# 5.2: Equations of Parallel and Perpendicular Lines

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Determine whether lines are parallel or perpendicular.
• Write equations of perpendicular lines.
• Write equations of parallel lines.
• Investigate families of lines.

## Introduction

In this section, you will learn how parallel lines are related to each other on the coordinate plane. You will also learn how perpendicular lines are related to each other. Let’s start by looking at a graph of two parallel lines.

The two lines will never meet because they are parallel. We can clearly see that the two lines have different \begin{align*}y-\end{align*}intercepts, more specifically 6 and -4.

How about the slopes of the lines? Are they related in any way? Because the lines never meet, they must rise at the same rate. This means that the slopes of the two lines are the same.

Indeed, if we calculate the slopes of the lines, we find the following results.

Line A: \begin{align*} m = \frac{6 - 2} {0 - (-2)} = \frac{4} {2} = 2\end{align*}

Line B: \begin{align*} m = \frac{0 - (-4)} {2 - 0} = \frac{4} {2} = 2\end{align*}

For Parallel Lines: the slopes are the same, \begin{align*}m_1 = m_2\end{align*}, and the \begin{align*}y-\end{align*}intercepts are different.

Now let’s look at a graph of two perpendicular lines.

We find that we can’t say anything about the \begin{align*}y-\end{align*}intercepts. In this example, they are different, but they would be the same if the lines intersected at the \begin{align*}y-\end{align*}intercept.

Now we want to figure out if there is any relationship between the slopes of the two lines.

First of all we see that the slopes must have opposite signs, one negative and one positive.

To find the slope of line \begin{align*}A\end{align*}, we pick two points on the line and draw the blue (upper) right triangle. The legs of the triangle represent the rise and the run.

Looking at the figure \begin{align*} m_1 = \frac{b} {a}\end{align*}

To find the slope of line \begin{align*}B\end{align*}, we pick two points on the line and draw the red (lower) right triangle. If we look at the figure, we see that the two triangles are identical, only rotated by \begin{align*}90^{\circ}\end{align*}.

Looking at the diagram \begin{align*} m_2 = -\frac{a} {b}\end{align*}

For Perpendicular Lines: the slopes are negative reciprocals of each other. \begin{align*} m_1 = - \frac{1} {m_2}\end{align*} or \begin{align*} m_1 m_2 = -1\end{align*}

## Determine Whether Lines are Parallel or Perpendicular

You can find whether lines are parallel or perpendicular by comparing the slopes of the lines. If you are given points on the line you can find the slope using the formula. If you are given the equations of the lines, rewrite each equation in a form so that it is easy to read the slope, such as the slope-intercept form.

Example 1

Determine whether the lines are parallel or perpendicular or neither.

a) One line passes through points (2, 11) and (-1, 2); another line passes through points (0, -4) and (-2, -10).

b) One line passes through points (-2, -7) and (1, 5); another line passes through points (4, 1) and (-8, 4).

c) One lines passes through points (3, 1) and (-2, -2); another line passes through points (5, 5) and (4, -6).

Solution

Find the slope of each line and compare them.

a) \begin{align*} m_1 = \frac{2 - 11} {-1 -2} = \frac{-9} {-3} = 3\end{align*} and \begin{align*} m_2 = \frac{-10 - (-4)} {-2-0} = \frac{-6} {-2} = 3\end{align*}

The slopes are equal, so the lines are parallel.

b) \begin{align*} m_1 = \frac{5 - (-7)} {1 - (-2)} = \frac{12} {3} = 4\end{align*} and \begin{align*} m_2 = \frac{4 - 1} {-8-4} = \frac{3} {-12} = - \frac{1} {4}\end{align*}

The slopes are negative reciprocals of each other, so the lines are perpendicular.

c) \begin{align*} m_1 = \frac{-2 -1} {-2-3} = \frac{-3} {-5} = \frac{3} {5}\end{align*} and \begin{align*} m_2 = \frac{-6-5} {4 - 5} = \frac{-13} { -1} = 13\end{align*}

The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.

