Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Find the Equation of a Line

Learning Objectives

By the end of this section, you will be able to:

Find an equation of the line given the slope and $y -intercept y -intercept$
Find an equation of the line given the slope and a point
Find an equation of the line given two points
Find an equation of a line parallel to a given line
Find an equation of a line perpendicular to a given line

Be Prepared 3.3

Before you get started, take this readiness quiz.

Solve: $\frac{2}{5} (x + 15) . \frac{2}{5} (x + 15) .$
If you missed this problem, review Example 1.50.
Simplify: $−3 (x - (−2)) . −3 (x - (−2)) .$
If you missed this problem, review Example 1.53.
Solve for y: $y - 3 = −2 (x + 1) . y - 3 = −2 (x + 1) .$
If you missed this problem, review Example 2.31.

How do online companies know that “you may also like” a particular item based on something you just ordered? How can economists know how a rise in the minimum wage will affect the unemployment rate? How do medical researchers create drugs to target cancer cells? How can traffic engineers predict the effect on your commuting time of an increase or decrease in gas prices? It’s all mathematics.

The physical sciences, social sciences, and the business world are full of situations that can be modeled with linear equations relating two variables. To create a mathematical model of a linear relation between two variables, we must be able to find the equation of the line. In this section, we will look at several ways to write the equation of a line. The specific method we use will be determined by what information we are given.

Find an Equation of the Line Given the Slope and y-Intercept

We can easily determine the slope and intercept of a line if the equation is written in slope-intercept form, $y = m x + b . y = m x + b .$ Now we will do the reverse—we will start with the slope and y-intercept and use them to find the equation of the line.

Example 3.24

Find the equation of a line with slope $−9 −9$ and y-intercept $(0, −4) . (0, −4) .$

Show solution

Solution

Since we are given the slope and y-intercept of the line, we can substitute the needed values into the slope-intercept form, $y = m x + b . y = m x + b .$

Name the slope.
Name the y-intercept.
Substitute the values into $y = m x + b . y = m x + b .$

Try It 3.47

Find the equation of a line with slope $\frac{2}{5} \frac{2}{5}$ and y-intercept $(0, 4) . (0, 4) .$

Try It 3.48

Find the equation of a line with slope $−1 −1$ and y-intercept $(0, −3) . (0, −3) .$

Sometimes, the slope and intercept need to be determined from the graph.

Example 3.25

Find the equation of the line shown in the graph.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (negative 3, negative 6), (0, negative 4), (3, negative 2), and (6, 0).

Show solution

Solution

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope-intercept form, $y = m x + b . y = m x + b .$

To find the slope, we choose two points on the graph.

The y-intercept is $(0, −4) (0, −4)$ and the graph passes through $(3, −2) . (3, −2) .$

Find the slope, by counting the rise and run.

Find the y-intercept.
Substitute the values into $y = m x + b . y = m x + b .$

Try It 3.49

Find the equation of the line shown in the graph.

Try It 3.50

Find the equation of the line shown in the graph.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 5), (3, negative 1), and (6, 3).

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope-intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. But what happens when you have another point instead of the y-intercept?

We are going to use the slope formula to derive another form of an equation of the line.

Suppose we have a line that has slope m and that contains some specific point $(x_{1}, y_{1}) (x_{1}, y_{1})$ and some other point, which we will just call $(x, y) . (x, y) .$ We can write the slope of this line and then change it to a different form.

$\begin{matrix} m & = & \frac{y - y_{1}}{x - x_{1}} \\ Multiply both sides of the equation by x - x_{1} . & m (x - x_{1}) & = & (\frac{y - y_{1}}{x - x_{1}}) (x - x_{1}) \\ Simplify. & m (x - x_{1}) & = & y - y_{1} \\ Rewrite the equation with the y terms on the left. & y - y_{1} & = & m (x - x_{1}) \end{matrix} \begin{matrix} m = \frac{y - y_{1}}{x - x_{1}} \\ Multiply both sides of the equation by & m (x - x_{1}) = (\frac{y - y_{1}}{x - x_{1}}) (x - x_{1}) \\ Simplify. & m (x - x_{1}) = y - y_{1} \\ Rewrite the equation with the & y - y_{1} = m (x - x_{1}) \end{matrix}$

This format is called the point-slope form of an equation of a line.

Point-slope Form of an Equation of a Line

The point-slope form of an equation of a line with slope m and containing the point $(x_{1}, y_{1}) (x_{1}, y_{1})$ is:

y - y_{1} = m (x - x_{1}) y - y_{1} = m (x - x_{1})

We can use the point-slope form of an equation to find an equation of a line when we know the slope and at least one point. Then, we will rewrite the equation in slope-intercept form. Most applications of linear equations use the the slope-intercept form.

Example 3.26

How to Find an Equation of a Line Given a Point and the Slope

Find an equation of a line with slope $m = - \frac{1}{3} m = - \frac{1}{3}$ that contains the point $(6, −4) . (6, −4) .$ Write the equation in slope-intercept form.

Show solution

Solution

Step 1 is to identify the slope. The slope is given. m equals negative 1 divided by 3.

Step 2 is to identify the point. The point is given. x 1 is 6 and y 1 is negative 4.

Step 3 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus negative 4 equals negative 1 divided by 3 times the quantity x minus 6 in parentheses. This simplifies to y plus 4 equals negative 1 divided by 3 times x plus 2.

Step 4 is to write the equation in slope-intercept form. y equals negative 1 divided by 3 times x minus 2.

Try It 3.51

Find the equation of a line with slope $m = - \frac{2}{5} m = - \frac{2}{5}$ and containing the point $(10, −5) . (10, −5) .$

Try It 3.52

Find the equation of a line with slope $m = - \frac{3}{4}, m = - \frac{3}{4},$ and containing the point $(4, −7) . (4, −7) .$

We list the steps for easy reference.

How To

To find an equation of a line given the slope and a point.

Step 1. Identify the slope.
Step 2. Identify the point.
Step 3. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$
Step 4. Write the equation in slope-intercept form.

Example 3.27

Find an equation of a horizontal line that contains the point $(−2, −6) . (−2, −6) .$ Write the equation in slope-intercept form.

Show solution

Solution

Every horizontal line has slope 0. We can substitute the slope and points into the point-slope form, $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$

Identify the slope.
Identify the point.
Substitute the values into $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$

Simplify.

Write in slope-intercept form.	It is in y-form, but could be written $y = 0 x - 6 . y = 0 x - 6 .$

Did we end up with the form of a horizontal line, $y = a ? y = a ?$

Try It 3.53

Find the equation of a horizontal line containing the point $(−3, 8) . (−3, 8) .$

Try It 3.54

Find the equation of a horizontal line containing the point $(−1, 4) . (−1, 4) .$

Find an Equation of the Line Given Two Points

When real-world data is collected, a linear model can be created from two data points. In the next example we’ll see how to find an equation of a line when just two points are given.

So far, we have two options for finding an equation of a line: slope-intercept or point-slope. When we start with two points, it makes more sense to use the point-slope form.

But then we need the slope. Can we find the slope with just two points? Yes. Then, once we have the slope, we can use it and one of the given points to find the equation.

Example 3.28

How to Find the Equation of a Line Given Two Points

Find an equation of a line that contains the points $(−3, −1) (−3, −1)$ and $(2, −2) (2, −2)$ Write the equation in slope-intercept form.

Show solution

Solution

Step 1 is to find the slope using the given points. Find the slope of the line through (negative 3, negative 1) and (2, and negative 2). m equals the quotient of y 2 minus y 1 in parentheses and x 2 minus x 1 in parentheses. m equals the quotient of negative 2 minus negative 1 in parentheses and 2 minus negative 3 in parentheses. m equals negative 1 divided by 5.

Step 2 is to identify the point. Choose either point. x 1 is 2 and y 1 is negative 2.

Step 3 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus negative 2 equals negative 1 divided by 5 times the quantity x minus 2 in parentheses. This simplifies to y plus 2 equals negative 1 divided by 5 times x plus 2 divided by 5.

Step 4 is to write the equation in slope-intercept form. y equals negative 1 divided by 5 times x minus 8 divided by 5.

Try It 3.55

Find the equation of a line containing the points $(−2, −4) (−2, −4)$ and $(1, −3) . (1, −3) .$

Try It 3.56

Find the equation of a line containing the points $(−4, −3) (−4, −3)$ and $(1, −5) . (1, −5) .$

The steps are summarized here.

How To

To find an equation of a line given two points.

Step 1. Find the slope using the given points. $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Step 2. Choose one point.
Step 3. Substitute the values into the point-slope form: $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$
Step 4. Write the equation in slope-intercept form.

Example 3.29

Find an equation of a line that contains the points $(−3, 5) (−3, 5)$ and $(−3, 4) . (−3, 4) .$ Write the equation in slope-intercept form.

Show solution

Solution

Again, the first step will be to find the slope.

$\begin{matrix} \begin{matrix} Find the slope of the line through (−3, 5) and (−3, 4) . \end{matrix} & \begin{matrix} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{4 - 5}{−3 - (−3)} \\ m & = & \frac{−1}{0} \end{matrix} \\ The slope is undefined. \end{matrix} \begin{matrix} \begin{matrix} Find the slope of the line through \end{matrix} \\ The slope is undefined. \end{matrix}$

This tells us it is a vertical line. Both of our points have an x-coordinate of $−2 . −2 .$ So our equation of the line is $x = −2 . x = −2 .$ Since there is no y, we cannot write it in slope-intercept form.

You may want to sketch a graph using the two given points. Does your graph agree with our conclusion that this is a vertical line?

Try It 3.57

Find the equation of a line containing the points $(5, 1) (5, 1)$ and $(5, −4) . (5, −4) .$

Try It 3.58

Find the equaion of a line containing the points $(−4, 4) (−4, 4)$ and $(−4, 3) . (−4, 3) .$

We have seen that we can use either the slope-intercept form or the point-slope form to find an equation of a line. Which form we use will depend on the information we are given.

To Write an Equation of a Line
If given:	Use:	Form:
Slope and y-intercept	slope-intercept	$y = m x + b y = m x + b$
Slope and a point	point-slope	$y - y_{1} = m (x - x_{1}) y - y_{1} = m (x - x_{1})$
Two points	point-slope	$y - y_{1} = m (x - x_{1}) y - y_{1} = m (x - x_{1})$

Find an Equation of a Line Parallel to a Given Line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point-slope equation.

First, let’s look at this graphically.

This graph shows $y = 2 x - 3 . y = 2 x - 3 .$ We want to graph a line parallel to this line and passing through the point $(−2, 1) . (−2, 1) .$

This figure has a graph of a straight line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The point (negative 2, 1) is plotted. The line does not go through the point (negative 2, 1).

We know that parallel lines have the same slope. So the second line will have the same slope as $y = 2 x - 3 . y = 2 x - 3 .$ That slope is $m_{∥} = 2 . m_{∥} = 2 .$ We’ll use the notation $m_{∥} m_{∥}$ to represent the slope of a line parallel to a line with slope m. (Notice that the subscript || looks like two parallel lines.)

The second line will pass through $(−2, 1) (−2, 1)$ and have $m = 2 . m = 2 .$

To graph the line, we start at $(−2, 1) (−2, 1)$ and count out the rise and run.

With $m = 2 m = 2$ (or $m = \frac{2}{1} m = \frac{2}{1}$ ), we count out the rise 2 and the run 1. We draw the line, as shown in the graph.

This figure has a graph of a two straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The first line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The points (negative 2, 1) and (negative 1, 3) are plotted. The second line goes through the points (negative 2, 1) and (negative 1, 3).

Do the lines appear parallel? Does the second line pass through $(−2, 1) ? (−2, 1) ?$

We were asked to graph the line, 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. Here we know one point and can find the slope. So we will use the point-slope form.

Example 3.30

How to Find the Equation of a Line Parallel to a Given Line and a Point

Find an equation of a line parallel to $y = 2 x - 3 y = 2 x - 3$ that contains the point $(−2, 1) . (−2, 1) .$ Write the equation in slope-intercept form.

Show solution

Solution

Step 1 is to find the slope of the given line. The line is in slope-intercept form, y equals 2 x minus 3. m equals 2.

Step 2 is to find the slope of the parallel line. Parallel lines have the same slope. m equals 2.

Step 4 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus 1 equals 2 times the quantity x minus negative 2 in parentheses. This simplifies to y minus 1 equals 2 x plus 4.

Step 5 is to write the equation in slope-intercept form. y equals 2 x plus 5.

Look at graph with the parallel lines shown previously. Does this equation make sense? What is the y-intercept of the line? What is the slope?

Try It 3.59

Find an equation of a line parallel to the line $y = 3 x + 1 y = 3 x + 1$ that contains the point $(4, 2) . (4, 2) .$ Write the equation in slope-intercept form.

Try It 3.60

Find an equation of a line parallel to the line $y = \frac{1}{2} x - 3 y = \frac{1}{2} x - 3$ that contains the point $(6, 4) . (6, 4) .$

Write the equation in slope-intercept form.

How To

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

Step 1. Find the slope of the given line.
Step 2. Find the slope of the parallel line.
Step 3. Identify the point.
Step 4. Substitute the values into the point-slope form: $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$
Step 5. Write the equation in slope-intercept form.

Find 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.

This graph shows $y = 2 x - 3 . y = 2 x - 3 .$ Now, we want to graph a line perpendicular to this line and passing through $(−2, 1) . (−2, 1) .$

We know that perpendicular lines have slopes that are negative reciprocals.

We’ll use the notation $m_{⊥} m_{⊥}$ 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{matrix} y = 2 x - 3 & perpendicular line \\ m = 2 & m_{⊥} = - \frac{1}{2} \end{matrix} \begin{matrix} y = 2 x - 3 & perpendicular line \\ m = 2 & m_{⊥} = - \frac{1}{2} \end{matrix}

We now know the perpendicular line will pass through $(−2, 1) (−2, 1)$ with $m_{⊥} = - \frac{1}{2} . m_{⊥} = - \frac{1}{2} .$

To graph the line, we will start at $(−2, 1) (−2, 1)$ and count out the rise $−1 −1$ and the run 2. Then we draw the line.

This figure has a graph of two perpendicular straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The first line goes through the points (0, negative 3), (1, negative 1), and (2, 1). The points (negative 2, 1) and (0, 0) are plotted. A right triangle is drawn connecting the points (negative 2, 1), (negative 2, 0), and (0, 0). The second line goes through the points (negative 2, 1) and (0, 0).

Do the lines appear perpendicular? Does the second line pass through $(−2, 1) ? (−2, 1) ?$

We were asked to graph the line, 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 3.31

How to Find the Equation of a Line Perpendicular to a Given Line and a Point

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

Show solution

Solution

Step 2 is to find the slope of the perpendicular line. The slopes of perpendicular lines are negative reciprocals. m equals negative 1 divided by 2

Step 3 is to identify the point. The given point is (negative 2, 1). x 1 is negative 2 and y 1 is 1.

Step 4 is to substitute the values into the point-slope form y minus y 1 equals m times the quantity x minus x 1 in parentheses. y minus 1 equals negative 1 divided by 2 times the quantity x minus negative 2 in parentheses. This simplifies to y minus 1 equals negative 1 divided by 2 times the quantity x plus 2 in parentheses. This further simplifies to y minus 1 equals negative 1 divided by 2 times x minus 1.

Step 5 is to write the equation in slope-intercept form. y equals negative 1 divided by 2 times x.

Try It 3.61

Find an equation of a line perpendicular to the line $y = 3 x + 1 y = 3 x + 1$ that contains the point $(4, 2) . (4, 2) .$ Write the equation in slope-intercept form.

Try It 3.62

Find an equation of a line perpendicular to the line $y = \frac{1}{2} x - 3 y = \frac{1}{2} x - 3$ that contains the point $(6, 4) . (6, 4) .$ Write the equation in slope-intercept form.

How To

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

Step 1. Find the slope of the given line.
Step 2. Find the slope of the perpendicular line.
Step 3. Identify the point.
Step 4. Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1}) . y - y_{1} = m (x - x_{1}) .$
Step 5. Write the equation in slope-intercept form.

Example 3.32

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

Show solution

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 $x = 5 . x = 5 .$ This line is vertical, so its perpendicular will be horizontal. This tells us the $m_{⊥} = 0 . m_{⊥} = 0 .$

$\begin{matrix} Identify the point. & (3, −2) \\ \begin{matrix} Identify the slope of the perpendicular line. \\ Substitute the values into y - y_{1} = m (x - x_{1}) . \\ Simplify. \end{matrix} & \begin{matrix} m_{⊥} & = & 0 \\ y - y_{1} & = & m (x - x_{1}) \\ y - (−2) & = & 0 (x - 3) \\ y + 2 & = & 0 \\ y & = & −2 \end{matrix} \end{matrix} \begin{matrix} Identify the point. & (3, −2) \\ \begin{matrix} Identify the slope of the perpendicular line. \\ Substitute the values into \\ Simplify. \end{matrix} & \begin{matrix} m_{⊥} & = & 0 \\ y - y_{1} & = & m (x - x_{1}) \\ y - (−2) & = & 0 (x - 3) \\ y + 2 & = & 0 \\ y & = & −2 \end{matrix} \end{matrix}$

Sketch the graph of both lines. On your graph, do the lines appear to be perpendicular?

Try It 3.63

Find an equation of a line that is perpendicular to the line $x = 4 x = 4$ that contains the point $(4, −5) . (4, −5) .$ . Write the equation in slope-intercept form.

Try It 3.64

Find an equation of a line that is perpendicular to the line $x = 2 x = 2$ that contains the point $(2, −1) . (2, −1) .$ Write the equation in slope-intercept form.

In Example 3.32, 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 $x = 5 x = 5$ that contains the point $(3, −2) . (3, −2) .$ This graph shows us the line $x = 5 x = 5$ and the point $(3, −2) . (3, −2) .$

This figure has a graph of a straight vertical line and a point on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The line goes through the points (5, 0), (5, 1), and (5, 2). The point (3, negative 2) is plotted. The line does not go through the point (3, negative 2).

We know every line perpendicular to a vertical line is horizontal, so we will sketch the horizontal line through $(3, −2) . (3, −2) .$

This figure has a graph of a straight vertical line and a straight horizontal line on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. The vertical line goes through the points (5, 0), (5, 1), and (5, 2). The horizontal line goes through the points (negative 2, negative 2), (0, negative 2), (3, negative 2), and (6, negative 2).

Do the lines appear perpendicular?

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

Example 3.33

Find an equation of a line that is perpendicular to $y = −3 y = −3$ that contains the point $(−3, 5) . (−3, 5) .$ Write the equation in slope-intercept form.

Show solution

Solution

The line $y = −3 y = −3$ is a horizontal line. Any line perpendicular to it must be vertical, in the form $x = a . x = a .$ Since the perpendicular line is vertical and passes through $(−3, 5), (−3, 5),$ every point on it has an x-coordinate of $−3 . −3 .$ The equation of the perpendicular line is $x = −3 x = −3$

You may want to sketch the lines. Do they appear perpendicular?

Try It 3.65

Find an equation of a line that is perpendicular to the line $y = 1 y = 1$ that contains the point $(−5, 1) . (−5, 1) .$ Write the equation in slope-intercept form.

Try It 3.66

Find an equation of a line that is perpendicular to the line $y = −5 y = −5$ that contains the point $(−4, −5) . (−4, −5) .$ Write the equation in slope-intercept form.

Media

Access these online resources for additional instruction and practice with finding the equation of a line.

Practice Makes Perfect

Find an Equation of the Line Given the Slope and y-Intercept

In the following exercises, find the equation of a line with given slope and y-intercept. Write the equation in slope-intercept form.

155.

slope 3 and
$y y$ -intercept $(0, 5) (0, 5)$

156.

slope 8 and
y-intercept $(0, −6) (0, −6)$

157.

slope $−3 −3$ and
$y y$ -intercept $(0, −1) (0, −1)$

158.

slope $−1 −1$ and
$y y$ -intercept $(0, 3) (0, 3)$

159.

slope $\frac{1}{5} \frac{1}{5}$ and
$y y$ -intercept $(0, −5) (0, −5)$

160.

slope $- \frac{3}{4} - \frac{3}{4}$ and
$y y$ -intercept $(0, −2) (0, −2)$

161.

slope 0 and
$y y$ -intercept $(0, −1) (0, −1)$

162.

slope $−4 −4$ and
$y y$ -intercept $(0, 0) (0, 0)$

In the following exercises, find the equation of the line shown in each graph. Write the equation in slope-intercept form.

163.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 5), (1, negative 2), and (2, 1).

164.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 4), (1, 2), and (2, 0).

165.

166.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 2), (4, 5), and (8, 8).

167.

This figure has a graph of a straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, 3), (3, negative 1), and (6, negative 5).

168.

169.

This figure has a graph of a horizontal straight line on the x y-coordinate plane. The x and y-axes run from negative 10 to 10. The line goes through the points (0, negative 2), (1, negative 2), and (2, negative 2).

170.

Find an Equation of the Line Given the Slope and a Point

In the following exercises, find the equation of a line with given slope and containing the given point. Write the equation in slope-intercept form.

171.

$m = \frac{5}{8}, m = \frac{5}{8},$ point $(8, 3) (8, 3)$

172.

$m = \frac{5}{6}, m = \frac{5}{6},$ point $(6, 7) (6, 7)$

173.

$m = - \frac{3}{5}, m = - \frac{3}{5},$ point $(10, −5) (10, −5)$

174.

$m = - \frac{3}{4}, m = - \frac{3}{4},$ point $(8, −5) (8, −5)$

175.

$m = - \frac{3}{2}, m = - \frac{3}{2},$ point $(−4, −3) (−4, −3)$

176.

$m = - \frac{5}{2}, m = - \frac{5}{2},$ point $(−8, −2) (−8, −2)$

177.

$m = −7, m = −7,$ point $(−1, −3) (−1, −3)$

178.

$m = −4, m = −4,$ point $(−2, −3) (−2, −3)$

179.

Horizontal line containing $(−2, 5) (−2, 5)$

180.

Horizontal line containing $(−2, −3) (−2, −3)$

181.

Horizontal line containing $(−1, −7) (−1, −7)$

182.

Horizontal line containing $(4, −8) (4, −8)$

Find an Equation of the Line Given Two Points

In the following exercises, find the equation of a line containing the given points. Write the equation in slope-intercept form.

183.

$(2, 6) (2, 6)$ and $(5, 3) (5, 3)$

184.

$(4, 3) (4, 3)$ and $(8, 1) (8, 1)$

185.

$(−3, −4) (−3, −4)$ and $(5 - 2) . (5 - 2) .$

186.

$(−5, −3) (−5, −3)$ and $(4, −6) . (4, −6) .$

187.

$(−1, 3) (−1, 3)$ and $(−6, −7) . (−6, −7) .$

188.

$(−2, 8) (−2, 8)$ and $(−4, −6) . (−4, −6) .$

189.

$(0, 4) (0, 4)$ and $(2, −3) . (2, −3) .$

190.

$(0, −2) (0, −2)$ and $(−5, −3) . (−5, −3) .$

191.

$(7, 2) (7, 2)$ and $(7, −2) . (7, −2) .$

192.

$(−2, 1) (−2, 1)$ and $(−2, −4) . (−2, −4) .$

193.

$(3, −4) (3, −4)$ and $(5, −4) . (5, −4) .$

194.

$(−6, −3) (−6, −3)$ and $(−1, −3) (−1, −3)$

Find an Equation of a Line Parallel to a Given Line

In the following exercises, find an equation of a line parallel to the given line and contains the given point. Write the equation in slope-intercept form.

195.

line $y = 4 x + 2, y = 4 x + 2,$
point $(1, 2) (1, 2)$

196.

line $y = −3 x - 1, y = −3 x - 1,$
point $(2, −3) . (2, −3) .$

197.

line $2 x - y = 6, 2 x - y = 6,$
point $(3, 0) . (3, 0) .$

198.

line $2 x + 3 y = 6, 2 x + 3 y = 6,$
point $(0, 5) . (0, 5) .$

199.

line $x = −4, x = −4,$
point $(−3, −5) . (−3, −5) .$

200.

line $x - 2 = 0, x - 2 = 0,$
point $(1, −2) (1, −2)$

201.

line $y = 5, y = 5,$
point $(2, −2) (2, −2)$

202.

line $y + 2 = 0, y + 2 = 0,$
point $(3, −3) (3, −3)$

Find an Equation of a Line Perpendicular to a Given Line

In the following exercises, find an equation of a line perpendicular to the given line and contains the given point. Write the equation in slope-intercept form.

203.

line $y = −2 x + 3, y = −2 x + 3,$
point $(2, 2) (2, 2)$

204.

line $y = − x + 5, y = − x + 5,$
point $(3, 3) (3, 3)$

205.

line $y = \frac{3}{4} x - 2, y = \frac{3}{4} x - 2,$
point $(−3, 4) (−3, 4)$

206.

line $y = \frac{2}{3} x - 4, y = \frac{2}{3} x - 4,$
point $(2, −4) (2, −4)$

207.

line $2 x - 3 y = 8, 2 x - 3 y = 8,$
point $(4, −1) (4, −1)$

208.

line $4 x - 3 y = 5, 4 x - 3 y = 5,$
point $(−3, 2) (−3, 2)$

209.

line $2 x + 5 y = 6, 2 x + 5 y = 6,$
point $(0, 0) (0, 0)$

210.

line $4 x + 5 y = −3, 4 x + 5 y = −3,$
point $(0, 0) (0, 0)$

211.

line $x = 3, x = 3,$
point $(3, 4) (3, 4)$

212.

line $x = −5, x = −5,$
point $(1, −2) (1, −2)$

213.

line $x = 7, x = 7,$
point $(−3, −4) (−3, −4)$

214.

line $x = −1, x = −1,$
point $(−4, 0) (−4, 0)$

215.

line $y - 3 = 0, y - 3 = 0,$
point $(−2, −4) (−2, −4)$

216.

line $y - 6 = 0, y - 6 = 0,$
point $(−5, −3) (−5, −3)$

217.

line y-axis,
point $(3, 4) (3, 4)$

218.

line y-axis,
point $(2, 1) (2, 1)$

Mixed Practice

In the following exercises, find the equation of each line. Write the equation in slope-intercept form.

219.

Containing the points $(4, 3) (4, 3)$ and $(8, 1) (8, 1)$

220.

Containing the points $(−2, 0) (−2, 0)$ and $(−3, −2) (−3, −2)$

221.

$m = \frac{1}{6}, m = \frac{1}{6},$ containing point $(6, 1) (6, 1)$

222.

$m = \frac{5}{6}, m = \frac{5}{6},$ containing point $(6, 7) (6, 7)$

223.

Parallel to the line $4 x + 3 y = 6, 4 x + 3 y = 6,$ containing point $(0, −3) (0, −3)$

224.

Parallel to the line $2 x + 3 y = 6, 2 x + 3 y = 6,$ containing point $(0, 5) (0, 5)$

225.

$m = - \frac{3}{4}, m = - \frac{3}{4},$ containing point $(8, −5) (8, −5)$

226.

$m = - \frac{3}{5}, m = - \frac{3}{5},$ containing point $(10, −5) (10, −5)$

227.

Perpendicular to the line $y - 1 = 0, y - 1 = 0,$ point $(−2, 6) (−2, 6)$

228.

Perpendicular to the line y-axis, point $(−6, 2) (−6, 2)$

229.

Parallel to the line $x = −3, x = −3,$ containing point $(−2, −1) (−2, −1)$

230.

Parallel to the line $x = −4, x = −4,$ containing point $(−3, −5) (−3, −5)$

231.

Containing the points $(−3, −4) (−3, −4)$ and $(2, −5) (2, −5)$

232.

Containing the points $(−5, −3) (−5, −3)$ and $(4, −6) (4, −6)$

233.

Perpendicular to the line $x - 2 y = 5, x - 2 y = 5,$ point $(−2, 2) (−2, 2)$

234.

Perpendicular to the line $4 x + 3 y = 1, 4 x + 3 y = 1,$ point $(0, 0) (0, 0)$

Writing Exercises

235.

Why are all horizontal lines parallel?

236.

Explain in your own words why the slopes of two perpendicular lines must have opposite signs.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

The figure shows a table with six rows and four columns. The first row is a header row and it labels each column. The first column header is “I can…”, the second is “confidently”, the third is “with some help”, “no minus I don’t get it!”. Under the first column are the phrases “find the equation of the line given the slope and y-intercept”, “find an equation of the line given the slope and a point”, “find an equation of the line given two points”, “find an equation of a line parallel to a given line”, and “find an equation of a line perpendicular to a given line”. Under the second, third, fourth columns are blank spaces where the learner can check what level of mastery they have achieved.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?