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3.6: Parallel and Perpendicular Lines

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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.

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 \(y=−5x+3\).

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.”

\(m_{∥}=−5\)

Example \(\PageIndex{2}\)

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

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 \(m_{⊥}=−\frac{7}{3}\).

\(m_{⊥}=−\frac{7}{3}\)

Exercise \(\PageIndex{1}\)

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

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

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

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.

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

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

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 \(y=−\frac{1}{4}x+2\).

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.

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

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 \(6x+3y=1\).

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 \(y\).

\(y=\frac{1}{2}x−6\)

Example \(\PageIndex{6}\)

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

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.

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

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

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 \(y=4\).

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.

\(x=−3\)

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

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.

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

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.

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

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 \((5, −10)\).

1. \(y=\frac{1}{2}x−4\)

3. \(y=−\frac{1}{3}x+1\)

5. \(x=−1\)

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. \(x=−6\)

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High school geometry

Course: high school geometry   >   unit 6.

• Parallel lines from equation
• Parallel lines from equation (example 2)
• Parallel lines from equation (example 3)
• Perpendicular lines from equation
• Parallel & perpendicular lines from equation
• Writing equations of perpendicular lines
• Writing equations of perpendicular lines (example 2)

Write equations of parallel & perpendicular lines

• Proof: parallel lines have the same slope
• Proof: perpendicular lines have opposite reciprocal slopes
• Analytic geometry FAQ

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Equations of Parallel and Perpendicular Lines

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• A Former Brilliant Member

As we have seen in the wiki Slopes and Intercepts of a Line , every line in the \(xy\)-plane has a slope , which can be thought of as the rate of change of \(y\) with respect to \(x\). Given two lines, comparing the slopes of the two lines gives us important information about how the lines meet in the \(xy\)-plane (if at all).

Parallel Lines

Perpendicular lines, problem solving.

Parallel lines are lines that do not meet at any point in the \(xy\)-plane. Another way to characterize parallel lines are distinct lines with the same slope. Suppose we are given two non-vertical lines in slope-intercept form:

\[ \begin{align} y &= m_1 x + b_1\\ y &= m_2 x + b_2. \end{align} \]

Then the two lines are parallel if \(m_1 = m_2\) and \(b_1 \ne b_2\).

Intuitively, if two distinct lines have the same rate of change, then the lines always point in the same direction and thus will never meet. In the above image, the slope-intercept form for the two lines are

\[ \begin{align} y &= \frac{1}{2} x + 0\\ y &= \frac{1}{2} x + 3. \end{align} \]

Since the two lines have the same slope and different \(y\)-intercepts, the two lines are parallel.

What is the equation of the line that is parallel to the line \(2x-3y-8=0\) and passes through the point \((3, 5)?\) Let \(y=ax+b\) be the equation of the line of interest. Then since this line is parallel to the line \(2x-3y-8=0\) or \(y=\frac{2x}{3}-\frac{8}{3},\) the slope of which is \(\frac{2}{3},\) so it must be true that \(a=\frac{2}{3}.\) So, the equation now becomes \(y=\frac{2}{3}x+b.\) Substituting in the coordinates \((3, 5),\) we have \[5=\frac{2}{3}\times 3+b \implies b=3.\] Therefore, the equation of the line of interest is \[y=\frac{2}{3}x+3. \ _\square\]

A pair of lines is perpendicular if the lines meet at \(90^\circ\) angle. Given two non-vertical lines in slope-intercept form

\[ \begin{align} y &= m_1 x + b_1\\ y &= m_2 x + b_2, \end{align} \]

the two lines are perpendicular if \(m_1 = - \frac{1}{m_2}\), that is, if the slopes are negative reciprocals of each other:

In the above image, the slope-intercept form of the two lines are

\[ \begin{align} y &= \frac{1}{2} x + 3\\ y &= -2x -2, \end{align} \]

and since the two slopes are negative reciprocals of each other, the lines are perpendicular.

