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Finding an Equation of a Line Perpendicular to a Given Line

Finding an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point–slope equation, like we did with parallel lines.

The graph shows the graph of \(y=2x-3\). Now, we want to graph a line perpendicular to this line and passing through \(\left(-2,1\right)\).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere on the graph, the point (negative 2, 1) is plotted.

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation \({m}_{\text{⊥}}\) to represent the slope of a line perpendicular to a line with slope \(m\). (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

\(\begin{array}{ccc}\hfill y=2x-3\hfill & & \text{perpendicular line}\hfill \\ \hfill m=2\hfill & & {m}_{\text{⊥}}=-\frac{1}{2}\hfill \end{array}\)

We now know the perpendicular line will pass through \(\left(-2,1\right)\) with \({m}_{\text{⊥}}=-\frac{1}{2}\).

To graph the line, we will start at \(\left(-2,1\right)\) and count out the rise \(-1\) and the run 2. Then we draw the line.

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere, the point (negative 2, 1) is plotted. Another line perpendicular to the first line passes through the point (negative 2, 1) and intercepts the x and y-axes at (0, 0). A red line with an arrow extends left from (0, 0) to (negative 2, 0), then extends up and terminates at (negative 2, 1), forming a right triangle with the second line as a hypotenuse.

Do the lines appear perpendicular? Does the second line pass through \(\left(-2,1\right)\)?

Now, let’s see how to do this algebraically. We can use either the slope–intercept form or the point–slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point–slope form.

Example: How to Find an Equation of a Line Perpendicular to a Given Line

Find an equation of a line perpendicular to \(y=2x-3\) that contains the point \(\left(-2,1\right)\). Write the equation in slope–intercept form.

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Find the slope of the given line.” The second cell reads: “The line is in slope-intercept form. y equals 2x minus 3.” The third cell contains the slope of a line, defined as m equals 2.In the second row, the first cell reads: “Step 2. Find the slope of the perpendicular line.” The second cell reads “The slopes of perpendicular lines are negative reciprocals.” The third cell contains the slope of the perpendicular line, defined as m perpendicular equals negative 1 half.In the third row, the first cell reads “Step 3. Identify the point.” The second cell reads “The given point is (negative 2, 1).” The third cell contains the ordered pair (negative 2, 1) with a superscript x subscript 1 above negative 2 and a superscript y subscript 1 above 1.In the fourth row, the first cell reads “Step 4. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top of the second cell is blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the form with negative 2 substituted for x subscript 1, 1 substituted for y subscript 1, and negative 1 half substituted for m: y minus 1 equals negative 1 half times x minus negative 2 in parentheses. One line down, the text in the second cell says “Simplify.” The right column contains y minus 1 equals negative 1 half times x plus 2. Below this is y minus 1 equals negative 1 half x plus minus 1.In the fifth row, the first cell says “Step 5. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative 1 half x.

Find an equation of a line perpendicular to a given line.

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line.
  3. Identify the point.
  4. Substitute the values into the point–slope form, \(y-{y}_{1}=m\left(x-{x}_{1}\right)\).
  5. Write the equation in slope–intercept form.

Example

Find an equation of a line perpendicular to \(x=5\) that contains the point \(\left(3,-2\right)\). Write the equation in slope–intercept form.

Solution

Again, since we know one point, the point–slope option seems more promising than the slope–intercept option. We need the slope to use this form, and we know the new line will be perpendicular to \(x=5\). This line is vertical, so its perpendicular will be horizontal. This tells us the \({m}_{\text{⊥}}=0\).

\(\begin{array}{cccc}\text{Identify the point.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\left(3,-2\right)\hfill \\ \text{Identify the slope of the perpendicular line.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{m}_{\text{⊥}}=0\hfill \\ \text{Substitute the values into}\phantom{\rule{0.2em}{0ex}}y-{y}_{1}=m\left(x-{x}_{1}\right).\hfill & & & \hfill \phantom{\rule{6.2em}{0ex}}y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ & & & \hfill \phantom{\rule{4.7em}{0ex}}y-\left(-2\right)=0\left(x-3\right)\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{3.6em}{0ex}}y+2=0\hfill \\ & & & \hfill \phantom{\rule{5.6em}{0ex}}y=-2\hfill \end{array}\)

Sketch the graph of both lines. Do they appear to be perpendicular?

In the example above, we used the point–slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to \(x=5\) that contains the point \(\left(3,-2\right)\). The graph shows us the line\(x=5\) and the point \(\left(3,-2\right)\).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is x equals 5 intercepts the x-axis at (5, 0) and runs parallel to the y-axis. Elsewhere on the graph, the point (3, negative 2) is plotted.

We know every line perpendicular to a vetical line is horizontal, so we will sketch the horizontal line through \(\left(3,-2\right)\).

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is x equals 5 intercepts the x-axis at (5, 0) and runs parallel to the y-axis. Elsewhere on the graph, the points (negative 2, negative 2), (0, negative 2), (3, negative 2), and (6, negative 2) are plotted. A line perpendicular to the previous line passes through those points and runs parallel to the x-axis.

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of \(-2\). So, the equation of the line perpendicular to the vertical line \(x=5\) is \(y=-2\).

Example

Find an equation of a line that is perpendicular to \(y=-4\) that contains the point \(\left(-4,2\right)\). Write the equation in slope–intercept form.

Solution

The line \(y=-4\) is a horizontal line. Any line perpendicular to it must be vertical, in the form \(x=a\). Since the perpendicular line is vertical and passes through \(\left(-4,2\right)\), every point on it has an x-coordinate of \(-4\). The equation of the perpendicular line is \(x=-4\). You may want to sketch the lines. Do they appear perpendicular?

Access this online resource for additional instruction and practice with finding the equation of a line.

This lesson is part of:

Graphs and Equations

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