The slope of a line is . what is the slope of a line that is perpendicular to this line? - -

In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.

Learning Objectives

• Determine the slopes of parallel and perpendicular lines.
• Find equations of parallel and perpendicular lines

Parallel lines are lines in the same plane that never intersect. Two nonvertical lines in the same plane, with slopes \(m_{1}\) and \(m_{2}\), are parallel if their slopes are the same, \(m_{1}=m_{2}\). Consider the following two lines:

Consider their corresponding graphs:

Figure \(\PageIndex{1}\)

Both lines have a slope \(m=\frac{3}{4}\) and thus are parallel.

Perpendicular lines are lines in the same plane that intersect at right angles (\(90\) degrees). Two nonvertical lines in the same plane, with slopes \(m_{1}\) and \(m_{2}\), are perpendicular if the product of their slopes is \(−1: m1⋅m2=−1\). We can solve for \(m_{1}\) and obtain \(m_{1}=\frac{−1}{m_{2}}\). In this form, we see that perpendicular lines have slopes that are negative reciprocals, or opposite reciprocals. For example, if given a slope

\(m=-\frac{5}{8}\)

then the slope of a perpendicular line is the opposite reciprocal:

\(m_{\perp}=\frac{8}{5}\)

The mathematical notation \(m_{⊥}\) reads “\(m\) perpendicular.” We can verify that two slopes produce perpendicular lines if their product is \(−1\).

\(m\cdot m_{\perp}=-\frac{5}{8}\cdot\frac{8}{5}=-\frac{40}{40}=-1\quad\color{Cerulean}{\checkmark}\)

Geometrically, we note that if a line has a positive slope, then any perpendicular line will have a negative slope. Furthermore, the rise and run between two perpendicular lines are interchanged.

Figure \(\PageIndex{2}\)

Perpendicular lines have slopes that are opposite reciprocals, so remember to find the reciprocal and change the sign. In other words,

If \(m=\frac{a}{b}\), then \(m_{\perp}=-\frac{b}{a}\)

Determining the slope of a perpendicular line can be performed mentally. Some examples follow

Example \(\PageIndex{1}\)

Determine the slope of a line parallel to \(y=−5x+3\).

Solution:

Since the given line is in slope-intercept form, we can see that its slope is \(m=−5\). Thus the slope of any line parallel to the given line must be the same, \(m_{∥}=−5\). The mathematical notation \(m_{∥}\) reads “\(m\) parallel.”

Answer:

\(m_{∥}=−5\)

Example \(\PageIndex{2}\)

Determine the slope of a line perpendicular to \(3x−7y=21\).

Solution:

First, solve for \(y\) and express the line in slope-intercept form.

In this form, we can see that the slope of the given line is \(m=\frac{3}{7}\), and thus \(m_{⊥}=−\frac{7}{3}\).

Answer:

\(m_{⊥}=−\frac{7}{3}\)

Exercise \(\PageIndex{1}\)

Find the slope of the line perpendicular to \(15x+5y=20\).

\(m_{\perp}=\frac{1}{3}\)

We have seen that the graph of a line is completely determined by two points or one point and its slope. Often you will be asked to find the equation of a line given some geometric relationship—for instance, whether the line is parallel or perpendicular to another line.

Example \(\PageIndex{3}\)

Find the equation of the line passing through \((6, −1)\) and parallel to \(y=\frac{1}{2}x+2\)

Solution

Here the given line has slope \(m=\frac{1}{2}\), and the slope of a line parallel is \(m_{∥}=\frac{1}{2}\). Since you are given a point and the slope, use the point-slope form of a line to determine the equation.

\(\begin{array}{cc}{\color{Cerulean}{Point}}&{\color{Cerulean}{Slope}}\\{(6,-1)}&{m_{\parallel}=\frac{1}{2}} \end{array}\)

Answer:

\(y=\frac{1}{2}x-4\)

It is important to have a geometric understanding of this question. We were asked to find the equation of a line parallel to another line passing through a certain point.

Figure \(\PageIndex{3}\)

Through the point \((6, −1)\) we found a parallel line, \(y=\frac{1}{2}x−4\), shown dashed. Notice that the slope is the same as the given line, but the \(y\)-intercept is different. If we keep in mind the geometric interpretation, then it will be easier to remember the process needed to solve the problem.

Example \(\PageIndex{4}\)

Find the equation of the line passing through \((−1, −5)\) and perpendicular to \(y=−\frac{1}{4}x+2\).

Solution:

The given line has slope \(m=−\frac{1}{4}\), and thus \(m_{⊥}=+\frac{4}{1}=4\). Substitute this slope and the given point into point-slope form.

\(\begin{array}{cc} {\color{Cerulean}{Point}}&{\color{Cerulean}{Slope}}\\{(-1,-5)}&{m_{\perp}=4}\end{array}\)

Answer:

\(y=4x-1\)

Geometrically, we see that the line \(y=4x−1\), shown dashed below, passes through \((−1, −5)\) and is perpendicular to the given line.

Figure \(\PageIndex{4}\)

It is not always the case that the given line is in slope-intercept form. Often you have to perform additional steps to determine the slope. The general steps for finding the equation of a line are outlined in the following example.

Example \(\PageIndex{5}\)

Find the equation of the line passing through \((8, −2)\) and perpendicular to \(6x+3y=1\).

Solution:

Step 1: Find the slope \(m\). First, find the slope of the given line. To do this, solve for \(y\) to change standard form to slope-intercept form, \(y=mx+b\).

\(\begin{aligned} 6x+3y&=1 \\ 6x+3y\color{Cerulean}{-6x}&=1\color{Cerulean}{-6x} \\ 3y&=-6x+1 \\ \frac{3y}{\color{Cerulean}{3}}&=\frac{-6x+1}{\color{Cerulean}{3}} \\ y&=\frac{-6x}{3}+\frac{1}{3}\\y&=-2x+\frac{1}{3} \end{aligned}\)

In this form, you can see that the slope is \(m=−2=−\frac{2}{1}\), and thus \(m_{⊥}=\frac{−1}{−2}=+\frac{1}{2}\).

Step 2: Substitute the slope you found and the given point into the point-slope form of an equation for a line. In this case, the slope is \(m_{⊥}=\frac{1}{2}\) and the given point is \((8, −2)\).

\(\begin{aligned} y-y_{1}&=m(x-x_{1}) \\ y-(-2)&=\frac{1}{2}(x-8) \end{aligned}\)

Step 3: Solve for \(y\).

Answer:

\(y=\frac{1}{2}x−6\)

Example \(\PageIndex{6}\)

Find the equation of the line passing through \((\frac{7}{2}, 1)\) and parallel to \(2x+14y=7\).

Solution:

Find the slope \(m\) by solving for \(y\).

\(\begin{aligned} 2x+14y&=7 \\ 2x+14y\color{Cerulean}{-2x}&=7\color{Cerulean}{-2x} \\ 14y&=-2x+7 \\ \frac{14y}{\color{Cerulean}{14}}&=\frac{-2x+7}{\color{Cerulean}{14}} \\ y&=\frac{-2x}{14}+\frac{7}{14} \\ y&=-\frac{1}{7}x+\frac{1}{2} \end{aligned}\)

The given line has the slope \(m=−\frac{1}{7}\), and so \(m_{∥}=−\frac{1}{7}\). We use this and the point \((\frac{7}{2}, 1)\) in point-slope form.

\(\begin{aligned} y-y_{1}&=m(x-x_{1}) \\ y-1&=-\frac{1}{7}\left(x-\frac{7}{2} \right) \\ y-1&=-\frac{1}{7}x+\frac{1}{2} \\ y-1\color{Cerulean}{+1}&=-\frac{1}{7}x+\frac{1}{2}\color{Cerulean}{+1} \\ y&=-\frac{1}{7}x+\frac{1}{2}+\color{Cerulean}{\frac{2}{2}} \\ y&=-\frac{1}{7}x+\frac{3}{2} \end{aligned}\)

Answer:

\(y=-\frac{1}{7}x+\frac{3}{2}\)

Exercise \(\PageIndex{2}\)

Find the equation of the line perpendicular to \(x−3y=9\) and passing through \((−\frac{1}{2}, 2)\).

\(y=-3x+\frac{1}{2}\)

When finding an equation of a line perpendicular to a horizontal or vertical line, it is best to consider the geometric interpretation.

Example \(\PageIndex{7}\)

Find the equation of the line passing through \((−3, −2)\) and perpendicular to \(y=4\).

Solution:

We recognize that \(y=4\) is a horizontal line and we want to find a perpendicular line passing through \((−3, −2)\).

Figure \(\PageIndex{5}\)

If we draw the line perpendicular to the given horizontal line, the result is a vertical line.

Figure \(\PageIndex{6}\)

Equations of vertical lines look like \(x=k\). Since it must pass through \((−3, −2)\), we conclude that \(x=−3\) is the equation. All ordered pair solutions of a vertical line must share the same \(x\)-coordinate.

Answer:

\(x=−3\)

We can rewrite the equation of any horizontal line, \(y=k\), in slope-intercept form as follows:

\(y=0x+k\)

Written in this form, we see that the slope is \(m=0=\frac{0}{1}\). If we try to find the slope of a perpendicular line by finding the opposite reciprocal, we run into a problem: \(m_{⊥}=−\frac{1}{0}\), which is undefined. This is why we took care to restrict the definition to two nonvertical lines. Remember that horizontal lines are perpendicular to vertical lines.

• Parallel lines have the same slope.
• Perpendicular lines have slopes that are opposite reciprocals. In other words, if \(m=\frac{a}{b}\), then \(m_{⊥}=−\frac{b}{a}\).
• To find an equation of a line, first use the given information to determine the slope. Then use the slope and a point on the line to find the equation using point-slope form.
• Horizontal and vertical lines are perpendicular to each other.

Exercise \(\PageIndex{3}\) Parallel and Perpendicular Lines

Determine the slope of parallel lines and perpendicular lines.

1. \(y=−\frac{3}{4}x+8\)
2. \(y=\frac{1}{2}x−3\)
3. \(y=4x+4\)
4. \(y=−3x+7\)
5. \(y=−\frac{5}{8}x−12\)
6. \(y=\frac{7}{3}x+\frac{3}{2}\)
7. \(y=9x−25\)
8. \(y=−10x+15\)
9. \(y=5\)
10. \(x=−12\)
11. \(x−y=0\)
12. \(x+y=0\)
13. \(4x+3y=0\)
14. \(3x−5y=10\)
15. \(−2x+7y=14\)
16. \(−x−y=\frac{1}{5}\)
17. \(\frac{1}{2}x−\frac{1}{3}y=−1\)
18. \(−\frac{2}{3}x+\frac{4}{5}y=8\)
19. \(2x−\frac{1}{5}y=\frac{1}{10}\)
20. \(−\frac{4}{5}x−2y=7\)

1. \(m_{∥}=−\frac{3}{4}\) and \(m_{⊥}=\frac{4}{3}\)

3. \(m_{∥}=4\) and \(m_{⊥}=−\frac{1}{4}\)

5. \(m_{∥}=−\frac{5}{8}\) and \(m_{⊥}=\frac{8}{5}\)

7. \(m_{∥}=9\) and \(m_{⊥}=−\frac{1}{9}\)

9. \(m_{∥}=0\) and \(m_{⊥}\) undefined

11. \(m_{∥}=1\) and \(m_{⊥}=−1\)

13. \(m_{∥}=−\frac{4}{3}\) and \(m_{⊥}=\frac{3}{4}\)

15. \(m_{∥}=\frac{2}{7}\) and \(m_{⊥}=−\frac{7}{2}\)

17. \(m_{∥}=\frac{3}{2}\) and \(m_{⊥}=−\frac{2}{3}\)

19. \(m_{∥}=10\) and \(m_{⊥}=−\frac{1}{10}\)

Exercise \(\PageIndex{4}\) Parallel and Perpendicular Lines

Determine if the lines are parallel, perpendicular, or neither.

1. \(\left\{\begin{aligned}y&=\frac{2}{3}x+3\\y&=\frac{2}{3}x−3\end{aligned}\right.\)
2. \(\left\{\begin{aligned}y&=\frac{3}{4}x−1\\y&=\frac{4}{3}x+3\end{aligned}\right.\)
3. \(\left\{\begin{aligned}y&=−2x+1\\ y&=\frac{1}{2}x+8\end{aligned}\right.\)
4. \(\left\{\begin{aligned}y&=3x−\frac{1}{2}\\ y&=3x+2\end{aligned}\right.\)
5. \(\left\{\begin{aligned}y&=5\\x&=−2\end{aligned}\right.\)
6. \(\left\{\begin{aligned}y&=7\\y&=−\frac{1}{7}\end{aligned}\right.\)
7. \(\left\{\begin{aligned}3x−5y&=15\\ 5x+3y&=9\end{aligned}\right.\)
8. \(\left\{\begin{aligned}x−y&=7\\3x+3y&=2\end{aligned}\right.\)
9. \(\left\{\begin{aligned}2x−6y&=4\\−x+3y&=−2 \end{aligned}\right.\)
10. \(\left\{\begin{aligned}−4x+2y&=3\\6x−3y&=−3 \end{aligned}\right.\)
11. \(\left\{\begin{aligned}x+3y&=9\\2x+3y&=6 \end{aligned}\right.\)
12. \(\left\{\begin{aligned}y−10&=0\\x−10&=0 \end{aligned}\right.\)
13. \(\left\{\begin{aligned}y+2&=0\\2y−10&=0 \end{aligned}\right.\)
14. \(\left\{\begin{aligned}3x+2y&=6\\2x+3y&=6 \end{aligned}\right.\)
15. \(\left\{\begin{aligned}−5x+4y&=20\\10x−8y&=16 \end{aligned}\right.\)
16. \(\left\{\begin{aligned}\frac{1}{2}x−\frac{1}{3}y&=1\\\frac{1}{6}x+\frac{1}{4}y&=−2\end{aligned}\right.\)

1. Parallel

3. Perpendicular

5. Perpendicular

7. Perpendicular

9. Parallel

11. Neither

13. Parallel

15. Parallel

Exercise \(\PageIndex{5}\) Equations in Point-Slope Form

Find the equation of the line

1. Parallel to \(y=\frac{1}{2}x+2\) and passing through \((6, −1)\).
2. Parallel to \(y=−\frac{3}{4}x−3\) and passing through \((−8, 2)\).
3. Perpendicular to \(y=3x−1\) and passing through \((−3, 2)\).
4. Perpendicular to \(y=−\frac{1}{3}x+2\) and passing through \((4, −3)\).
5. Perpendicular to \(y=−2\) and passing through \((−1, 5)\).
6. Perpendicular to \(x=\frac{1}{5}\) and passing through \((5, −3)\).
7. Parallel to \(y=3\) and passing through \((2, 4)\).
8. Parallel to \(x=2\) and passing through (7, −3)\).
9. Perpendicular to \(y=x\) and passing through \((7, −13)\).
10. Perpendicular to \(y=2x+9\) and passing through \((3, −1)\).
11. Parallel to \(y=\frac{1}{4}x−5\) and passing through \((−2, 1)\).
12. Parallel to \(y=−\frac{3}{4}x+1\) and passing through \((4, \frac{1}{4})\).
13. Parallel to \(2x−3y=6\) and passing through \((6, −2)\).
14. Parallel to \(−x+y=4\) and passing through \((9, 7)\).
15. Perpendicular to \(5x−3y=18\) and passing through \((−9, 10)\).
16. Perpendicular to \(x−y=11\) and passing through \((6, −8)\).
17. Parallel to \(\frac{1}{5}x−\frac{1}{3}y=2\) and passing through \((−15, 6)\).
18. Parallel to \(−10x−\frac{5}{7}y=12\) and passing through \((−1, \frac{1}{2})\).
19. Perpendicular to \(\frac{1}{2}x−\frac{1}{3}y=1\) and passing through \((−10, 3)\).
20. Perpendicular to \(−5x+y=−1\) and passing through \((−4, 0)\).
21. Parallel to \(x+4y=8\) and passing through \((−1, −2)\).
22. Parallel to \(7x−5y=35\) and passing through \((2, −3)\).
23. Perpendicular to \(6x+3y=1\) and passing through \((8, −2)\).
24. Perpendicular to \(−4x−5y=1\) and passing through \((−1, −1)\).
25. Parallel to \(−5x−2y=4\) and passing through \((\frac{1}{5}, −\frac{1}{4})\).
26. Parallel to \(6x−\frac{3}{2}y=9\) and passing through \((\frac{1}{3}, \frac{2}{3})\).
27. Perpendicular to \(y−3=0\) and passing through \((−6, 12)\).
28. Perpendicular to \(x+7=0\) and passing through \((5, −10)\).

1. \(y=\frac{1}{2}x−4\)

3. \(y=−\frac{1}{3}x+1\)

5. \(x=−1\)

7. \(y=4\)

9. \(y=−x−6\)

11. \(y=\frac{1}{4}x+\frac{3}{2}\)

13. \(y=\frac{2}{3}x−6\)

15. \(y=−\frac{3}{5}x+\frac{23}{5}\)

17. \(y=\frac{3}{5}x+15\)

19. \(y=−\frac{2}{3}x−\frac{11}{3}\)

21. \(y=−\frac{1}{4}x−\frac{9}{4}\)

23. \(y=\frac{1}{2}x−6\)

25. \(y=−\frac{5}{2}x+\frac{1}{4}\)

27. \(x=−6\)