What is the equation of the line that passes through the point \((-7, 3)\) and is perpendicular to the line \(y=\frac{1}{5}x-2?\) Let \(y=ax+b\) be the equation of the line of interest. Then since this line is perpendicular to the line \(y=\frac{1}{5}x-2\) the slope of which is \(\frac{1}{5},\) it must be true that \(a=-5.\) So, the equation now becomes \(y=-5x+b.\) Substituting in the coordinates \((-7, 3),\) we have \[3=-5\times (-7)+b \implies b=-32.\] Therefore, the equation of the line of interest is \[y=-5x-32. \ _\square\]

What is the sum of all the constants \(k\) such that the two lines \[\begin{array} &(k+1)x-3y+2=0, &(k-2)x+4y-1=0\end{array}\] are perpendicular to each other? For the two lines to be perpendicular, it must be true that \[(k+1)\times(k-2)+(-3)\times 4=0.\] Hence, \[\begin{align} (k+1)\times(k-2)+(-3)\times 4&=0\\ k^2-k-2-12&=0\\ k^2-k-14&=0. \end{align}\] Therefore, by Vieta's formula the sum of all the possible values of \(k\) is \(1.\) \(_\square\)

In some problems, we may be given properties of the slopes and intercepts of two lines and wish to calculate the values for the slopes and intercepts.

Consider two lines \(y=-2x+3\) and \(y=(K+1)x+4.\) When \(K=a,\) the two lines are parallel. When \(K=b,\) the two lines are perpendicular. What is \(a+b?\) Observe that the slope of the line \(y=-2x+3\) is \(-2\) and the slope of the line \(y=(K+1)x+4\) is \(K+1.\) Then, since the two lines are parallel when \(K=a,\) it follows that \[a+1=-2 \implies a=-3.\] Similarly, since the two lines are perpendicular when \(K=b,\) it follows that \[b+1=\frac{1}{2} \implies b=-\frac{1}{2}.\] Therefore, our answer is \[a+b=-3-\frac{1}{2}=-\frac{7}{2}. \ _\square\]

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Parallel and Perpendicular Lines

How to use Algebra to find parallel and perpendicular lines

Parallel Lines

How do we know when two lines are parallel ?

Their slopes are the same!

Find the equation of the line that is:

• parallel to y = 2x + 1
• and passes though the point (5,4)

The slope of y = 2x + 1 is 2

The parallel line needs to have the same slope of 2.

We can solve it by using the "point-slope" equation of a line :

y − y 1 = 2(x − x 1 )

And then put in the point (5,4):

y − 4 = 2(x − 5)

That is an answer!

But it might look better in y = mx + b form. Let's expand 2(x − 5) and then rearrange:

y − 4 = 2x − 10

Vertical Lines

But this does not work for vertical lines ... I explain why at the end.

Not The Same Line

Be careful! They may be the same line (but with a different equation), and so are not parallel .

How do we know if they are really the same line? Check their y-intercepts (where they cross the y-axis) as well as their slope:

Example: is y = 3x + 2 parallel to y − 2 = 3x ?

For y = 3x + 2 : the slope is 3, and y-intercept is 2

For y − 2 = 3x : the slope is 3, and y-intercept is 2

So they are the same line and so are not parallel

Perpendicular Lines

Two lines are perpendicular when they meet at a right angle (90°).

To find a perpendicular slope:

When one line has a slope of m , a perpendicular line has a slope of −1 m

In other words the negative reciprocal

Find the equation of the line that is

• perpendicular to y = −4x + 10
• and passes though the point (7,2)

The slope of y = −4x + 10 is −4

The negative reciprocal of that slope is:

m = −1 −4 = 1 4

So the perpendicular line will have a slope of 1/4:

y − y 1 = (1/4)(x − x 1 )

And now we put in the point (7,2):

y − 2 = (1/4)(x − 7)

That answer is OK, but let's also put it in "y=mx+b" form:

y − 2 = x/4 − 7/4

y = x/4 + 1/4

Quick Check of Perpendicular

When we multiply a slope m by its perpendicular slope −1 m we get simply −1 .

So to quickly check if two lines are perpendicular:

When we multiply their slopes, we get −1

Are these two lines perpendicular?

When we multiply the two slopes we get:

2 × (−0.5) = −1

Yes, we got −1, so they are perpendicular.

