Contents

Use the Slope–Intercept Form of an Equation of a Line

Learning Objectives

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

Recognize the relation between the graph and the slope–intercept form of an equation of a line
Identify the slope and y-intercept form of an equation of a line
Graph a line using its slope and intercept
Choose the most convenient method to graph a line
Graph and interpret applications of slope–intercept
Use slopes to identify parallel lines
Use slopes to identify perpendicular lines

Be Prepared 4.5

Before you get started, take this readiness quiz.

Add: $\frac{x}{4} + \frac{1}{4} . \frac{x}{4} + \frac{1}{4} .$
If you missed this problem, review Example 1.77.
Find the reciprocal of $\frac{3}{7} . \frac{3}{7} .$
If you missed this problem, review Example 1.70.
Solve $2 x - 3 y = 12 for y 2 x - 3 y = 12$ .
If you missed this problem, review Example 2.63.

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using the point–slope method. Once we see how an equation in slope–intercept form and its graph are related, we’ll have one more method we can use to graph lines.

In Graph Linear Equations in Two Variables, we graphed the line of the equation $y = \frac{1}{2} x + 3 y = \frac{1}{2} x + 3$ by plotting points. See Figure 4.24. Let’s find the slope of this line.

This figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line is labeled with the equation y equals one half x, plus 3. The points (0, 3), (2, 4) and (4, 5) are labeled also. A red vertical line begins at the point (2, 4) and ends one unit above the point. It is labeled “Rise equals 1”. A red horizontal line begins at the end of the vertical line and ends at the point (4, 5). It is labeled “Run equals 2. The red lines create a right triangle with the line y equals one half x, plus 3 as the hypotenuse.

Figure 4.24

The red lines show us the rise is 1 and the run is 2. Substituting into the slope formula:

\begin{matrix} m & = & \frac{rise}{run} \\ m & = & \frac{1}{2} \end{matrix} \begin{matrix} m & = & \frac{rise}{run} \\ m & = & \frac{1}{2} \end{matrix}

What is the y-intercept of the line? The y-intercept is where the line crosses the y-axis, so y-intercept is $(0, 3) (0, 3)$ . The equation of this line is:

$The figure shows the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue.$

Notice, the line has:

When a linear equation is solved for $y y$ , the coefficient of the $x x$ term is the slope and the constant term is the y-coordinate of the y-intercept. We say that the equation $y = \frac{1}{2} x + 3 y = \frac{1}{2} x + 3$ is in slope–intercept form.

$The figure shows the statement “m equals one half; y-intercept is (0, 3). The slope, one half, is colored red and the number 3 in the y-intercept is colored blue. Below that statement is the equation y equals one half x, plus 3. The fraction one half is colored red and the number 3 is colored blue. Below the equation is another equation y equals m x, plus b. The variable m is colored red and the variable b is colored blue.$

Slope-Intercept Form of an Equation of a Line

The slope–intercept form of an equation of a line with slope $m m$ and y-intercept, $(0, b) (0, b)$ is,

y = m x + b y = m x + b

Sometimes the slope–intercept form is called the “y-form.”

Example 4.40

Use the graph to find the slope and y-intercept of the line, $y = 2 x + 1 y = 2 x + 1$ .

Compare these values to the equation $y = m x + b y = m x + b$ .

Show solution

Solution

To find the slope of the line, we need to choose two points on the line. We’ll use the points $(0, 1) (0, 1)$ and $(1, 3) (1, 3)$ .


Find the rise and run.


Find the y-intercept of the line.	The y-intercept is the point (0, 1).

The slope is the same as the coefficient of $x x$ and the y-coordinate of the y-intercept is the same as the constant term.

Try It 4.79

Use the graph to find the slope and y-intercept of the line $y = \frac{2}{3} x - 1 y = \frac{2}{3} x - 1$ . Compare these values to the equation $y = m x + b y = m x + b$ .

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The line goes through the points (0, negative 1) and (6, 3).

Try It 4.80

Use the graph to find the slope and y-intercept of the line $y = \frac{1}{2} x + 3 y = \frac{1}{2} x + 3$ . Compare these values to the equation $y = m x + b y = m x + b$ .

Identify the Slope and y-Intercept From an Equation of a Line

In Understand Slope of a Line, we graphed a line using the slope and a point. When we are given an equation in slope–intercept form, we can use the y-intercept as the point, and then count out the slope from there. Let’s practice finding the values of the slope and y-intercept from the equation of a line.

Example 4.41

Identify the slope and y-intercept of the line with equation $y = −3 x + 5 y = −3 x + 5$ .

Show solution

Solution

We compare our equation to the slope–intercept form of the equation.


Write the equation of the line.
Identify the slope.
Identify the y-intercept.

Try It 4.81

Identify the slope and y-intercept of the line $y = \frac{2}{5} x - 1 y = \frac{2}{5} x - 1$ .

Try It 4.82

Identify the slope and y-intercept of the line $y = - \frac{4}{3} x + 1 y = - \frac{4}{3} x + 1$ .

When an equation of a line is not given in slope–intercept form, our first step will be to solve the equation for $y y$ .

Example 4.42

Identify the slope and y-intercept of the line with equation $x + 2 y = 6 x + 2 y = 6$ .

Show solution

Solution

This equation is not in slope–intercept form. In order to compare it to the slope–intercept form we must first solve the equation for $y y$ .

Solve for y.	$x + 2 y = 6 x + 2 y = 6$
Subtract x from each side.
Divide both sides by 2.
Simplify.
$(Remember: \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}) (Remember:$
Simplify.
Write the slope–intercept form of the equation of the line.
Write the equation of the line.
Identify the slope.
Identify the y-intercept.

Try It 4.83

Identify the slope and y-intercept of the line $x + 4 y = 8 x + 4 y = 8$ .

Try It 4.84

Identify the slope and y-intercept of the line $3 x + 2 y = 12 3 x + 2 y = 12$ .

Graph a Line Using its Slope and Intercept

Now that we know how to find the slope and y-intercept of a line from its equation, we can graph the line by plotting the y-intercept and then using the slope to find another point.

Example 4.43

How to Graph a Line Using its Slope and Intercept

Graph the line of the equation $y = 4 x - 2 y = 4 x - 2$ using its slope and y-intercept.

Show solution

Solution

The figure shows the steps to graph the equation y equals 4x minus 2. Step 1 is to find the slope intercept form of the equation. The equation is already in slope intercept form.

Step 2 is to identify the slope and y-intercept. Use the equation y equals m x, plus b. The equation y equals m x, plus b is shown with the variable m colored red and the variable b colored blue. Below that is the equation y equals 4 x, plus -2. The number 4 is colored red and -2 is colored blue. From this equation we can see that m equals 4 and b equals -2 so the slope is 4 and the y-intercept is the point (0, negative 2).

Step 3 is to plot the y-intercept. An x y-coordinate plane is shown with the x-axis of the plane running from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. The point (0, negative 2) is plotted.

Step 4 is to use the slope formula m equals rise over run to identify the rise and the run. Since m equals 4, rise over run equals 4 over 1. From this we can determine that the rise is 4 and the run is 1.

Step 5 is to start at they-intercept, count out the rise and run to mark the second point. So start at the point (0, negative 2) and count the rise and the run. The rise is up 4 and the run is right 1. On the x y-coordinate plane is a red vertical line starts at the point (0, negative 2) and rises 4 units at its end a red horizontal line runs 1 unit to end at the point (1, 2). The point (1, 2) is plotted.

Step 6 is to connect the points with a line. On the x y-coordinate plane the points (0, negative 2) and (1, 2) are plotted and a line runs through the two points. The line is the graph of y equals 4 x, minus 2.

Try It 4.85

Graph the line of the equation $y = 4 x + 1 y = 4 x + 1$ using its slope and y-intercept.

Try It 4.86

Graph the line of the equation $y = 2 x - 3 y = 2 x - 3$ using its slope and y-intercept.

How To

Graph a line using its slope and y-intercept.

Step 1. Find the slope-intercept form of the equation of the line.
Step 2. Identify the slope and y-intercept.
Step 3. Plot the y-intercept.
Step 4. Use the slope formula $m = \frac{rise}{run} m = \frac{rise}{run}$ to identify the rise and the run.
Step 5. Starting at the y-intercept, count out the rise and run to mark the second point.
Step 6. Connect the points with a line.

Example 4.44

Graph the line of the equation $y = − x + 4 y = − x + 4$ using its slope and y-intercept.

Show solution

Solution

	$y = m x + b y = m x + b$
The equation is in slope–intercept form.	$y = − x + 4 y = − x + 4$
Identify the slope and y-intercept.	$m = −1 m = −1$
	y-intercept is (0, 4)
Plot the y-intercept.	See graph below.
Identify the rise and the run.	$m = \frac{−1}{1} m = \frac{−1}{1}$
Count out the rise and run to mark the second point.	rise −1, run 1
Draw the line.
To check your work, you can find another point on the line and make sure it is a solution of the equation. In the graph we see the line goes through (4, 0).
Check. $\begin{matrix} y & = & − x + 4 \\ 0 & \overset{?}{=} & - 4 + 4 \\ 0 & = & 0 ✓ \end{matrix} \begin{matrix} y & = & − x + 4 & 0 & \overset{?}{=} & - 4 + 4 & 0 & = & 0 ✓ \end{matrix}$

Try It 4.87

Graph the line of the equation $y = − x - 3 y = − x - 3$ using its slope and y-intercept.

Try It 4.88

Graph the line of the equation $y = − x - 1 y = − x - 1$ using its slope and y-intercept.

Example 4.45

Graph the line of the equation $y = - \frac{2}{3} x - 3 y = - \frac{2}{3} x - 3$ using its slope and y-intercept.

Show solution

Solution

	$y = m x + b y = m x + b$
The equation is in slope–intercept form.	$y = - \frac{2}{3} x - 3 y = - \frac{2}{3} x - 3$
Identify the slope and y-intercept.	$m = - \frac{2}{3} m = - \frac{2}{3}$ ; y-intercept is (0, −3)
Plot the y-intercept.	See graph below.
Identify the rise and the run.
Count out the rise and run to mark the second point.
Draw the line.

Try It 4.89

Graph the line of the equation $y = - \frac{5}{2} x + 1 y = - \frac{5}{2} x + 1$ using its slope and y-intercept.

Try It 4.90

Graph the line of the equation $y = - \frac{3}{4} x - 2 y = - \frac{3}{4} x - 2$ using its slope and y-intercept.

Example 4.46

Graph the line of the equation $4 x - 3 y = 12 4 x - 3 y = 12$ using its slope and y-intercept.

Show solution

Solution

	$4 x - 3 y = 12 4 x - 3 y = 12$
Find the slope–intercept form of the equation.	$−3 y = −4 x + 12$
	$- \frac{3 y}{3} = \frac{−4 x + 12}{−3}$
The equation is now in slope–intercept form.	$y = \frac{4}{3} x - 4$
Identify the slope and y-intercept.	$m = \frac{4}{3}$
	y-intercept is (0, −4)
Plot the y-intercept.	See graph below.
Identify the rise and the run; count out the rise and run to mark the second point.
Draw the line.

Try It 4.91

Graph the line of the equation $2 x - y = 6 2 x - y = 6$ using its slope and y-intercept.

Try It 4.92

Graph the line of the equation $3 x - 2 y = 8 3 x - 2 y = 8$ using its slope and y-intercept.

We have used a grid with $x x$ and $y y$ both going from about $−10 −10$ to 10 for all the equations we’ve graphed so far. Not all linear equations can be graphed on this small grid. Often, especially in applications with real-world data, we’ll need to extend the axes to bigger positive or smaller negative numbers.

Example 4.47

Graph the line of the equation $y = 0.2 x + 45 y = 0.2 x + 45$ using its slope and y-intercept.

Show solution

Solution

We’ll use a grid with the axes going from about $−80 −80$ to 80.

	$y = m x + b y = m x + b$
The equation is in slope–intercept form.	$y = 0.2 x + 45 y = 0.2 x + 45$
Identify the slope and y-intercept.	$m = 0.2 m = 0.2$
	The y-intercept is (0, 45)
Plot the y-intercept.	See graph below.
Count out the rise and run to mark the second point. The slope is $m = 0.2 m = 0.2$ ; in fraction form this means $m = \frac{2}{10} m = \frac{2}{10}$ . Given the scale of our graph, it would be easier to use the equivalent fraction $m = \frac{10}{50} m = \frac{10}{50}$ .
Draw the line.

Try It 4.93

Graph the line of the equation $y = 0.5 x + 25 y = 0.5 x + 25$ using its slope and y-intercept.

Try It 4.94

Graph the line of the equation $y = 0.1 x - 30 y = 0.1 x - 30$ using its slope and y-intercept.

Now that we have graphed lines by using the slope and y-intercept, let’s summarize all the methods we have used to graph lines. See Figure 4.25.

The table has two rows and four columns. The first row spans all four columns and is a header row. The header is “Methods to Graph Lines”. The second row is made up of four columns. The first column is labeled “Plotting Points” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The rest of the table is blank. Below the table it reads “Find three points. Plot the points, make sure they line up, then draw the line.” The Second column is labeled “Slope–Intercept” and shows the equation y equals m x, plus b. Below the equation it reads “Find the slope and y-intercept. Start at the y-intercept, then count the slope to get a second point.” The third column is labeled “Intercepts” and shows a smaller table with four rows and two columns. The first row is a header row with the first column labeled “x” and the second labeled “y”. The second row has a 0 in the “x” column and the “y” column is blank. The second row is blank in the “x” column and has a 0 in the “y” column. The third row is blank. Below the table it reads “Find the intercepts and a third point. Plot the points, make sure they line up, then draw the line.” The fourth column is labeled “Recognize Vertical and Horizontal Lines”. Below that it reads “The equation has only one variable.” The equation x equals a is a vertical line and the equation y equals b is a horizontal line.

Figure 4.25

Choose the Most Convenient Method to Graph a Line

Now that we have seen several methods we can use to graph lines, how do we know which method to use for a given equation?

While we could plot points, use the slope–intercept form, or find the intercepts for any equation, if we recognize the most convenient way to graph a certain type of equation, our work will be easier. Generally, plotting points is not the most efficient way to graph a line. We saw better methods in sections 4.3, 4.4, and earlier in this section. Let’s look for some patterns to help determine the most convenient method to graph a line.

Here are six equations we graphed in this chapter, and the method we used to graph each of them.

\begin{matrix} Equation & Method \\ #1 & x = 2 & Vertical line \\ #2 & y = 4 & Horizontal line \\ #3 & − x + 2 y = 6 & Intercepts \\ #4 & 4 x - 3 y = 12 & Intercepts \\ #5 & y = 4 x - 2 & Slope–intercept \\ #6 & y = − x + 4 & Slope–intercept \end{matrix} \begin{matrix} Equation Method \\ #1 & x = 2 Vertical line \\ #2 & y = 4 Horizontal line \\ #3 & − x + 2 y = 6 Intercepts \\ #4 & 4 x - 3 y = 12 Intercepts \\ #5 & y = 4 x - 2 Slope–intercept \\ #6 & y = − x + 4 Slope–intercept \end{matrix}

Equations #1 and #2 each have just one variable. Remember, in equations of this form the value of that one variable is constant; it does not depend on the value of the other variable. Equations of this form have graphs that are vertical or horizontal lines.

In equations #3 and #4, both $x x$ and $y y$ are on the same side of the equation. These two equations are of the form $A x + B y = C A x + B y = C$ . We substituted $y = 0 y = 0$ to find the x-intercept and $x = 0 x = 0$ to find the y-intercept, and then found a third point by choosing another value for $x x$ or $y y$ .

Equations #5 and #6 are written in slope–intercept form. After identifying the slope and y-intercept from the equation we used them to graph the line.

This leads to the following strategy.

Strategy for Choosing the Most Convenient Method to Graph a Line

Consider the form of the equation.

If it only has one variable, it is a vertical or horizontal line.
- $x = a x = a$ is a vertical line passing through the x-axis at $a a$ .
- $y = b y = b$ is a horizontal line passing through the y-axis at $b b$ .
If yy is isolated on one side of the equation, in the form y=mx+by=mx+b, graph by using the slope and y-intercept.
- Identify the slope and y-intercept and then graph.
If the equation is of the form Ax+By=CAx+By=C, find the intercepts.
- Find the x- and y-intercepts, a third point, and then graph.

Example 4.48

Determine the most convenient method to graph each line.

ⓐ $y = −6 y = −6$ ⓑ $5 x - 3 y = 15 5 x - 3 y = 15$ ⓒ $x = 7 x = 7$ ⓓ $y = \frac{2}{5} x - 1 y = \frac{2}{5} x - 1$ .

Show solution

Solution

ⓐ $y = −6 y = −6$
This equation has only one variable, $y y$ . Its graph is a horizontal line crossing the y-axis at $−6 −6$ .
ⓑ $5 x - 3 y = 15 5 x - 3 y = 15$
This equation is of the form $A x + B y = C A x + B y = C$ . The easiest way to graph it will be to find the intercepts and one more point.
ⓒ $x = 7 x = 7$
There is only one variable, $x x$ . The graph is a vertical line crossing the x-axis at 7.
ⓓ $y = \frac{2}{5} x - 1 y = \frac{2}{5} x - 1$
Since this equation is in $y = m x + b y = m x + b$ form, it will be easiest to graph this line by using the slope and y-intercept.

Try It 4.95

Determine the most convenient method to graph each line: ⓐ $3 x + 2 y = 12 3 x + 2 y = 12$ ⓑ $y = 4 y = 4$ ⓒ $y = \frac{1}{5} x - 4 y = \frac{1}{5} x - 4$ ⓓ $x = −7 x = −7$ .

Try It 4.96

Determine the most convenient method to graph each line: ⓐ $x = 6 x = 6$ ⓑ $y = - \frac{3}{4} x + 1 y = - \frac{3}{4} x + 1$ ⓒ $y = −8 y = −8$ ⓓ $4 x - 3 y = −1 4 x - 3 y = −1$ .

Graph and Interpret Applications of Slope–Intercept

Many real-world applications are modeled by linear equations. We will take a look at a few applications here so you can see how equations written in slope–intercept form relate to real-world situations.

Usually when a linear equation models a real-world situation, different letters are used for the variables, instead of x and y. The variable names remind us of what quantities are being measured.

Example 4.49

The equation $F = \frac{9}{5} C + 32 F = \frac{9}{5} C + 32$ is used to convert temperatures, $C C$ , on the Celsius scale to temperatures, $F F$ , on the Fahrenheit scale.

ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0.
ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20.
ⓒ Interpret the slope and F-intercept of the equation.
ⓓ Graph the equation.

Show solution

Solution

ⓐ Find the Fahrenheit temperature for a Celsius temperature of 0. Find $F F$ when $C = 0 C = 0$ . Simplify.	$\begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (0) + 32 \\ F = 32 \end{matrix} \begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (0) + 32 \\ F = 32 \end{matrix}$
ⓑ Find the Fahrenheit temperature for a Celsius temperature of 20. Find $F F$ when $C = 20 C = 20$ . Simplify. Simplify.	$\begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (20) + 32 \\ F = 36 + 32 \\ F = 68 \end{matrix} \begin{matrix} F = \frac{9}{5} C + 32 \\ F = \frac{9}{5} (20) + 32 \\ F = 36 + 32 \\ F = 68 \end{matrix}$

Table 4.41

Even though this equation uses $F F$ and $C C$ , it is still in slope–intercept form.

This image shows three lines of equations. The first line reads y equals m x plus b. The second line reads F equals m C plus b and the third line reads F equals nine fifths times C plus 32.

The slope, $\frac{9}{5} \frac{9}{5}$ , means that the temperature Fahrenheit (F) increases 9 degrees when the temperature Celsius (C) increases 5 degrees.

The F-intercept means that when the temperature is $0 ° 0 °$ on the Celsius scale, it is $32 ° 32 °$ on the Fahrenheit scale.

ⓓ Graph the equation.

We’ll need to use a larger scale than our usual. Start at the F-intercept $(0, 32) (0, 32)$ then count out the rise of 9 and the run of 5 to get a second point. See Figure 4.26.

Figure 4.26

Try It 4.97

The equation $h = 2 s + 50 h = 2 s + 50$ is used to estimate a woman’s height in inches, h, based on her shoe size, s.

ⓐ Estimate the height of a child who wears women’s shoe size 0.
ⓑ Estimate the height of a woman with shoe size 8.
ⓒ Interpret the slope and h-intercept of the equation.
ⓓ Graph the equation.

Try It 4.98

The equation $T = \frac{1}{4} n + 40 T = \frac{1}{4} n + 40$ is used to estimate the temperature in degrees Fahrenheit, T, based on the number of cricket chirps, n, in one minute.

ⓐEstimate the temperature when there are no chirps.
ⓑ Estimate the temperature when the number of chirps in one minute is 100.
ⓒ Interpret the slope and T-intercept of the equation.
ⓓ Graph the equation.

The cost of running some types business has two components—a fixed cost and a variable cost. The fixed cost is always the same regardless of how many units are produced. This is the cost of rent, insurance, equipment, advertising, and other items that must be paid regularly. The variable cost depends on the number of units produced. It is for the material and labor needed to produce each item.

Example 4.50

Stella has a home business selling gourmet pizzas. The equation $C = 4 p + 25 C = 4 p + 25$ models the relation between her weekly cost, C, in dollars and the number of pizzas, p, that she sells.

ⓐ Find Stella’s cost for a week when she sells no pizzas.
ⓑ Find the cost for a week when she sells 15 pizzas.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Show solution

Solution

ⓐ Find Stella's cost for a week when she sells no pizzas.
Find C when $p = 0 p = 0$ .
Simplify.
	Stella's fixed cost is $25 when she sells no pizzas.
ⓑ Find the cost for a week when she sells 15 pizzas.
Find C when $p = 15 p = 15$ .
Simplify.

	Stella's costs are $85 when she sells 15 pizzas.
ⓒ Interpret the slope and C-intercept of the equation.
	The slope, 4, means that the cost increases by $4 for each pizza Stella sells. The C-intercept means that even when Stella sells no pizzas, her costs for the week are $25.
ⓓ Graph the equation. We'll need to use a larger scale than our usual. Start at the C-intercept (0, 25) then count out the rise of 4 and the run of 1 to get a second point.

Try It 4.99

Sam drives a delivery van. The equation $C = 0.5 m + 60 C = 0.5 m + 60$ models the relation between his weekly cost, C, in dollars and the number of miles, m, that he drives.

ⓐ Find Sam’s cost for a week when he drives 0 miles.
ⓑ Find the cost for a week when he drives 250 miles.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Try It 4.100

Loreen has a calligraphy business. The equation $C = 1.8 n + 35 C = 1.8 n + 35$ models the relation between her weekly cost, C, in dollars and the number of wedding invitations, n, that she writes.

ⓐFind Loreen’s cost for a week when she writes no invitations.
ⓑ Find the cost for a week when she writes 75 invitations.
ⓒ Interpret the slope and C-intercept of the equation.
ⓓ Graph the equation.

Use Slopes to Identify Parallel Lines

The slope of a line indicates how steep the line is and whether it rises or falls as we read it from left to right. Two lines that have the same slope are called parallel lines. Parallel lines never intersect.

The figure shows three pairs of lines side-by-side. The pair of lines on the left run diagonally rising from left to right. The pair run side-by-side, not crossing. The pair of lines in the middle run diagonally dropping from left to right. The pair run side-by-side, not crossing. The pair of lines on the right run diagonally also dropping from left to right, but with a lesser slope. The pair run side-by-side, not crossing.

We say this more formally in terms of the rectangular coordinate system. Two lines that have the same slope and different y-intercepts are called parallel lines. See Figure 4.27.

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (negative 5,1) and (5,5). The other line goes through the points (negative 5, negative 4) and (5,0).

Figure 4.27 Verify that both lines have the same slope,

m = \frac{2}{5} m = \frac{2}{5}

, and different y-intercepts.

What about vertical lines? The slope of a vertical line is undefined, so vertical lines don’t fit in the definition above. We say that vertical lines that have different x-intercepts are parallel. See Figure 4.28.

The figure shows two vertical lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line goes through the points (2,1) and (2,5). The other line goes through the points (5, negative 4) and (5,0).

Figure 4.28 Vertical lines with diferent x-intercepts are parallel.

Parallel Lines

Parallel lines are lines in the same plane that do not intersect.

Parallel lines have the same slope and different y-intercepts.
If $m_{1} m_{1}$ and $m_{2} m_{2}$ are the slopes of two parallel lines then $m_{1} = m_{2} m_{1} = m_{2}$ .
Parallel vertical lines have different x-intercepts.

Let’s graph the equations $y = −2 x + 3 y = −2 x + 3$ and $2 x + y = −1 2 x + y = −1$ on the same grid. The first equation is already in slope–intercept form: $y = −2 x + 3 y = −2 x + 3$ . We solve the second equation for $y y$ :

\begin{matrix} 2 x + y & = & −1 \\ y & = & −2 x - 1 \end{matrix} \begin{matrix} 2 x + y & = & −1 \\ y & = & −2 x - 1 \end{matrix}

Graph the lines.

Notice the lines look parallel. What is the slope of each line? What is the y-intercept of each line?

\begin{matrix} y & = & m x + b & y & = & m x + b \\ y & = & −2 x + 3 & y & = & −2 x - 1 \\ m & = & −2 & m & = & −2 \\ b & = & 3, (0, 3) & b & = & −1, (0, −1) \end{matrix} \begin{matrix} y & = & m x + b & y = m x + b \\ y & = & −2 x + 3 & y = −2 x - 1 \\ m & = & −2 & m = −2 \\ b & = & 3, (0, 3) & b = −1, (0, −1) \end{matrix}

The slopes of the lines are the same and the y-intercept of each line is different. So we know these lines are parallel.

Since parallel lines have the same slope and different y-intercepts, we can now just look at the slope–intercept form of the equations of lines and decide if the lines are parallel.

Example 4.51

Use slopes and y-intercepts to determine if the lines $3 x - 2 y = 6 3 x - 2 y = 6$ and $y = \frac{3}{2} x + 1 y = \frac{3}{2} x + 1$ are parallel.

Show solution

Solution

Solve the first equation for $y y$ .	$\begin{matrix} 3 x - 2 y & = & 6 \\ - 2 y & = & −3 x + 6 \\ \frac{−2 y}{−2} & = & \frac{−3 x + 6}{−2} \end{matrix} \begin{matrix} 3 x - 2 y & = & 6 \\ - 2 y & = & −3 x + 6 \\ \frac{−2 y}{−2} & = & \frac{−3 x + 6}{−2} \end{matrix}$	and	$y = \frac{3}{2} x + 1 y = \frac{3}{2} x + 1$
The equation is now in slope-intercept form.	$y = \frac{3}{2} x - 3 y = \frac{3}{2} x - 3$
The equation of the second line is already in slope-intercept form.			$y = \frac{3}{2} x + 1 y = \frac{3}{2} x + 1$
Identify the slope and $y y$ -intercept of both lines.	$\begin{matrix} y & = & \frac{3}{2} x - 3 \\ y & = & m x + b \\ m & = & \frac{3}{2} \end{matrix} \begin{matrix} y & = & \frac{3}{2} x - 3 & y & = & m x + b & m & = & \frac{3}{2} \end{matrix}$		$\begin{matrix} y = \frac{3}{2} x + 1 \\ y = m x + b \\ m = \frac{3}{2} \end{matrix} \begin{matrix} m = \frac{3}{2} \end{matrix}$
	$y -intercept is (0, −3) y -intercept is (0, −3)$		$y -intercept is (0, 1) y -intercept is (0, 1)$

Table 4.42

The lines have the same slope and different y-intercepts and so they are parallel. You may want to graph the lines to confirm whether they are parallel.

Try It 4.101

Use slopes and y-intercepts to determine if the lines $2 x + 5 y = 5 and y = - \frac{2}{5} x - 4 2 x + 5 y = 5$ are parallel.

Try It 4.102

Use slopes and y-intercepts to determine if the lines $4 x - 3 y = 6 and y = \frac{4}{3} x - 1 4 x - 3 y = 6$ are parallel.

Example 4.52

Use slopes and y-intercepts to determine if the lines $y = −4 y = −4$ and $y = 3 y = 3$ are parallel.

Show solution

Solution

	$\begin{matrix} y = −4 \\ y = 0 x - 4 \end{matrix} \begin{matrix} y = −4 \\ y = 0 x - 4 \end{matrix}$	and	$\begin{matrix} y = 3 \\ y = 0 x + 3 \end{matrix} \begin{matrix} y = 3 \\ y = 0 x + 3 \end{matrix}$
Write each equation in slope-intercept form.	$y = 0 x - 4 y = 0 x - 4$		$y = 0 x + 3 y = 0 x + 3$
Since there is no $x x$ term we write $0 x 0 x$ .	$y = m x + b y = m x + b$		$y = m x + b y = m x + b$
Identify the slope and $y y$ -intercept of both lines.	$m = 0 m = 0$		$m = 0 m = 0$
	$y -intercept is (0, 4) y -intercept is (0, 4)$		$y -intercept is (0, 3) y -intercept is (0, 3)$

Table 4.43

The lines have the same slope and different y-intercepts and so they are parallel.

There is another way you can look at this example. If you recognize right away from the equations that these are horizontal lines, you know their slopes are both 0. Since the horizontal lines cross the y-axis at $y = −4 y = −4$ and at $y = 3 y = 3$ , we know the y-intercepts are $(0, −4) (0, −4)$ and $(0, 3) (0, 3)$ . The lines have the same slope and different y-intercepts and so they are parallel.

Try It 4.103

Use slopes and y-intercepts to determine if the lines $y = 8 and y = −6 y = 8$ are parallel.

Try It 4.104

Use slopes and y-intercepts to determine if the lines $y = 1 and y = −5 y = 1$ are parallel.

Example 4.53

Use slopes and y-intercepts to determine if the lines $x = −2 x = −2$ and $x = −5 x = −5$ are parallel.

Show solution

Solution

x = −2 and x = −5 x = −2

Since there is no $y y$ , the equations cannot be put in slope–intercept form. But we recognize them as equations of vertical lines. Their x-intercepts are $−2 −2$ and $−5 −5$ . Since their x-intercepts are different, the vertical lines are parallel.

Try It 4.105

Use slopes and y-intercepts to determine if the lines $x = 1 x = 1$ and $x = −5 x = −5$ are parallel.

Try It 4.106

Use slopes and y-intercepts to determine if the lines $x = 8 x = 8$ and $x = −6 x = −6$ are parallel.

Example 4.54

Use slopes and y-intercepts to determine if the lines $y = 2 x - 3 y = 2 x - 3$ and $−6 x + 3 y = −9 −6 x + 3 y = −9$ are parallel. You may want to graph these lines, too, to see what they look like.

Show solution

Solution

	$y = 2 x - 3 y = 2 x - 3$	and	$−6 x + 3 y = −9 −6 x + 3 y = −9$
The first equation is already in slope-intercept form.	$y = 2 x - 3 y = 2 x - 3$
Solve the second equation for $y y$ .			$\begin{matrix} - 6 x + 3 y & = & −9 \\ 3 y & = & 6 x - 9 \\ \frac{3 y}{3} & = & \frac{6 x - 9}{3} \\ y & = & 2 x - 3 \end{matrix} \begin{matrix} - 6 x + 3 y & = & −9 \\ 3 y & = & 6 x - 9 \\ \frac{3 y}{3} & = & \frac{6 x - 9}{3} \\ y & = & 2 x - 3 \end{matrix}$
The second equation is now in slope-intercept form.	$y = 2 x - 3 y = 2 x - 3$
Identify the slope and $y y$ -intercept of both lines.	$\begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix} \begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix}$		$\begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix} \begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix}$
	$y -intercept is (0, −3) y -intercept is (0, −3)$		$y -intercept is (0, −3) y -intercept is (0, −3)$

Table 4.44

The lines have the same slope, but they also have the same y-intercepts. Their equations represent the same line. They are not parallel; they are the same line.

Try It 4.107

Use slopes and y-intercepts to determine if the lines $y = - \frac{1}{2} x - 1 y = - \frac{1}{2} x - 1$ and $x + 2 y = 2 x + 2 y = 2$ are parallel.

Try It 4.108

Use slopes and y-intercepts to determine if the lines $y = \frac{3}{4} x - 3 y = \frac{3}{4} x - 3$ and $3 x - 4 y = 12 3 x - 4 y = 12$ are parallel.

Use Slopes to Identify Perpendicular Lines

Let’s look at the lines whose equations are $y = \frac{1}{4} x - 1 y = \frac{1}{4} x - 1$ and $y = −4 x + 2 y = −4 x + 2$ , shown in Figure 4.29.

Figure 4.29

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular.

What do you notice about the slopes of these two lines? As we read from left to right, the line $y = \frac{1}{4} x - 1 y = \frac{1}{4} x - 1$ rises, so its slope is positive. The line $y = −4 x + 2 y = −4 x + 2$ drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, $m_{1} = \frac{1}{4} m_{1} = \frac{1}{4}$ , and the slope of the second line, $m_{2} = −4 m_{2} = −4$ , we can see that they are negative reciprocals of each other. If we multiply them, their product is $−1 . −1 .$

\begin{matrix} m_{1} \cdot m_{2} \\ \frac{1}{4} (−4) \\ - 1 \end{matrix} \begin{matrix} m_{1} \cdot m_{2} \\ \frac{1}{4} (−4) \\ - 1 \end{matrix}

This is always true for perpendicular lines and leads us to this definition.

Perpendicular Lines

Perpendicular lines are lines in the same plane that form a right angle.

If $m_{1} and m_{2} m_{1}$ are the slopes of two perpendicular lines, then:

m_{1} \cdot m_{2} = −1 and m_{1} = \frac{−1}{m_{2}} m_{1} \cdot m_{2} = −1

Vertical lines and horizontal lines are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is $−1 −1$ , the lines are perpendicular. Perpendicular lines may have the same y-intercepts.

Example 4.55

Use slopes to determine if the lines, $y = −5 x - 4 y = −5 x - 4$ and $x - 5 y = 5 x - 5 y = 5$ are perpendicular.

Show solution

Solution

The first equation is already in slope-intercept form.	$\begin{matrix} y & = & −5 x - 4 \end{matrix} \begin{matrix} y & = & −5 x - 4 \end{matrix}$
Solve the second equation for $y y$ .	$\begin{matrix} x - 5 y & = & 5 \\ - 5 y & = & − x + 5 \\ \frac{−5 y}{−5} & = & \frac{− x + 5}{−5} \\ y & = & \frac{1}{5} x - 1 \end{matrix} \begin{matrix} x - 5 y & = & 5 \\ - 5 y & = & − x + 5 \\ \frac{−5 y}{−5} & = & \frac{− x + 5}{−5} \\ y & = & \frac{1}{5} x - 1 \end{matrix}$
Identify the slope of each line.	$\begin{matrix} y & = & −5 x - 4 \\ y & = & m x + b \\ m_{1} & = & −5 \end{matrix} \begin{matrix} y & = & −5 x - 4 \\ y & = & m x + b \\ m_{1} & = & −5 \end{matrix}$	$\begin{matrix} y & = & \frac{1}{5} x - 1 \\ y & = & m x + b \\ m_{2} & = & \frac{1}{5} \end{matrix} \begin{matrix} y & = & \frac{1}{5} x - 1 \\ y & = & m x + b \\ m_{2} & = & \frac{1}{5} \end{matrix}$

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

\begin{matrix} m_{1} \cdot m_{2} \\ - 5 (\frac{1}{5}) \\ - 1 ✓ \end{matrix} \begin{matrix} m_{1} \cdot m_{2} \\ - 5 (\frac{1}{5}) & - 1 ✓ \end{matrix}

Try It 4.109

Use slopes to determine if the lines $y = −3 x + 2 y = −3 x + 2$ and $x - 3 y = 4 x - 3 y = 4$ are perpendicular.

Try It 4.110

Use slopes to determine if the lines $y = 2 x - 5 y = 2 x - 5$ and $x + 2 y = −6 x + 2 y = −6$ are perpendicular.

Example 4.56

Use slopes to determine if the lines, $7 x + 2 y = 3 7 x + 2 y = 3$ and $2 x + 7 y = 5 2 x + 7 y = 5$ are perpendicular.

Show solution

Solution

Solve the equations for $y y$ .	$\begin{matrix} 7 x + 2 y & = & 3 \\ 2 y & = & −7 x + 3 \\ \frac{2 y}{2} & = & \frac{−7 x + 3}{2} \\ y & = & - \frac{7}{2} x + \frac{3}{2} \end{matrix} \begin{matrix} = & 3 \\ = & −7 x + 3 \\ = & \frac{−7 x + 3}{2} \\ = & - \frac{7}{2} x + \frac{3}{2} \end{matrix}$	$\begin{matrix} 2 x + 7 y & = & 5 \\ 7 y & = & −2 x + 5 \\ \frac{7 y}{7} & = & \frac{−2 x + 5}{7} \\ y & = & - \frac{2}{7} x + \frac{5}{7} \end{matrix} \begin{matrix} 2 x + 7 y & = & 5 \\ 7 y & = & −2 x + 5 \\ \frac{7 y}{7} & = & \frac{−2 x + 5}{7} \\ y & = & - \frac{2}{7} x + \frac{5}{7} \end{matrix}$
Identify the slope of each line.	$\begin{matrix} y & = & m x + b \\ m_{1} & = & - \frac{7}{2} \end{matrix} \begin{matrix} y & = & m x + b \\ = & - \frac{7}{2} \end{matrix}$	$\begin{matrix} y & = & m x + b \\ m_{2} & = & - \frac{2}{7} \end{matrix} \begin{matrix} y & = & m x + b \\ = & - \frac{2}{7} \end{matrix}$

Table 4.45

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

Try It 4.111

Use slopes to determine if the lines $5 x + 4 y = 1 5 x + 4 y = 1$ and $4 x + 5 y = 3 4 x + 5 y = 3$ are perpendicular.

Try It 4.112

Use slopes to determine if the lines $2 x - 9 y = 3 2 x - 9 y = 3$ and $9 x - 2 y = 1 9 x - 2 y = 1$ are perpendicular.

Media

Access this online resource for additional instruction and practice with graphs.

Explore the Relation Between a Graph and the Slope–Intercept Form of an Equation of a Line

Practice Makes Perfect

Recognize the Relation Between the Graph and the Slope–Intercept Form of an Equation of a Line

In the following exercises, use the graph to find the slope and y-intercept of each line. Compare the values to the equation $y = m x + b y = m x + b$ .

288.

The figure shows a line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The line goes through the points (0, negative 5) and (1, negative 2).

$y = 3 x - 5 y = 3 x - 5$

289.

$y = 4 x - 2 y = 4 x - 2$

290.

$y = − x + 4 y = − x + 4$

291.

$y = −3 x + 1 y = −3 x + 1$

292.

$y = - \frac{4}{3} x + 1 y = - \frac{4}{3} x + 1$

293.

$y = - \frac{2}{5} x + 3 y = - \frac{2}{5} x + 3$

Identify the Slope and y-Intercept From an Equation of a Line

In the following exercises, identify the slope and y-intercept of each line.

294.

$y = −7 x + 3 y = −7 x + 3$

295.

$y = −9 x + 7 y = −9 x + 7$

296.

$y = 6 x - 8 y = 6 x - 8$

297.

$y = 4 x - 10 y = 4 x - 10$

298.

$3 x + y = 5 3 x + y = 5$

299.

$4 x + y = 8 4 x + y = 8$

300.

$6 x + 4 y = 12 6 x + 4 y = 12$

301.

$8 x + 3 y = 12 8 x + 3 y = 12$

302.

$5 x - 2 y = 6 5 x - 2 y = 6$

303.

$7 x - 3 y = 9 7 x - 3 y = 9$

Graph a Line Using Its Slope and Intercept

In the following exercises, graph the line of each equation using its slope and y-intercept.

304.

$y = x + 3 y = x + 3$

305.

$y = x + 4 y = x + 4$

306.

$y = 3 x - 1 y = 3 x - 1$

307.

$y = 2 x - 3 y = 2 x - 3$

308.

$y = − x + 2 y = − x + 2$

309.

$y = − x + 3 y = − x + 3$

310.

$y = − x - 4 y = − x - 4$

311.

$y = − x - 2 y = − x - 2$

312.

$y = - \frac{3}{4} x - 1 y = - \frac{3}{4} x - 1$

313.

$y = - \frac{2}{5} x - 3 y = - \frac{2}{5} x - 3$

314.

$y = - \frac{3}{5} x + 2 y = - \frac{3}{5} x + 2$

315.

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

316.

$3 x - 4 y = 8 3 x - 4 y = 8$

317.

$4 x - 3 y = 6 4 x - 3 y = 6$

318.

$y = 0.1 x + 15 y = 0.1 x + 15$

319.

$y = 0.3 x + 25 y = 0.3 x + 25$

Choose the Most Convenient Method to Graph a Line

In the following exercises, determine the most convenient method to graph each line.

320.

$x = 2 x = 2$

321.

$y = 4 y = 4$

322.

$y = 5 y = 5$

323.

$x = −3 x = −3$

324.

$y = −3 x + 4 y = −3 x + 4$

325.

$y = −5 x + 2 y = −5 x + 2$

326.

$x - y = 5 x - y = 5$

327.

$x - y = 1 x - y = 1$

328.

$y = \frac{2}{3} x - 1 y = \frac{2}{3} x - 1$

329.

$y = \frac{4}{5} x - 3 y = \frac{4}{5} x - 3$

330.

$y = −3 y = −3$

331.

$y = −1 y = −1$

332.

$3 x - 2 y = −12 3 x - 2 y = −12$

333.

$2 x - 5 y = −10 2 x - 5 y = −10$

334.

$y = - \frac{1}{4} x + 3 y = - \frac{1}{4} x + 3$

335.

$y = - \frac{1}{3} x + 5 y = - \frac{1}{3} x + 5$

Graph and Interpret Applications of Slope–Intercept

336.

The equation $P = 31 + 1.75 w P = 31 + 1.75 w$ models the relation between the amount of Tuyet’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find Tuyet’s payment for a month when 0 units of water are used.
ⓑ Find Tuyet’s payment for a month when 12 units of water are used.
ⓓ Graph the equation.

337.

The equation $P = 28 + 2.54 w P = 28 + 2.54 w$ models the relation between the amount of Randy’s monthly water bill payment, P, in dollars, and the number of units of water, w, used.

ⓐ Find the payment for a month when Randy used 0 units of water.
ⓑ Find the payment for a month when Randy used 15 units of water.
ⓓ Graph the equation.

338.

Bruce drives his car for his job. The equation $R = 0.575 m + 42 R = 0.575 m + 42$ models the relation between the amount in dollars, R, that he is reimbursed and the number of miles, m, he drives in one day.

ⓐ Find the amount Bruce is reimbursed on a day when he drives 0 miles.
ⓑ Find the amount Bruce is reimbursed on a day when he drives 220 miles.
ⓓ Graph the equation.

339.

Janelle is planning to rent a car while on vacation. The equation $C = 0.32 m + 15 C = 0.32 m + 15$ models the relation between the cost in dollars, C, per day and the number of miles, m, she drives in one day.

ⓐ Find the cost if Janelle drives the car 0 miles one day.
ⓑ Find the cost on a day when Janelle drives the car 400 miles.
ⓓ Graph the equation.

340.

Cherie works in retail and her weekly salary includes commission for the amount she sells. The equation $S = 400 + 0.15 c S = 400 + 0.15 c$ models the relation between her weekly salary, S, in dollars and the amount of her sales, c, in dollars.

ⓐ Find Cherie’s salary for a week when her sales were 0.
ⓑ Find Cherie’s salary for a week when her sales were 3600.
ⓓ Graph the equation.

341.

Patel’s weekly salary includes a base pay plus commission on his sales. The equation $S = 750 + 0.09 c S = 750 + 0.09 c$ models the relation between his weekly salary, S, in dollars and the amount of his sales, c, in dollars.

ⓐ Find Patel’s salary for a week when his sales were 0.
ⓑ Find Patel’s salary for a week when his sales were 18,540.
ⓓ Graph the equation.

342.

Costa is planning a lunch banquet. The equation $C = 450 + 28 g C = 450 + 28 g$ models the relation between the cost in dollars, C, of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 40.
ⓑ Find the cost if the number of guests is 80.
ⓓ Graph the equation.

343.

Margie is planning a dinner banquet. The equation $C = 750 + 42 g C = 750 + 42 g$ models the relation between the cost in dollars, C of the banquet and the number of guests, g.

ⓐ Find the cost if the number of guests is 50.
ⓑ Find the cost if the number of guests is 100.
ⓓ Graph the equation.

Use Slopes to Identify Parallel Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel.

344.

$y = \frac{3}{4} x - 3; 3 x - 4 y = - 2 y = \frac{3}{4} x - 3;$

345.

$y = \frac{2}{3} x - 1; 2 x - 3 y = - 2 y = \frac{2}{3} x - 1;$

346.

$2 x - 5 y = - 3; y = \frac{2}{5} x + 1 2 x - 5 y = - 3;$

347.

$3 x - 4 y = - 2; y = \frac{3}{4} x - 3 3 x - 4 y = - 2;$

348.

$2 x - 4 y = 6; x - 2 y = 3 2 x - 4 y = 6;$

349.

$6 x - 3 y = 9; 2 x - y = 3 6 x - 3 y = 9;$

350.

$4 x + 2 y = 6; 6 x + 3 y = 3 4 x + 2 y = 6;$

351.

$8 x + 6 y = 6; 12 x + 9 y = 12 8 x + 6 y = 6;$

352.

$x = 5; x = - 6 x = 5;$

353.

$x = 7; x = - 8 x = 7;$

354.

$x = - 4; x = - 1 x = - 4;$

355.

$x = - 3; x = - 2 x = - 3;$

356.

$y = 2; y = 6 y = 2;$

357.

$y = 5; y = 1 y = 5;$

358.

$y = - 4; y = 3 y = - 4;$

359.

$y = - 1; y = 2 y = - 1;$

360.

$x - y = 2; 2 x - 2 y = 4 x - y = 2;$

361.

$4 x + 4 y = 8; x + y = 2 4 x + 4 y = 8;$

362.

$x - 3 y = 6; 2 x - 6 y = 12 x - 3 y = 6;$

363.

$5 x - 2 y = 11; 5 x - y = 7 5 x - 2 y = 11;$

364.

$3 x - 6 y = 12; 6 x - 3 y = 3 3 x - 6 y = 12;$

365.

$4 x - 8 y = 16; x - 2 y = 4 4 x - 8 y = 16;$

366.

$9 x - 3 y = 6; 3 x - y = 2 9 x - 3 y = 6;$

367.

$x - 5 y = 10; 5 x - y = - 10 x - 5 y = 10;$

368.

$7 x - 4 y = 8; 4 x + 7 y = 14 7 x - 4 y = 8;$

369.

$9 x - 5 y = 4; 5 x + 9 y = - 1 9 x - 5 y = 4;$

Use Slopes to Identify Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are perpendicular.

370.

$3 x - 2 y = 8; 2 x + 3 y = 6 3 x - 2 y = 8; 2 x + 3 y = 6$

371.

$x - 4 y = 8; 4 x + y = 2 x - 4 y = 8; 4 x + y = 2$

372.

$2 x + 5 y = 3; 5 x - 2 y = 6 2 x + 5 y = 3; 5 x - 2 y = 6$

373.

$2 x + 3 y = 5; 3 x - 2 y = 7 2 x + 3 y = 5; 3 x - 2 y = 7$

374.

$3 x - 2 y = 1; 2 x - 3 y = 2 3 x - 2 y = 1; 2 x - 3 y = 2$

375.

$3 x - 4 y = 8; 4 x - 3 y = 6 3 x - 4 y = 8; 4 x - 3 y = 6$

376.

$5 x + 2 y = 6; 2 x + 5 y = 8 5 x + 2 y = 6; 2 x + 5 y = 8$

377.

$2 x + 4 y = 3; 6 x + 3 y = 2 2 x + 4 y = 3; 6 x + 3 y = 2$

378.

$4 x - 2 y = 5; 3 x + 6 y = 8 4 x - 2 y = 5; 3 x + 6 y = 8$

379.

$2 x - 6 y = 4; 12 x + 4 y = 9 2 x - 6 y = 4; 12 x + 4 y = 9$

380.

$6 x - 4 y = 5; 8 x + 12 y = 3 6 x - 4 y = 5; 8 x + 12 y = 3$

381.

$8 x - 2 y = 7; 3 x + 12 y = 9 8 x - 2 y = 7; 3 x + 12 y = 9$

Everyday Math

382.

The equation $C = \frac{5}{9} F - 17.8 C = \frac{5}{9} F - 17.8$ can be used to convert temperatures F, on the Fahrenheit scale to temperatures, C, on the Celsius scale.

ⓐ Explain what the slope of the equation means.
ⓑ Explain what the C–intercept of the equation means.

383.

The equation $n = 4 T - 160 n = 4 T - 160$ is used to estimate the number of cricket chirps, n, in one minute based on the temperature in degrees Fahrenheit, T.

ⓐ Explain what the slope of the equation means.
ⓑ Explain what the n–intercept of the equation means. Is this a realistic situation?

Writing Exercises

384.

Explain in your own words how to decide which method to use to graph a line.

385.

Why are all horizontal lines parallel?

Self Check

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

This table has eight rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “recognize the relation between the graph and the slope-intercept form of an equation of a line.” The third row reads “identify the Slope and y-intercept from an equation of a line”. The fourth row reads “graph a line using its slope and intercept”. The fifth row reads “choose the most convenient method to graph a line.” The sixth row reads “graph and interpret applications of slope-intercept”. The seventh row reads “use slopes to identify parallel lines” and the last row reads “use slopes to identify perpendicular lines.” The remaining columns are blank.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?