2y-12&=-3x-12\\ 2y-12{\color{red}+12}&=-3x-12{\color{red}+12}\\ 2y&=-3x\\ {\color{red}3x}+2y&=-3x{\color{red}+3x}\\ 3x+2y&=0\end{align*}

The equation of the line is \begin{align*}\boxed{3x+2y=0}\end{align*}

Example C

Determine the equation of the line that passes through the point (6, –2) and is perpendicular to the graph of \begin{align*}3x=2y-4\end{align*}. Write the equation in standard form.

Begin by writing the equation \begin{align*}3x=2y-4\end{align*} in slope-intercept form.

\begin{align*}3x&=2y-4\\ 2y-4&=3x\\ 2y-4{\color{red}+4}&=3x{\color{red}+4}\\ 2y&=3x+4\\ \frac{\cancel{2}y}{\cancel{2}}&=\frac{3x}{2}+\frac{4}{2}\\ y&=\frac{3}{2}x+2\\ & \quad \ {\color{red}\updownarrow}\\ y&=mx+b\end{align*}

The slope of the given line is \begin{align*}\frac{3}{2}\end{align*}. The slope of the perpendicular line is \begin{align*}\boxed{-\frac{2}{3}}\end{align*}. The line passes through the point (6, –2).

\begin{align*}y-y_1&=m(x-x_1)\\ y-{\color{red}-2}&={\color{red}-\frac{2}{3}}(x-{\color{red}6})\\ y{\color{red}+}2&=-\frac{2}{3}(x-6)\\ y+2&=-{\color{red}\frac{2x}{3}+\frac{12}{3}}\\ 3(y)+3(2)&=3\left(-\frac{2x}{3}\right)+3\left(\frac{12}{3}\right)\\ 3(y)+3(2)&=\cancel{3}\left(-\frac{2x}{\cancel{3}}\right)+\cancel{3}\left(\frac{12}{\cancel{3}}\right)\\ 3y+6&=-2x+12\\ 3y+6{\color{red}-12}&=-2x+12{\color{red}-12}\\ 3y-6&=-2x\\ {\color{red}2x}+3y-6&=-2x{\color{red}+2x}\\ 2x+3y-6&=0\end{align*}

The equation of the line is \begin{align*}\boxed{2x+3y-6=0}\end{align*}.

Vocabulary

Parallel Lines
Parallel lines are lines in the same plane that have the same slope or no slope. The lines never intersect and always maintain the same distance apart.
Perpendicular Lines
Perpendicular lines are lines in the same plane that intersect each other at right angles. The slopes of perpendicular lines are negative reciprocals. The product of the slopes of two perpendicular lines is –1.

Guided Practice

1. Determine whether the lines that pass through the two pairs of points are parallel, perpendicular or neither parallel nor perpendicular.

i) (–2, 8), (3, 7) and (4, 3), (9, 2)

ii) (2, 5), (8, 7) and (–3, 1), (–2, –2)

iii) (4, 6), (–3, –1) and (6, –3), (4, 5)

2. Write the equation for the line that passes through the point (–3, 6) and is perpendicular to the graph of \begin{align*}3x=5y+6\end{align*}. Write the equation of the line in slope-intercept form.

3. Write the equation for the line that passes through the point (–2, –3) and is parallel to the graph of \begin{align*}y+2x=8\end{align*}. Write the equation of the line in standard form.

1. i) (–2, 8), (3, 7) and (4, 3), (9, 2)

\begin{align*}& m=\frac{y_2-y_1}{x_2-x_1} && m=\frac{y_2-y_1}{x_2-x_1} && \text{Use the formula to calculate the slopes}\\ & m=\frac{7-8}{3--2} && m=\frac{2-3}{9-4} && \text{Calculate the slopes for each pair of points}\\ & m=\frac{7-8}{3+2} && \boxed{m=\frac{-1}{5}}\\ & \boxed{m=\frac{-1}{5}}\end{align*}

The slopes of the lines are the same. The lines are parallel.

ii) (2, 5), (8, 7) and (–3, 1), (–2, –2)

\begin{align*}& m=\frac{y_2-y_1}{x_2-x_1} && m=\frac{y_2-y_1}{x_2-x_1} && \text{Use the formula to calculate the slopes}\\ & m=\frac{7-5}{8-2} && m=\frac{-2-1}{-2--3} && \text{Calculate the slopes for each pair of points}\\ & m=\frac{2}{6} && m=\frac{-2-1}{-2+3}\\ & \boxed{m=\frac{1}{3}} && \boxed{m=\frac{-3}{1}}\end{align*}

