# 4.6: Equations of Parallel and Perpendicular Lines

**Advanced**Created by: CK-12

^{%}

**Practice**Comparing Equations of Parallel and Perpendicular Lines

Can you 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*}

### Watch This

Khan Academy Perpendicular Lines

### Guidance

**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. 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.

Two lines in the same plane that intersect or cross each other at right angles are **perpendicular lines**. Perpendicular lines have **slopes that are opposite reciprocals**. The following graph shows two lines with slopes that are opposite reciprocals. The slope of one line is \begin{align*}\frac{3}{4}\end{align*}

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

#### Example A

Given the slopes of two lines, tell whether the lines are parallel, perpendicular or neither.

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*}**not** opposite reciprocals. The lines are neither parallel nor perpendicular.

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*}

#### 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*}

**Solution:**

If the equation of the line you are looking for is parallel to the given line, then the two lines have the same slope. Begin by expressing \begin{align*}3x+2y-7=0\end{align*}

\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.

**Solution:** 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*}

\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*}

#### Concept Problem Revisited

Can you 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*}? Can you write the equation of the line in standard form?

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*}

### Guided Practice

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

1. (–2, 8), (3, 7) and (4, 3), (9, 2)

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

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

4. 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.

**Answers:**

1. (–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.

2. (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 opposite reciprocals. The lines are perpendicular.

3. (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.

4. 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 opposite 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*}

\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*}

### Explore More

For each pair of given equations, determine if the lines are parallel, perpendicular or neither.

- \begin{align*}y=2x-5\end{align*} and \begin{align*}y=2x+3\end{align*}
- \begin{align*}y=\frac{1}{3}x+5\end{align*} and \begin{align*}y=-3x-5\end{align*}
- \begin{align*}x=8\end{align*} and \begin{align*}x=-2\end{align*}
- \begin{align*}y=4x+7\end{align*} and \begin{align*}y=-4x-7\end{align*}
- \begin{align*}y=-x-3\end{align*} and \begin{align*}y=x+6\end{align*}
- \begin{align*}3y=9x+8\end{align*} and \begin{align*}y=3x-4\end{align*}

Determine the equation of the line satisfying the following conditions:

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

- the equation of the line through \begin{align*}D\end{align*} and parallel to \begin{align*}EF\end{align*}.
- the equation of the line containing the altitude from \begin{align*}D\end{align*} to \begin{align*}EF\end{align*} (the line perpendicular to \begin{align*}EF\end{align*} that contains \begin{align*}D\end{align*}).

Parallel

Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.Perpendicular

Perpendicular lines are lines that intersect at a angle. The product of the slopes of two perpendicular lines is -1.standard form

The standard form of a quadratic function is .### Image Attributions

## Description

## Learning Objectives

Here you will learn about parallel and perpendicular lines and how to determine whether or not two lines are parallel or perpendicular using slope.

## Related Materials

## Difficulty Level:

Advanced## Subjects:

## Date Created:

Apr 30, 2013## Last Modified:

Sep 21, 2015## Vocabulary

**You can only attach files to Modality which belong to you**

If you would like to associate files with this Modality, please make a copy first.