Example 2

Determine whether the lines are parallel or perpendicular or neither.

a) Line 1: \begin{align*}3x + 4y = 2\end{align*} Line 2: \begin{align*}8x -6y = 5 \end{align*}

b) Line 1: \begin{align*}2x = y-10\end{align*} Line 2: \begin{align*}y = -2x + 5 \end{align*}

c) Line 1: \begin{align*}7y + 1 = 7x\end{align*} Line 2: \begin{align*}x + 5 = y\end{align*}

Solution

Write each equation in slope-intercept form.

a) Line 1 \begin{align*} 3x + 4y = 2 \Rightarrow 4y = -3x + 2 \Rightarrow y = - \frac{3} {4}x + \frac{1} {2} \Rightarrow \text{slope} = - \frac{3} {4}\end{align*}

Line 2: \begin{align*} 8x - 6y = 5 \Rightarrow 8x - 5 = 6y \Rightarrow y = \frac{8} {6} x - \frac{5} {6} \Rightarrow y = \frac{4} {3}x - \frac{5} {6} \Rightarrow \text{slope} = \frac{4} {3}\end{align*}

The slopes are negative reciprocals of each other, so the lines are perpendicular to each other.

b) Line 1 \begin{align*}2x= y- 10\Rightarrow y=2x + 10 \Rightarrow \text{slope} = 2\end{align*}

Line 2 \begin{align*}y=-2x + 5 \Rightarrow \text{slope} = -2\end{align*}

The slopes are not the same or negative reciprocals of each other, so the lines are neither parallel nor perpendicular.

c) Line 1 \begin{align*} 7y + 1 = 7x \Rightarrow 7y = 7x - 1 \Rightarrow y = x - \frac{1} {7} \Rightarrow \text{slope} = 1\end{align*}

Line 2: \begin{align*} x + 5 = y \Rightarrow y = x + 5 \Rightarrow \text{slope} = 1\end{align*}

The slopes are the same so the lines are parallel.

## Write Equations of Perpendicular Lines

We can use the properties of perpendicular lines to write an equation of a line perpendicular to a given line. You will be given the equation of a line and asked to find the equation of the perpendicular line passing through a specific point. Here is the general method for solving a problem like this.

• Find the slope of the given line from its equation. You might need to rewrite the equation in a form such as the slope-intercept form.
• Find the slope of the perpendicular line by writing the negative reciprocal of the slope of the given line.
• Use the slope and the point to write the equation of the perpendicular line in point-slope form.

Example 3

Find the equation perpendicular to the line \begin{align*}y=-3x + 5\end{align*} that passes through point (2, 6).

Solution

Find the slope of the given line \begin{align*}y = -3x + 5\end{align*} has a \begin{align*}\text{slope} = -3\end{align*}.

The slope of the perpendicular line is the negative reciprocal \begin{align*}m = \frac{1}{3}\end{align*}

Now, we are trying to find the equation of a line with slope \begin{align*}m = \frac{1}{3}\end{align*} that passes through point (2, 6).

Use the point-slope form with the slope and the point \begin{align*} y - 6 = \frac{1} {3} (x - 2)\end{align*}

The equation of the line could also be written as \begin{align*} y = \frac{1} {3}x + \frac{16} {3}\end{align*}

Example 4

Find the equation of the line perpendicular to \begin{align*}x-5y = 15\end{align*} that passes through the point (-2, 5).

Solution

Rewrite the equation in slope-intercept form \begin{align*} x - 5y = 15 \Rightarrow - 5y = - x + 15 \Rightarrow y = \frac{1} {5} x - 3\end{align*}

The slope of the given line is \begin{align*}m= \frac{1}{5}\end{align*} and the slope of the perpendicular is the negative reciprocal or \begin{align*}m = -5\end{align*}. We are looking for a line with a slope \begin{align*}m=-5\end{align*} that passes through the point (-2, 5).

Use the point-slope form with the slope and the point \begin{align*}y-5=-5(x+2)\end{align*}

The equation of the line could also be written as \begin{align*}y = -5x- 5\end{align*}

Example 5

Find the equation of the line perpendicular to \begin{align*}y= -2\end{align*} that passes through the point (4, -2).

Solution

The equation is already in slope intercept form but it has an \begin{align*}x\end{align*} term of 0 \begin{align*}y = 0x -2\end{align*}. This means the slope is \begin{align*}m = 0\end{align*}.

We’d like a line with slope that is the negative reciprocal of 0. The reciprocal of 0 is \begin{align*}m = \frac{1}{0} = undefined\end{align*}. Hmmm... It seems like we have a problem. But, look again at the desired slope in terms of the definition of slope \begin{align*}m = \frac{rise}{run} = \frac{1}{0}\end{align*}. So our desired line will move 0 units in \begin{align*}x\end{align*} for every 1 unit it rises in \begin{align*}y\end{align*}. This is a vertical line, so the solution is the vertical line that passes through (4, -2). This is a line with an \begin{align*}x\end{align*} coordinate of 4 at every point along it.

The equation of the line is: \begin{align*}x = 4\end{align*}

## Write Equations of Parallel Lines

We can use the properties of parallel lines to write an equation of a line parallel to a given line. You will be given the equation of a line and asked to find the equation of the parallel line passing through a specific point. Here is the general method for solving a problem like this.

• Find the slope of the given line from its equation. You might need to rewrite the equation in a form such as the slope-intercept form.
• The slope of the parallel line is the same as that of the given line.
• Use the slope and the point to write the equation of the perpendicular line in slope-intercept form or point-slope form.

Example 6

Find the equation parallel to the line \begin{align*}y= 6x-9\end{align*} that passes through point (-1, 4).

Solution

Find the slope of the given line \begin{align*}y = 6x- 9\end{align*} has a \begin{align*}\text{slope} = 6\end{align*}.

Since parallel lines have the same slope, we are trying to find the equation of a line with slope \begin{align*}m = 6\end{align*} that passes through point (-1, 4).

\begin{align*}\text{Start with the slope-intercept form}. & & y & = mx + b\\ \text{Plug in the slope}& & y & = 6x + b\\ \text{Plug in point} \ (-1, 4).& & 4 & = 6(-1) + b \Rightarrow b= 4 + 6 \Rightarrow b = 10\end{align*}

The equation of the line is \begin{align*}y = 6x + 10\end{align*}.

Example 7

Find the equation of the line parallel to \begin{align*}7 -4y = 0\end{align*} that passes through the point (9, 2).

Solution

Rewrite the equation in slope-intercept form.

\begin{align*} 7 - 4y = 0 \Rightarrow 4y - 7 = 0 \Rightarrow 4y = 7 \Rightarrow y = \frac{7} {4} \Rightarrow y = 0x + \frac{7} {4}\end{align*}

The slope of the given line is \begin{align*}m = 0\end{align*}. This is a horizontal line.

Since the slopes of parallel lines are the same, we are looking for a line with slope \begin{align*}m = 0\end{align*} that passes through the point (9, 2).

\begin{align*}\text{Start with the slope-intercept form}. & & y & = 0x + b\\ \text{Plug in the slope}. & & y & = 0x+b\\ \text{Plug in point} \ (9, 2). & & 2 & = 0 (9) + b\\ & & \Rightarrow b & =2\end{align*}

The equation of the line is \begin{align*}y=2\end{align*}.

Example 8

Find the equation of the line parallel to \begin{align*}6x- 5y = 12\end{align*} that passes through the point (-5, -3).

Solution

Rewrite the equation in slope-intercept form.

\begin{align*} 3x - 5y = 12 \Rightarrow 5y = 6x - 12 \Rightarrow y = \frac{6} {5}x - \frac{12} {5}\end{align*}

The slope of the given line is \begin{align*}m = \frac{6}{5}\end{align*}.

Since the slopes of parallel lines are the same, we are looking for a line with slope \begin{align*}m = \frac{6}{5}\end{align*} that passes through the point (-5, -3).

\begin{align*}\text{Start with the slope-intercept form}. & & y & = mx + b\\ \text{Plug in the slope}. & & y & = \frac{6} {5} x + b\\ \text{Plug in point} \ (-5, -3). & & -3 & = \frac{6} {5} (-5) + b \Rightarrow -3 = -6 + b \Rightarrow b = 3\end{align*}

The equation of the line is: \begin{align*} y = \frac{6} {5} x + 3\end{align*}

## Investigate Families of Lines

A straight line has two very important properties, its slope and its \begin{align*}y-\end{align*}intercept. The slope tells us how steeply the line rises or falls, and the \begin{align*}y-\end{align*}intercept tells us where the line intersects the \begin{align*}y-\end{align*}axis. In this section, we will look at two families of lines. A family of lines is a set of lines that have something in common with each other. Straight lines can belong to two types of families. One where the slope is the same and one where the \begin{align*}y-\end{align*} intercept is the same.

Family 1

Keep slope unchanged and vary the \begin{align*}y-\end{align*}intercept.

The figure to the right shows the family of lines \begin{align*}y = -2x +b\end{align*}.

All the lines have a slope of -2 but the value of \begin{align*}b\end{align*} is different for each of the lines.

Notice that in such a family all the lines are parallel. All the lines look the same but they are shifted up and down the \begin{align*}y-\end{align*}axis. As \begin{align*}b\end{align*} gets larger the line rises on the \begin{align*}y-\end{align*}axis and as \begin{align*}b\end{align*} gets smaller the line goes lower on the \begin{align*}y-\end{align*}axis. This behavior is often called a vertical shift.

Family 2

Keep the \begin{align*}y-\end{align*}intercept unchanged and vary the slope.

The figure to the right shows the family of lines \begin{align*}y = mx + 2\end{align*}.

All lines have a \begin{align*}y-\end{align*}intercept of two but the value of the slope is different for each of the lines. The lines “start” with \begin{align*}y=2\end{align*} (red line) which has a slope of zero. They get steeper as the slope increases until it gets to the line \begin{align*}x= 0\end{align*} (purple line) which has an undefined slope.

Example 9

Write the equation of the family of lines satisfying the given condition:

a) Parallel to the \begin{align*}x-\end{align*}axis

b) Through the point (0, -1)

c) Perpendicular to \begin{align*}2x + 7y-9 = 0\end{align*}

d) Parallel to \begin{align*}x + 4y-12 = 0\end{align*}

Solution

a) All lines parallel to the \begin{align*}x-\end{align*}axis will have a slope of zero. It does not matter what the \begin{align*}y-\end{align*}intercept is.

The family of lines is \begin{align*}y= 0\cdot x + b\end{align*} or \begin{align*}y=b\end{align*}.

b) All lines passing through the point (0, -1) have the same \begin{align*}y-\end{align*}intercept, \begin{align*}b=-1\end{align*}.

The family of lines is \begin{align*}y=mx-1\end{align*}.

c) First we need to find the slope of the given line.

Rewrite \begin{align*}2x + 7y- 9 = 0\end{align*} in slope-intercept form \begin{align*} y = - \frac{2} {7} x + \frac{9} {7}\end{align*}.

The slope is \begin{align*}-\frac{ 2}{7}\end{align*}.

The slope of our family of lines is the negative reciprocal of the given slope \begin{align*}m = \frac{7}{2}\end{align*}.

All the lines in this family have a slope of \begin{align*}m =\frac{7}{2}\end{align*} but different \begin{align*}y-\end{align*}intercepts.

The family of lines is \begin{align*}y = \frac{7} {2}x + b\end{align*}.

d) First we need to find the slope of the given line.

Rewrite \begin{align*}x+4y-12=0\end{align*} in slope-intercept form \begin{align*} y = - \frac{1} {4} x + 3\end{align*}.

The slope is \begin{align*}m = -\frac{1}{4}\end{align*}.

All the lines in the family have a slope of \begin{align*}m = -\frac{1}{4}\end{align*} but different \begin{align*}y-\end{align*}intercepts.

The family of lines is \begin{align*} y = - \frac{1} {4} x + b\end{align*}.

## Lesson Summary

• Parallel lines have the same slopes, \begin{align*}m_1 = m_2 \end{align*}, but different \begin{align*}y-\end{align*}intercepts.
• Perpendicular lines have slopes which are the negative reciprocals of each other.

\begin{align*} m_1 = - \frac{1} {m_2} \quad \text{or} \quad m_1 m_2 = -1\end{align*}

• To find the line parallel (or perpendicular) to a specific line which passes through a given point:
1. Find the slope of the given line from its equation.
2. Compute the slope parallel (or perpendicular) to the line.
3. Use the computed slope and the specified point to write the equation of the new line in point-slope form.
4. Transform from point-slope form to another form if required.
• A family of lines is a set of lines that have something in common with each other. There are two types of line families. One where the slope is the same and one where the \begin{align*}y-\end{align*}intercept is the same.

## Review Questions

Determine whether the lines are parallel, perpendicular or neither.

1. One line passes through points (-1, 4) and (2, 6); another line passes through points (2, -3) and (8, 1).
2. One line passes through points (4, -3) and (-8, 0); another line passes through points (-1, -1) and (-2, 6).
3. One line passes through points (-3, 14) and (1, -2); another line passes through points (0, -3) and (-2, 5).
4. One line passes through points (3, 3) and (-6, -3); another line passes through points (2, -8) and (-6, 4).
5. Line 1: \begin{align*}4y + x = 8 \end{align*} Line 2: \begin{align*}12y + 3x = 1\end{align*}
6. Line 1: \begin{align*}5y + 3x + 1 \end{align*} Line 2: \begin{align*}6y + 10x = -3\end{align*}
7. Line 1: \begin{align*}2y -3x + 5 = 0 \end{align*} Line 2: \begin{align*}y + 6x = -3\end{align*}
8. Find the equation of the line parallel to \begin{align*}5x-2y = 2\end{align*} that passes through point (3, -2).
9. Find the equation of the line perpendicular to \begin{align*}y = -\frac{2}{5} x - 3\end{align*} that passes through point (2, 8).
10. Find the equation of the line parallel to \begin{align*}7y + 2x-10 = 0\end{align*} that passes through the point (2, 2).
11. Find the equation of the line perpendicular to \begin{align*}y+5 = 3(x-2)\end{align*} that passes through the point (6, 2). Write the equation of the family of lines satisfying the given condition.
12. All lines pass through point (0, 4).
13. All lines are perpendicular to \begin{align*}4x + 3y- 1 = 0\end{align*}.
14. All lines are parallel to \begin{align*}y- 3 = 4x + 2\end{align*}.
15. All lines pass through point (0, -1).

1. parallel
2. neither
3. parallel
4. perpendicular
5. parallel
6. perpendicular
7. neither
8. \begin{align*}y = \frac{5}{2}x -\frac{19}{2} \end{align*}
9. \begin{align*}y = \frac{5}{2}x + 3 \end{align*}
10. \begin{align*}y = -\frac{2}{7}x + \frac{18}{7} \end{align*}
11. \begin{align*}y = -\frac{1}{3}x + 4 \end{align*}
12. \begin{align*}y = mx + 4 \end{align*}
13. \begin{align*} y = \frac{3}{4}x + b \end{align*}
14. \begin{align*} y = 4x + b\end{align*}
15. \begin{align*}y = mx-1\end{align*}

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