4.6: Parallel and Perpendicular Lines (2024)

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Learning Objectives

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

Definition of Parallel and Perpendicular

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.

Table \(\PageIndex{1}\)
Given slope	Slope of perpendicular line
\(m=\frac{1}{2}\)	\(m_{\perp}=-2\)
\(m=-\frac{3}{4}\)	\(m_{\perp}=\frac{4}{3}\)
\(m=3\)	\(m_{\perp}=-\frac{1}{3}\)
\(m=-4\)	\(m_{\perp}=\frac{1}{4}\)

Finding Equations of Parallel and Perpendicular Lines

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 \((−\frac{1}{2}, 2)\).

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

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.

Key Takeaways

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.

\(y=−\frac{3}{4}x+8\)
\(y=\frac{1}{2}x−3\)
\(y=4x+4\)
\(y=−3x+7\)
\(y=−\frac{5}{8}x−12\)
\(y=\frac{7}{3}x+\frac{3}{2}\)
\(y=9x−25\)
\(y=−10x+15\)
\(y=5\)
\(x=−12\)
\(x−y=0\)
\(x+y=0\)
\(4x+3y=0\)
\(3x−5y=10\)
\(−2x+7y=14\)
\(−x−y=\frac{1}{5}\)
\(\frac{1}{2}x−\frac{1}{3}y=−1\)
\(−\frac{2}{3}x+\frac{4}{5}y=8\)
\(2x−\frac{1}{5}y=\frac{1}{10}\)
\(−\frac{4}{5}x−2y=7\)

Answer

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.

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

Answer

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

Parallel to \(y=\frac{1}{2}x+2\) and passing through \((6, −1)\).
Parallel to \(y=−\frac{3}{4}x−3\) and passing through \((−8, 2)\).
Perpendicular to \(y=3x−1\) and passing through \((−3, 2)\).
Perpendicular to \(y=−\frac{1}{3}x+2\) and passing through \((4, −3)\).
Perpendicular to \(y=−2\) and passing through \((−1, 5)\).
Perpendicular to \(x=\frac{1}{5}\) and passing through \((5, −3)\).
Parallel to \(y=3\) and passing through \((2, 4)\).
Parallel to \(x=2\) and passing through (7, −3)\).
Perpendicular to \(y=x\) and passing through \((7, −13)\).
Perpendicular to \(y=2x+9\) and passing through \((3, −1)\).
Parallel to \(y=\frac{1}{4}x−5\) and passing through \((−2, 1)\).
Parallel to \(y=−\frac{3}{4}x+1\) and passing through \((4, \frac{1}{4})\).
Parallel to \(2x−3y=6\) and passing through \((6, −2)\).
Parallel to \(−x+y=4\) and passing through \((9, 7)\).
Perpendicular to \(5x−3y=18\) and passing through \((−9, 10)\).
Perpendicular to \(x−y=11\) and passing through \((6, −8)\).
Parallel to \(\frac{1}{5}x−\frac{1}{3}y=2\) and passing through \((−15, 6)\).
Parallel to \(−10x−\frac{5}{7}y=12\) and passing through \((−1, \frac{1}{2})\).
Perpendicular to \(\frac{1}{2}x−\frac{1}{3}y=1\) and passing through \((−10, 3)\).
Perpendicular to \(−5x+y=−1\) and passing through \((−4, 0)\).
Parallel to \(x+4y=8\) and passing through \((−1, −2)\).
Parallel to \(7x−5y=35\) and passing through \((2, −3)\).
Perpendicular to \(6x+3y=1\) and passing through \((8, −2)\).
Perpendicular to \(−4x−5y=1\) and passing through \((−1, −1)\).
Parallel to \(−5x−2y=4\) and passing through \((\frac{1}{5}, −\frac{1}{4})\).
Parallel to \(6x−\frac{3}{2}y=9\) and passing through \((\frac{1}{3}, \frac{2}{3})\).
Perpendicular to \(y−3=0\) and passing through \((−6, 12)\).
Perpendicular to \(x+7=0\) and passing through \((5, −10)\).

Answer

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. \(x=−6\)

FAQs

4.6: Parallel and Perpendicular Lines? ›

Parallel lines are lines in the same plane that have the same slope. The lines never intersect and always maintain the same distance apart. Perpendicular lines are lines in the same plane that intersect each other at right angles. The slopes of perpendicular lines are opposite reciprocals.

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How to calculate parallel and perpendicular lines? ›

Step 1: Identify the slope of both of your equations. Step 2: Compare the slope values for both of your equations. If the slope values are the same, then the two lines are parallel. If one of the slope values is the negative reciprocal of the other, then the two lines are perpendicular.

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What are parallel and perpendicular lines Grade 4? ›

Parallel lines are lines that never intersect, and they form the same angle when they cross another line. Perpendicular lines intersect at a 90-degree angle, forming a square corner. We can identify these lines using angles and symbols in diagrams.

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How to find the perpendicular line? ›

Perpendicular lines have opposite (signs) reciprocal (flip fraction). So if slope is -4/5, the perpendicular slope is 5/4. If y=mx+b, we can subtract mx on both sides to get b=y-mx, and subbing in x=4, y=-3, and m=5/4, we get b=-3-(5/4)(4), 4s cancel, so you have b=-3-5=-8.

How to solve for parallel lines? ›

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. f(x)=m1x+b1 and g(x)=m2x+b2 are parallel if m1=m2 f ( x ) = m 1 x + b 1 and g ( x ) = m 2 x + b 2 are parallel if m 1 = m 2 . If and only if b1=b2 b 1 = b 2 and m1=m2 m 1 = m 2 , we say the lines coincide.

How can a line be both parallel and perpendicular? ›

Parallel lines never touch each other while perpendicular lines meet each other at the intersecting point. Thus, you cannot have two lines that are both parallel and perpendicular to each other.

How to find if two lines are parallel? ›

To see whether or not two lines are parallel, we must compare their slopes. Two lines are parallel if and only if their slopes are equal. The line 2x – 3y = 4 is in standard form. In general, a line in the form Ax + By = C has a slope of –A/B; therefore, the slope of line q must be –2/–3 = 2/3.

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How to do parallel, perpendicular, and neither? ›

Parallel: The slopes are the same. Perpendicular: The slopes are opposite reciprocals. Neither: The slopes are not the same; the slopes are not opposite reciprocals.

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What is the rule for perpendicular lines? ›

What is Perpendicular? In Mathematics, a perpendicular is defined as a straight line that makes the right angle (90 degrees) with the other line. In other words, if two lines intersect each other at the right angle, then the lines are perpendicular to each other.

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What are parallel and perpendicular lines on a graph? ›

Parallel lines have slopes that are the same. 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 90∘ angle. The slope of one line is the negative reciprocal of the other.

Do perpendicular lines always intersect? ›

Perpendicular lines always intersect at a right angle. A right angle always measures 90°. If one line has two perpendicular lines crossing it, those intersecting lines are parallel and never intersect with one another.

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What is the equation of a line parallel and perpendicular to a given line? ›

Use the form y=mx+c to identify parallel {and perpendicular} lines; find the equation of the line through 2 given points, or through 1 point with a given gradient.

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What is the formula for parallel and perpendicular vectors? ›

The vectors are parallel if ⃑ 𝐴 = 𝑘 ⃑ 𝐵 , where 𝑘 is a nonzero real constant. The vectors are perpendicular if ⃑ 𝐴 ⋅ ⃑ 𝐵 = 0 . If neither of these conditions are met, then the vectors are neither parallel nor perpendicular to one another.

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How to find the equation of a line with two points? ›

To calculate the equation of the line, use the y=mx+b y = m x + b format. Slope is equal to the change in y over the change in x , or rise over run. The change in x is equal to the difference in x-coordinates (also called run), and the change in y is equal to the difference in y-coordinates (also called rise).

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