The previous methods work nicely except for a vertical line :

In this case the gradient is undefined (as we cannot divide by 0 ):

m = y A − y B x A − x B = 4 − 1 2 − 2 = 3 0 = undefined

So just rely on the fact that:

• a vertical line is parallel to another vertical line.
• a vertical line is perpendicular to a horizontal line (and vice versa).
• parallel lines: same slope
• perpendicular lines: negative reciprocal slope (−1/m)

Module 15: Graphs

Equations of parallel and perpendicular lines, learning outcomes.

• Write equations of parallel and perpendicular lines

Write the equations of parallel and perpendicular lines

The relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.

Let’s start with an example involving parallel lines.

Write the equation of a line that is parallel to the line [latex]x–y=5[/latex] and goes through the point [latex](−2,1)[/latex].

Rewrite the line you want to be parallel to into the [latex]y=mx+b[/latex] form, if needed.

[latex]\begin{array}{r}x–y=5\,\,\,\,\,\,\,\,\,\,\,\,\,\\−y=−x+5\\y=x–5\,\,\,\,\,\,\,\end{array}[/latex]

Identify the slope of the given line.

In the equation above, [latex]m=1[/latex] and [latex]b=−5[/latex].

Since [latex]m=1[/latex], the slope is [latex]1[/latex].

To find the slope of a parallel line, use the same slope.

The slope of the parallel line is [latex]1[/latex].

Use the method for writing an equation from the slope and a point on the line. Substitute 1 for m , and the point [latex](−2,1)[/latex] for [latex]x[/latex] and [latex]y[/latex].

[latex]\begin{array}{l}y=mx+b\\1=1(−2)+b\end{array}[/latex]

Solve for b.

[latex]\begin{array}{l}1=−2+b\\3=b\end{array}[/latex]

Write the equation using the new slope for m and the b you just found.

[latex]y=x+3[/latex]

Determine the Equation of a Line Parallel to Another Line Through a Given Point

Determine the Equation of a Line Perpendicular to Another Line Through a Given Point

When you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal.  In other words, flip it and change the sign.

Write the equation of a line that contains the point [latex](1,5)[/latex] and is perpendicular to the line [latex]y=2x– 6[/latex].

Identify the slope of the line you want to be perpendicular to.

The given line is written in [latex]y=mx+b[/latex] form, with [latex]m=2[/latex] and [latex]b=-6[/latex]. The slope is [latex]2[/latex].

To find the slope of a perpendicular line, find the reciprocal, [latex] \displaystyle \frac{1}{2}[/latex], then the opposite, [latex] \displaystyle -\frac{1}{2}[/latex].

The slope of the perpendicular line is [latex] \displaystyle -\frac{1}{2}[/latex].

Use the method for writing an equation from the slope and a point on the line. Substitute [latex] \displaystyle -\frac{1}{2}[/latex] for m, and the point [latex](1,5)[/latex] for [latex]x[/latex] and [latex]y[/latex].

[latex] \displaystyle \begin{array}{l}y=mx+b\\5=-\frac{1}{2}(1)+b\end{array}[/latex]

[latex] \displaystyle \begin{array}{l}\,\,\,5=-\frac{1}{2}+b\\\frac{11}{2}=b\end{array}[/latex]

[latex]y=-\frac{1}{2}x+\frac{11}{2}[/latex]

Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form

Write the equation of a line that is parallel to the line [latex]y=4[/latex] through the point [latex](0,10)[/latex].

Rewrite the line into [latex]y=mx+b[/latex] form, if needed.

You may notice without doing this that [latex]y=4[/latex] is a horizontal line 4 units above the x -axis. Because it is horizontal, you know its slope is zero.

[latex]\begin{array}{l}y=4\\y=0x+4\end{array}[/latex]

In the equation above, [latex]m=0[/latex] and [latex]b=4[/latex].

Since [latex]m=0[/latex], the slope is [latex]0[/latex]. This is a horizontal line.

The slope of the parallel line is also [latex]0[/latex].

Since the parallel line will be a horizontal line, its form is

[latex]y=\text{a constant}[/latex]

Since we want this new line to pass through the point [latex](0,10)[/latex], we will need to write the equation of the new line as:

[latex]y=10[/latex]

This line is parallel to [latex]y=4[/latex] and passes through [latex](0,10)[/latex].

Write the equation of a line that is perpendicular to the line [latex]y=-3[/latex] through the point [latex](-2,5)[/latex].

In the equation above, [latex]m=0[/latex] and [latex]b=-3[/latex].

A perpendicular line will have a slope that is the negative reciprocal of the slope of [latex]y=-3[/latex], but what does that mean in this case?

The reciprocal of [latex]0[/latex] is [latex]\frac{1}{0}[/latex], but we know that dividing by [latex]0[/latex] is undefined.

This means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.

The form of a vertical line is [latex]x=\text{a constant}[/latex], where every x-value on the line is equal to some constant.  Since we are looking for a line that goes through the point [latex](-2,5)[/latex], all of the [latex]x[/latex]-values on this line must be [latex]-2[/latex].

The equation of a line passing through [latex](-2,5)[/latex] that is perpendicular to the horizontal line [latex]y=-3[/latex] is therefore,

[latex]x=-2[/latex]

Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line

When lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a [latex]90°[/latex] angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.

• Determine the Equation of a Line Parallel to a Line in General Form. Authored by : Mathispower4u. Located at : https://youtu.be/TQKz2XHI09E . License : CC BY: Attribution
• Determine the Equation of a Line Perpendicular to a Line in Slope-Intercept Form. Authored by : Mathispower4u. Located at : https://youtu.be/QtvtzKjtowA . License : CC BY: Attribution
• Find the Equation of a Perpendicular and Horizontal Line to a Horizontal Line. Authored by : Mathispower4u. Located at : https://youtu.be/Qpn3f3wMeIs . License : CC BY: Attribution

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How to Solve Parallel Lines and Transversals Problems? (+FREE Worksheet!)

When a line (transversal) intersects two parallel lines in the same plane, eight angles are formed. In this article, we will teach you how to find the missing angles in this case by using the Parallel Lines and Transversals rules.

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A step-by-step guide to solving Parallel Lines and Transversals Problem

• When a line (transversal) intersects two parallel lines in the same plane, eight angles are formed. In the following diagram, a transversal intersects two parallel lines. Angles 1, 3, 5, and 7 are congruent. Angles 2, 4, 6, and 8 are also congruent.
• In the following diagram, the following angles are supplementary angles (their sum is 180):

• Angles 1 and 8 • Angles 2 and 7 • Angles 3 and 6 • Angles 4 and 5

Parallel Lines and Transversals – Example 1:

In the following diagram, two parallel lines are cut by a transversal. What is the value of \(x\)?

The two angles \(3x-15\) and \(2x+7\) are equivalent. That is: \(3x-15=2x+7\) Now, solve for \(x: 3x-15+15=2x+7+15 →\) \(3x=2x+22→3x-2x=2x+22-2x→x=22\)

Parallel Lines and Transversals – Example 2:

The two angles \(75^\circ\) and \(11x-2\) are equal. \(11x-2=75\) Now, solve for \(x: 11x-2+2=75+2→ 11x=77→x=\frac{77}{11}→x=7\)

Parallel Lines and Transversals – Example 3:

The two angles \(7x-35\) and \(3x+45\) are equivalents. That is: \(7x-35=3x+45\) Now, solve for \(x: 7x-35+35=3x+45+35 →\) \(7x=3x+80→7x-3x=3x+80-3x→4x=80→x=\frac{80}{4}→x=20\)

Parallel Lines and Transversals – Example 4:

The two angles \(3x-27\) and \(-x+33\) are equivalents. That is: \(3x-27=-x+33\) Now, solve for \(x: 3x-27+27=-x+33+27 →\) \(3x=-x+60→3x+x=-x+60+x→4x=60→x=\frac{60}{4}→x=15\)

Exercises for Parallel Lines and Transversals

Find missing angles with parallel lines and transversals..

1.Find the measure of the angle indicated.

2. Solve for \(x\).

3. Find the measure of the angle indicated.

4. Solve for \(x\).

1.\(\color{blue}{110^\circ}\)

2.\(\color{blue}{x=8}\)

3. \(\color{blue}{84^\circ}\)

4.\(\color{blue}{x=5}\)

by: Effortless Math Team about 3 years ago (category: Articles , Free Math Worksheets )

Effortless Math Team

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