The slopes of the lines are negative reciprocals. The lines are perpendicular.

iii) (4, 6), (–3, –1) and (6, –3), (4, 5)

\begin{align*}& m=\frac{y_2-y_1}{x_2-x_1} && m=\frac{y_2-y_1}{x_2-x_1} && \text{Use the formula to calculate the slopes}\\ & m=\frac{-1-6}{-3-4} && m=\frac{5--3}{4-6} && \text{Calculate the slopes for each pair of points}\\ & m=\frac{-7}{-7} && m=\frac{5+3}{4-6}\\ & \boxed{m=1} && m=\frac{8}{-2}\\ & && \boxed{m=-4}\end{align*}

The lines are neither parallel nor perpendicular.

2. Begin by writing the given equation in slope-intercept form. This will give the slope of this line. The slope of the perpendicular line is the negative reciprocal.

\begin{align*}3x&=5y+6\\ 5y+6&=3x\\ 5y+6{\color{red}-6}&=3x{\color{red}-6}\\ 5y&=3x-6\\ \frac{5y}{{\color{red}5}}&=\frac{3x}{{\color{red}5}}-\frac{6}{{\color{red}5}}\\ \frac{\cancel{5}y}{\cancel{5}}&=\frac{3x}{5}-\frac{6}{5}\\ y&=\frac{3}{5}x-\frac{6}{5}\end{align*}

The slope of the given line is \begin{align*}\frac{3}{5}\end{align*}. The slope of the perpendicular line is \begin{align*}\boxed{-\frac{5}{3}}\end{align*}.

The equation of the perpendicular line that passes through the point (–3, 6) is:

\begin{align*}y&=mx+b\\ {\color{red}6}&={\color{red}-\frac{5}{3}}({\color{red}-3})+b\\ 6&=-\frac{5}{\cancel{3}}\left(\cancel{-} \overset{{\color{red}-1}}{\cancel{3}}\right)+b\\ 6&=5+b\\ 6{\color{red}-5}&=5{\color{red}-5}+b\\ 1&=b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, 1) and the slope of the line is \begin{align*}\boxed{-\frac{5}{3}}\end{align*}.

The equation of the line is \begin{align*}\boxed{y=-\frac{5}{3}x+1}\end{align*}

3. Begin by writing the given equation in slope-intercept form. This will give the slope of this line. The slope of the parallel line is the same as the slope of the given line.

\begin{align*}y+2x&=8\\ y+2x{\color{red}-2x}&={\color{red}-2x}+8\\ y&=-2x+8\end{align*}

The slope of the given line is –2. The slope of the parallel line is also \begin{align*}\boxed{-2}\end{align*}.

\begin{align*}y-y_1&=m(x-x_1)\\ y-{\color{red}-3}&={\color{red}-2}(x-{\color{red}-2})\\ y+3&=-2(x+2)\\ y+3&=-2x{\color{red}-4}\\ y+3&=-2x-4\\ {\color{red}2x}+y+3&=-2x{\color{red}+2x}-4\\ 2x+y+3&=-4\\ 2x+y+3{\color{red}+4}&=-4{\color{red}+4}\\ 2x+y+7&=0\end{align*}

The equation of the line is \begin{align*}\boxed{2x+y+7=0}\end{align*}

Summary

In this lesson you have learned that parallel lines are lines in the same plane that have the same slope or no slope. The lines never intersect and always maintain the same distance apart. You also learned that perpendicular lines are lines in the same plane that intersect each other at right angles. The slopes of perpendicular lines are negative reciprocals. The product of the slopes of two perpendicular lines is –1.

You have learned to use the relationship between the slopes of parallel lines and the slopes of perpendicular lines to write the equations of other lines.

Problem Set

For each pair of given equations, determine if the lines are parallel, perpendicular or neither parallel nor perpendicular. Complete the table:

Given Equations Parallel Perpendicular Neither Parallel Nor Perpendicular
(1) \begin{align*}y=2x-5\end{align*} \begin{align*}y=2x+3\end{align*}
(2) \begin{align*}y=\frac{1}{3}x+5\end{align*} \begin{align*}y=-3x-5\end{align*}
(3) \begin{align*}x=8\end{align*} \begin{align*}x=-2\end{align*}
(4) \begin{align*}y=4x+7\end{align*} \begin{align*}y=-4x-7\end{align*}
(5) \begin{align*}y=-x-3\end{align*} \begin{align*}y=x+6\end{align*}
(6) \begin{align*}3y=9x+8\end{align*} \begin{align*}y=3x-4\end{align*}

Determine the equation of the line, in slope-intercept form, satisfying the following conditions:

1. through the point (5, –6) and parallel to the line \begin{align*}y=5x+4\end{align*}
2. through the point (–1, 7) and perpendicular to the line \begin{align*}y=-4x+5\end{align*}
3. containing the point (–1, –5) and parallel to \begin{align*}3x+2y=9\end{align*}
4. containing the point (0, –6) and perpendicular to \begin{align*}6x-3y+8=0\end{align*}

If \begin{align*}D(4, -1), E(-4, 5)\end{align*} and \begin{align*}F(3, 6)\end{align*} are the vertices of \begin{align*}\Delta DEF\end{align*} determine

1. the equation of the line, in standard form, through \begin{align*}D\end{align*} and parallel to \begin{align*}EF\end{align*}.
2. the equation of the line, in standard form, containing the altitude from \begin{align*}D\end{align*} to \begin{align*}EF\end{align*}.

For each pair of given equations...

Given Equations Parallel Perpendicular Neither Parallel Nor Perpendicular
(1) \begin{align*}y=2x-5\end{align*} \begin{align*}y=2x+3\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}
(2) \begin{align*}y=\frac{1}{3}x+5\end{align*} \begin{align*}y=-3x-5\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}
(3) \begin{align*}x=8\end{align*} \begin{align*}x=-2\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}
(4) \begin{align*}y=4x+7\end{align*} \begin{align*}y=-4x-7\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}
(5) \begin{align*}y=-x-3\end{align*} \begin{align*}y=x+6\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}
(6) \begin{align*}3y=9x+8\end{align*} \begin{align*}y=3x-4\end{align*} \begin{align*}{\color{blue}\sqrt{}}\end{align*}

Determine the equation of...

1. \begin{align*}y=5x+4\end{align*} The slope of the line is 5. The slope of the parallel line is also 5. The equation of the line through the point (5, –6) having a slope of 5 is:

\begin{align*}y&=mx+b\\ -6&=5(5)+b\\ -6&=25+b\\ -6-25&=25-25+b\\ -31&=b\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, –31).

The equation of the line is \begin{align*}\boxed{y=5x-31}\end{align*}

\begin{align*}6x-3y+8&=0\\ 6x-6x-3y+8&=0-6x\\ -3y+8&=-6x\\ -3y+8-8&=-6x-8\\ -3y&=-6x-8\\ \frac{-3y}{-3}&=\frac{-6x}{-3}-\frac{8}{-3}\\ y&=2x+\frac{8}{3}\end{align*}

The slope of the given line is 2. The slope of the perpendicular line is \begin{align*}-\frac{1}{2}\end{align*}.

The equation of the line that has a slope of \begin{align*}-\frac{1}{2}\end{align*} and a \begin{align*}y-\end{align*}intercept of (0, –6) is \begin{align*}\boxed{y=-\frac{1}{2}x-6}\end{align*}.

If \begin{align*}D(4, -1), E(-4, 5)\end{align*} and \begin{align*}F(3, 6)\end{align*} are the vertices...

1. The following graph shows the triangle: The slope of \begin{align*}EF\end{align*} is:

\begin{align*}m&=\frac{y_2-y_1}{x_2-x_1}\\ m&=\frac{6-5}{3--4}\\ m&=\frac{6-5}{3+4}\\ m&=\frac{1}{7}\end{align*}

The equation of the line through \begin{align*}D (4, -1)\end{align*} is:

\begin{align*}& \ y-y_1=m(x-x_1)\\ & y--1=\frac{1}{7}(x-4)\\ & y+1=\frac{1}{7}(x-4)\\ & y+1=\frac{1x}{7}-\frac{4}{7}\\ & 7(y)+7(1)=7\left(\frac{1x}{7}\right)-7\left(\frac{4}{7}\right)\\ & 7(y)+7(1)=\cancel{7}\left(\frac{1x}{\cancel{7}}\right)-\cancel{7}\left(\frac{4}{\cancel{7}}\right)\\ & 7y+7=1x-4\\ & -1x+7y+7=1x-1x-4\\ & -1x+7y+7=-4\\ & -1x+7y+7+4=-4+4\\ & -1x+7y+11=0\\ & \boxed{x-7y-11=0}\end{align*}

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

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: