# Parallel and Perpendicular Lines

### Learning Outcomes

- Determine whether two lines are parallel or perpendicular.
- Find the equations of parallel and perpendicular lines.
- Write the equations of lines that are parallel or perpendicular to a given line.

*y-*intercepts. Lines that are

**parallel**to each other will never intersect. For example, the figure below shows the graphs of various lines with the same slope, [latex]m=2[/latex].

*y-*intercepts. Lines that are

**perpendicular**intersect to form a [latex]{90}^{\circ }[/latex] angle. The slope of one line is the

**negative**

**reciprocal**of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\cdot {m}_{2}=-1[/latex]. For example, the figure below shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\frac{1}{3}[/latex].

[latex]\begin{array}{l}\text{ }{m}_{1}\cdot {m}_{2}=-1\hfill \\ \text{ }3\cdot \left(-\frac{1}{3}\right)=-1\hfill \end{array}[/latex]

### Example: Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[/latex] and [latex]3x - 4y=8[/latex].Answer: The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation:

[latex]\begin{array}{l}3y=-4x+3\hfill \\ y=-\frac{4}{3}x+1\hfill \end{array}[/latex]

Second equation:[latex]\begin{array}{l}3x - 4y=8\hfill \\ -4y=-3x+8\hfill \\ y=\frac{3}{4}x - 2\hfill \end{array}[/latex]

See the graph of both lines in the graph below. From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.[latex]\begin{array}{l}{m}_{1}=-\frac{4}{3}\hfill \\ {m}_{2}=\frac{3}{4}\hfill \\ {m}_{1}\cdot {m}_{2}=\left(-\frac{4}{3}\right)\left(\frac{3}{4}\right)=-1\hfill \end{array}[/latex]

The slopes are negative reciprocals of each other confirming that the lines are perpendicular.### Try It

Graph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[/latex] and [latex]2y=x+4[/latex].Answer: Parallel lines. Write the equations in slope-intercept form.

Check your work with an online graphing calculator.## 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 following line:[latex]y=2x+4[/latex]

The slope of the line is 2 and its negative reciprocal is [latex]-\frac{1}{2}[/latex]. Any function with a slope of [latex]-\frac{1}{2}[/latex] will be perpendicular to [latex]y=2x+4[/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[/latex].[latex]\begin{array}{lll}y=-\frac{1}{2}x+4\hfill & \\ y=-\frac{1}{2}x+2\hfill & \\ y=-\frac{1}{2}x-\frac{1}{2}\hfill \end{array}[/latex]

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 [latex]y=2x+4[/latex] and passes through the point (4, 0). We already know that the slope is [latex]-\frac{1}{2}[/latex]. Now we can use the point to find the*y*-intercept by substituting the given values into slope-intercept form and solving for

*b*.

[latex]\begin{array}{lllll}y=mx+b\hfill & \\ 0=-\frac{1}{2}\left(4\right)+b\hfill & \\ 0=-2+b\hfill \\ 2=b\hfill & \\ b=2\hfill \end{array}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a*y-*intercept of 2 is [latex]y=-\frac{1}{2}x+2[/latex]. So [latex]y=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]y=2x+4[/latex] 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. As you will learn later, a vertical line is not a function so the definition is not contradicted.*

### How To: Given the equation of a LINE, write the equation of a line Perpendicular to the given line that passes through A given point

- Find the slope of the given line.
- Determine the negative reciprocal of the slope.
- Substitute the slope and point into either point-slope form or slope-intercept form.
- Simplify.

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

Find the equation of a line perpendicular to [latex]y=3x+3[/latex] that passes through the point (3, 0).Answer:
The original line has slope *m *= 3, so the slope of the perpendicular line will be its negative reciprocal or [latex]-\frac{1}{3}[/latex]. Using this slope and the given point, we can find the equation for the line.

[latex]\begin{array}y=-\frac{1}{3}x+b\hfill & \\ \text{}0=-\frac{1}{3}\left(3\right)+b\hfill & \\ \text{}1=b\hfill \\ \text{ }b=1\hfill \end{array}[/latex]

The line perpendicular to [latex]y=3x+3[/latex] that passes through (3, 0) is [latex]y=-\frac{1}{3}x+1[/latex].#### Analysis of the Solution

A graph of the two lines is shown below.### Try It

Given the line [latex]y=2x - 4[/latex], write an equation for the line passing through (0, 0) that is- parallel to
*y* - perpendicular to
*y*

Answer:

- [latex]y=2x[/latex] is parallel
- [latex]y=-\frac{1}{2}x[/latex] is perpendicular

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

Find the equation of the line perpendicular to [latex]5x - 3y+4=0[/latex] which goes through the point [latex]\left(-4,1\right)[/latex].Answer: The first step is to write the equation in slope-intercept form.

[latex]\begin{array}{l}5x - 3y+4=0\hfill \\ -3y=-5x - 4\hfill \\ y=\frac{5}{3}x+\frac{4}{3}\hfill \end{array}[/latex]

We see that the slope is [latex]m=\frac{5}{3}[/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal or [latex]-\frac{3}{5}[/latex]. Next, we use point-slope form with this new slope and the given point.[latex]\begin{array}{l}y - 1=-\frac{3}{5}\left(x-\left(-4\right)\right)\hfill \\ y - 1=-\frac{3}{5}x-\frac{12}{5}\hfill \\ y=-\frac{3}{5}x-\frac{12}{5}+\frac{5}{5}\hfill \\ y=-\frac{3}{5}x-\frac{7}{5}\hfill \end{array}[/latex]

## Licenses & Attributions

### CC licensed content, Original

- Revision and Adaptation.
**Provided by:**Lumen Learning**License:**CC BY: Attribution.

### CC licensed content, Shared previously

- College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**Located at:**https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites.**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected]. - Question ID 1436.
**Authored by:**WebWork-Rochester.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL. - Question ID 110960, 110970, 110971.
**Authored by:**Lumen Learning.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL.