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Study Guides > College Algebra

Parallel and Perpendicular Lines

Learning Objectives

  • 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
Parallel lines have the same slope and different 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].
Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5. Parallel lines
All of the lines shown in the graph are parallel because they have the same slope and different 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 above 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]

Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x/3 minus 2. Their intersection is marked by a box to show that it is a right angle. Perpendicular lines

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. Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x/3 plus 1 and y = 3 times x/4 minus 2. A box is placed at the intersection to note that it forms a right angle. 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 now

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: equations are written in slope-intercept form. Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x/2 plus 5 and y = x/2 plus 2. The lines do not cross.

Check your work with Desmos.
 

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 function:

[latex]f\left(x\right)=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 f(x). So the lines formed by all of the following functions will be perpendicular to f(x).

[latex]\begin{cases}g\left(x\right)=-\frac{1}{2}x+4\hfill \\ h\left(x\right)=-\frac{1}{2}x+2\hfill \\ p\left(x\right)=-\frac{1}{2}x-\frac{1}{2}\hfill \end{cases}[/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 f(x) 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 the slope-intercept form of a line and solving for b.

[latex]\begin{cases}g\left(x\right)=mx+b\hfill \\ 0=-\frac{1}{2}\left(4\right)+b\hfill \\ 0=-2+b\hfill \\ 2=b\hfill \\ b=2\hfill \end{cases}[/latex]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a y-intercept of 2 is [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex]. So [latex]g\left(x\right)=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]f\left(x\right)=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. 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 [latex]g\left(x\right)=mx+b[/latex].
  4. Solve for b.
  5. Write the equation for the line.

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{cases}y_{2}=-\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{cases}[/latex]

The line perpendicular to y that passes through (3, 0) is [latex]y_{2}=-\frac{1}{3}x+1[/latex].

Analysis of the Solution

A graph of the two lines is shown in the graph below. Graph of two functions where the blue line is g(x) = -1/3x + 1, and the orange line is f(x) = 3x + 6.

Try It

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

Answer:

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

Example: Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to [latex]5x - 3y+4=0\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 the point-slope formula 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]

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  • 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].
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