# Parallel and Perpendicualr Lines

 Site: Saylor Academy Course: MA001: College Algebra Book: Parallel and Perpendicualr Lines
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## Description

Now, you will build on your knowledge of writing equations of linear functions, drawing their graphs, and performing transformations by exploring parallel and perpendicular lines. You will learn how to identify whether lines are parallel or perpendicular given their equations and graphs. You will also learn how to define the equation of a line parallel or perpendicular to another line given an equation and a point.

## Determining Whether Lines are Parallel or Perpendicular

The two lines in Figure 28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the other, they would become coincident. Figure 28 Parallel lines

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

$\left.\left.\begin{array}{l} f(x)=-2 x+6 \\ f(x)=-2 x-4 \end{array}\right\} \quad \text { parallel } \qquad \begin{array}{l} f(x)=3 x+2 \\ f(x)=2 x+2 \end{array}\right\} \text { not parallel }$

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular. Figure 29 Perpendicular lines

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if $m_{1}$ and $m_{2}$ are negative reciprocals of one another, they can be multiplied together to yield $-1$.

$m_{1} m_{2}=-1$

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is $\dfrac{1}{8}$, and the reciprocal of $\dfrac{1}{8}$ is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

$f(x)=\dfrac{1}{4} x+2 \quad$ negative reciprocal of $\dfrac{1}{4}$ is $-4$ $f(x)=-4 x+3 \quad$ negative reciprocal of $-4$ is $\dfrac{1}{4}$

The product of the slopes is $-1$.

$-4\left(\dfrac{1}{4}\right)=-1$

#### PARALLEL AND PERPENDICULAR LINES

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

$f(x)=m_{1} x+b_{1}$ and $g(x)=m_{2} x+b_{2} \quad \text{are parallel if and only if} \, m_{1}=m_{2}$

If and only if $b_{1}=b_{2}$ and $m_{1}=m_{2}$, we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect to form a right angle.

$f(x)=m_{1} x+b_{1}$ and $g(x)=m_{2} x+b_{2} \quad \text{are perpendicular if and only if}$

$m_{1} m_{2}=-1, \text { so } \quad m_{2}=-\dfrac{1}{m_{1}}$

#### EXAMPLE 18

##### Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

$\begin{array}{ll} f(x)=2 x+3 & \qquad h(x)=-2 x+2 \\ g(x)=\dfrac{1}{2} x-4 & \qquad j(x)=2 x-6 \end{array}$

##### Solution

Parallel lines have the same slope. Because the functions $f(x)=2 x+3$ and $j(x)=2 x-6$ each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes.

Because $-2$ and $\dfrac{1}{2}$ are negative reciprocals, the functions $g(x)=\dfrac{1}{2} x-4$ and $h(x)=-2 x+2$ represent perpendicular lines.

##### Analysis

A graph of the lines is shown in Figure 30. Figure 30

The graph shows that the lines $f(x)=2 x+3$ and $j(x)=2 x-6$ are parallel, and the lines $g(x)=\dfrac{1}{2} x-4$ and $h(x)=-2 x+2$ are perpendicular.

Source: Rice University, https://openstax.org/books/college-algebra/pages/4-1-linear-functions This work is licensed under a Creative Commons Attribution 4.0 License.

## Writing the Equation of a Line Parallel or Perpendicular to a Given Line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

#### Writing Equations of Parallel Lines

Suppose for example, we are given the equation shown.

$f(x)=3 x+1$

We know that the slope of the line formed by the function is $3$. We also know that the $y$-intercept is $(0,1)$. Any other line with a slope of $3$ will be parallel to $f(x)$. So the lines formed by all of the following functions will be parallel to $f(x)$.

\begin{aligned} &g(x)=3 x+6 \\ &h(x)=3 x+1 \\ &p(x)=3 x+\dfrac{2}{3} \end{aligned}

Suppose then we want to write the equation of a line that is parallel to $f$ and passes through the point $(1,7)$. This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is $3$. We need to determine which value of $b$ will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

\begin{aligned} y-y_{1} &=m\left(x-x_{1}\right) \\ y-7 &=3(x-1) \\ y-7 &=3 x-3 \\ y &=3 x+4 \end{aligned}

So $g(x)=3 x+4$ is parallel to $f(x)=3 x+1$ and passes through the point $(1,7)$.

#### HOW TO

Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

1. Find the slope of the function.
2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.
3. Simplify.

#### EXAMPLE 19

##### Finding a Line Parallel to a Given Line

Find a line parallel to the graph of $f(x)=3 x+6$ that passes through the point $(3,0)$.

##### Solution

The slope of the given line is $3$. If we choose the slope-intercept form, we can substitute $m=3, x=3$, and $f(x)=0$ into the slope-intercept form to find the $y$-intercept.

\begin{aligned} g(x) &=3 x+b \\ 0 &=3(3)+b \\ b &=-9 \end{aligned}

The line parallel to $f(x)$ that passes through $(3,0)$ is $g(x)=3 x-9$.

##### Analysis

We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect. Figure 31

#### Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown.

$f(x)=2 x+4$

The slope of the line is $2$, and its negative reciprocal is $-\dfrac{1}{2}$. Any function with a slope of $-\dfrac{1}{2}$ will be perpendicular to $f(x)$. So the lines formed by all of the following functions will be perpendicular to $f(x)$.

\begin{aligned} &g(x)=-\dfrac{1}{2} x+4 \\ &h(x)=-\dfrac{1}{2} x+2 \\ &p(x)=-\dfrac{1}{2} x-\dfrac{1}{2} \end{aligned}

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to $f(x)$ and passes through the point $(4,0)$. We already know that the slope is $-\dfrac{1}{2}$. Now we can use the point to find the $y$-intercept by substituting the given values into the slope-intercept form of a line and solving for $b$.

\begin{aligned} g(x) &=m x+b \\ 0 &=-\dfrac{1}{2}(4)+b \\ 0 &=-2+b \\ 2 &=b \\ b &=2 \end{aligned}

The equation for the function with a slope of $-\dfrac{1}{2}$ and a $y$-intercept of 2 is

$g(x)=-\dfrac{1}{2} x+2$

So $g(x)=-\dfrac{1}{2} x+2$ is perpendicular to $f(x)=2 x+4$ and passes through the point $(4,0)$. Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

#### Q&A

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not $-1$. Doesn't this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is $-1$. However, a vertical line is not a function so the definition is not contradicted.

#### HOW TO

##### Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

1. Find the slope of the function.
2. Determine the negative reciprocal of the slope.
3. Substitute the new slope and the values for $x$ and $y$ from the coordinate pair provided into $g(x)=m x+b.$
4. Solve for $b$.
5. Write the equation of the line.

#### EXAMPLE 20

##### Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to $f(x)=3 x+3$ that passes through the point $(3,0)$.

##### Solution

The original line has slope $m=3$, so the slope of the perpendicular line will be its negative reciprocal, or $-\dfrac{1}{3}$. Using this slope and the given point, we can find the equation of the line.

\begin{aligned} g(x) &=-\dfrac{1}{3} x+b \\ 0 &=-\dfrac{1}{3}(3)+b \\ 1 &=b \\ b &=1 \end{aligned}

The line perpendicular to $f(x)$ that passes through $(3,0)$ is $g(x)=-\dfrac{1}{3} x+1$.

##### Analysis

A graph of the two lines is shown in Figure 32. Figure 32

Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely.

#### TRY IT #9

Given the function $h(x)=2 x-4$, write an equation for the line passing through $(0,0)$ that is

(a) parallel to $h(x)$

(b) perpendicular to $h(x)$

#### HOW TO

##### Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

1. Determine the slope of the line passing through the points.
2. Find the negative reciprocal of the slope.
3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.
4. Simplify.

#### EXAMPLE 21

Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

##### Solution

From the two points of the given line, we can calculate the slope of that line.

\begin{aligned} m_{1} &=\dfrac{5-6}{4-(-2)} \\ &=\dfrac{-1}{6} \\ &=-\dfrac{1}{6} \end{aligned}

Find the negative reciprocal of the slope.

\begin{aligned} m_{2} &=\dfrac{-1}{-\dfrac{1}{6}} \\ &=-1\left(-\dfrac{6}{1}\right) \\ &=6 \end{aligned}

We can then solve for the $y$-intercept of the line passing through the point $(4,5)$.

\begin{aligned} g(x) &=6 x+b \\ 5 &=6(4)+b \\ 5 &=24+b \\ -19 &=b \\ b &=-19 \end{aligned}

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point $(4,5)$ is

$y=6 x-19$

#### TRY IT #10

A line passes through the points, $(-2,-15)$ and $(2,-3)$. Find the equation of a perpendicular line that passes through the point, $(6,4)$.

## Ex: Determine the Equation of a Line Parallel to A Given Line Passing Thru a Given Point (09x-34)

Source: Mathispower4u, https://youtu.be/7uucAn3ZB6U This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License.

## Ex 1: Find the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Source: Mathispower4u, https://youtu.be/H98iBqt1fFQ This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License.