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

Recognize the relation between the graph and the slope–intercept form of an equation of a line

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 \(y=\frac{1}{2}x+3\) by plotting points. See . Let’s find the slope of this line.

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

\[\begin{array}{lll}m & = & \frac{\text{rise}}{\text{run}} \\ m & = & \frac{1}{2}\end{array}\]

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

Notice, the line has:

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

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

Example

Try it.

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

Compare these values to the equation\(y=mx+b\).

Solution

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

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 \(x\) and the y-coordinate of the y-intercept is the same as the constant term.

Identify the Slope and

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

Try it.

Identify the slope and y-intercept of the line with equation \(y=-3x+5\).

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.

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

Example

Try it.

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

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\(y\).

Solve for y.\(x+2y=6\)
Subtract x from each side.
Divide both sides by 2.
Simplify.
\((\text{Remember:}\ \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c})\)
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.

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.

How to Graph a Line Using its Slope and Intercept

Try it.

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

Solution
Example

Try it.

Graph the line of the equation \(y=\text{-}x+4\) using its slope and y-intercept.

Solution

\(y=mx+b\)
The equation is in slope–intercept form.\(y=\text{-}x+4\)
Identify the slope and y-intercept.\(m=-1\)
y-intercept is (0, 4)
Plot the y-intercept.See graph below.
Identify the rise and the run.\(m=\frac{-1}{1}\)
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.
\(\begin{array}{lll} \\ y & = & \text{-}x+4 \\ 0 & \overset{?}{=} & -4+4 \\ 0 & = & 0✓\end{array}\)

Example

Try it.

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

Solution

\(y=mx+b\)
The equation is in slope–intercept form.\(y=-\frac{2}{3}x-3\)
Identify the slope and y-intercept.\(m=-\frac{2}{3}\); 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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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.

\[\begin{array}{llllllll} & & & \text{Equation} & & \ & & \text{Method} \\ \text{\#1} & & & x=2 & & \ & & \text{Vertical line} \\ \text{\#2} & & & y=4 & & \ & & \text{Horizontal line} \\ \text{\#3} & & & \text{-}x+2y=6 & & \ & & \text{Intercepts} \\ \text{\#4} & & & 4x-3y=12 & & \ & & \text{Intercepts} \\ \text{\#5} & & & y=4x-2 & & \ & & \text{Slope-intercept} \\ \text{\#6} & & & y=\text{-}x+4 & & \ & & \text{Slope-intercept}\end{array}\]

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 \(x\) and \(y\) are on the same side of the equation. These two equations are of the form \(Ax+By=C\). We substituted \(y=0\) to find the x-intercept and \(x=0\) to find the y-intercept, and then found a third point by choosing another value for \(x\) or \(y\).

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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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

Try it.

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

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

ⓒ Interpret the slope and F-intercept of the equation.

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

The slope, \(\frac{9}{5}\), 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 \(0\text{^{\circ}}\) on the Celsius scale, it is \(32\text{^{\circ}}\) on the Fahrenheit scale.

ⓓ Graph the equation.

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

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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.

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 .

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 .

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

\[\begin{array}{lll}2x+y & = & -1 \\ y & = & -2x-1\end{array}\]

Graph the lines.

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

\[\begin{array}{llllllllll}y & = & mx+b & & \ & & & y & = & mx+b \\ y & = & -2x+3 & & \ & & & y & = & -2x-1 \\ m & = & -2 & & \ & & & m & = & -2 \\ b & = & 3,\text{(0, 3)} & & \ & & & b & = & -1,\text{(0, -1)}\end{array}\]

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

Example

Try it.

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

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

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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Use Slopes to Identify Perpendicular Lines

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

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 \(y=\frac{1}{4}x-1\) rises, so its slope is positive. The line\(y=-4x+2\) 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, \({m}_{1}=\frac{1}{4}\), and the slope of the second line, \({m}_{2}=-4\), we can see that they are negative reciprocals of each other. If we multiply them, their product is \(-1.\)

\[\begin{array}{l}{m}_{1}\cdot {m}_{2} \\ \frac{1}{4}(-4) \\ -1\end{array}\]

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

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 \(-1\), the lines are perpendicular. Perpendicular lines may have the same y-intercepts.

Example

Try it.

Use slopes to determine if the lines, \(y=-5x-4\) and \(x-5y=5\) are perpendicular.

Solution
The first equation is already in slope-intercept form.\(\begin{array}{lll}y & = & -5x-4\end{array}\)
Solve the second equation for \(y\).\(\begin{array}{lll}x-5y & = & 5 \\ -5y & = & \text{-}x+5 \\ \frac{-5y}{-5} & = & \frac{\text{-}x+5}{-5} \\ y & = & \frac{1}{5}x-1\end{array}\)
Identify the slope of each line.\(\begin{array}{lll}y & = & -5x-4 \\ y & = & mx+b \\ {m}_{1} & = & -5\end{array}\)\(\begin{array}{lll}y & = & \frac{1}{5}x-1 \\ y & = & mx+b \\ {m}_{2} & = & \frac{1}{5}\end{array}\)

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

\[\begin{array}{l}{m}_{1}\cdot {m}_{2} \\ \\ -5(\frac{1}{5}) \\ -1✓\end{array}\]

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

  • The slope–intercept form of an equation of a line with slope \(m\) and y-intercept, \((0,b)\) is, \(y=mx+b\).
  • Graph a Line Using its Slope and y-Intercept
    1. Find the slope-intercept form of the equation of the line.
    2. Identify the slope and y-intercept.
    3. Plot the y-intercept.
    4. Use the slope formula \(m=\frac{\text{rise}}{\text{run}}\) to identify the rise and the run.
    5. Starting at the y-intercept, count out the rise and run to mark the second point.
    6. Connect the points with a line.



  • 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.
      \(x=a\) is a vertical line passing through the x-axis at \(a\).
      \(y=b\) is a horizontal line passing through the y-axis at \(b\).
    • If \(y\) is isolated on one side of the equation, in the form \(y=mx+b\), graph by using the slope and y-intercept.
      Identify the slope and y-intercept and then graph.
    • If the equation is of the form \(Ax+By=C\), find the intercepts.
      Find the x- and y-intercepts, a third point, and then graph.
  • Parallel lines are lines in the same plane that do not intersect.
    • Parallel lines have the same slope and different y-intercepts.
    • If m1 and m2 are the slopes of two parallel lines then \({m}_{1}={m}_{2}.\)
    • Parallel vertical lines have different x-intercepts.
  • Perpendicular lines are lines in the same plane that form a right angle.
    • If \({m}_{1}\ \text{and}\ {m}_{2}\) are the slopes of two perpendicular lines, then \({m}_{1}\cdot {m}_{2}=-1\) and \({m}_{1}=\frac{-1}{{m}_{2}}\).
    • Vertical lines and horizontal lines are always perpendicular to each other.

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

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 \(y=mx+b\).

Try it.

\(y=3x-5\)

Try it.

\(y=4x-2\)

Solution

slope \(m=4\) and y-intercept \((0,-2)\)

Try it.

\(y=\text{-}x+4\)

Try it.

\(y=-3x+1\)

Solution

slope \(m=-3\) and y-intercept \((0,1)\)

Try it.

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

Try it.

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

Solution

slope \(m=-\frac{2}{5}\) and y-intercept \((0,3)\)

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.

Try it.

\(y=-7x+3\)

Try it.

\(y=-9x+7\)

Solution

\(-9;(0,7)\)

Try it.

\(y=6x-8\)

Try it.

\(y=4x-10\)

Solution

\(4;(0,-10)\)

Try it.

\(3x+y=5\)

Try it.

\(4x+y=8\)

Solution

\(-4;(0,8)\)

Try it.

\(6x+4y=12\)

Try it.

\(8x+3y=12\)

Solution

\(-\frac{8}{3};(0,4)\)

Try it.

\(5x-2y=6\)

Try it.

\(7x-3y=9\)

Solution

\(\frac{7}{3};(0,-3)\)

Graph a Line Using Its Slope and Intercept

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

Try it.

\(y=x+3\)

Try it.

\(y=x+4\)

Solution

Try it.

\(y=3x-1\)

Try it.

\(y=2x-3\)

Solution

Try it.

\(y=\text{-}x+2\)

Try it.

\(y=\text{-}x+3\)

Solution

Try it.

\(y=\text{-}x-4\)

Try it.

\(y=\text{-}x-2\)

Solution

Try it.

\(y=-\frac{3}{4}x-1\)

Try it.

\(y=-\frac{2}{5}x-3\)

Solution

Try it.

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

Try it.

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

Solution

Try it.

\(3x-4y=8\)

Try it.

\(4x-3y=6\)

Solution

Try it.

\(y=0.1x+15\)

Try it.

\(y=0.3x+25\)

Solution

Choose the Most Convenient Method to Graph a Line

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

Try it.

\(x=2\)

Try it.

\(y=4\)

Solution

horizontal line

Try it.

\(y=5\)

Try it.

\(x=-3\)

Solution

vertical line

Try it.

\(y=-3x+4\)

Try it.

\(y=-5x+2\)

Solution

slope–intercept

Try it.

\(x-y=5\)

Try it.

\(x-y=1\)

Solution

intercepts

Try it.

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

Try it.

\(y=\frac{4}{5}x-3\)

Solution

slope–intercept

Try it.

\(y=-3\)

Try it.

\(y=-1\)

Solution

horizontal line

Try it.

\(3x-2y=-12\)

Try it.

\(2x-5y=-10\)

Solution

intercepts

Try it.

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

Try it.

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

Solution

slope–intercept

Graph and Interpret Applications of Slope–Intercept

Try it.

The equation \(P=31+1.75w\) 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.

  1. ⓐ Find Tuyet’s payment for a month when 0 units of water are used.
  2. ⓑ Find Tuyet’s payment for a month when 12 units of water are used.
  3. ⓒ Interpret the slope and P-intercept of the equation.
  4. ⓓ Graph the equation.

Try it.

The equation \(P=28+2.54w\) 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.

  1. ⓐ Find the payment for a month when Randy used 0 units of water.
  2. ⓑ Find the payment for a month when Randy used 15 units of water.
  3. ⓒ Interpret the slope and P-intercept of the equation.
  4. ⓓ Graph the equation.
Solution
  1. ⓐ $28
  2. ⓑ $66.10
  3. ⓒ The slope, 2.54, means that Randy’s payment, P, increases by $2.54 when the number of units of water he used, w, increases by 1. The P–intercept means that if the number units of water Randy used was 0, the payment would be $28.
  4. ⓓ

Try it.

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

  1. ⓐ Find the amount Bruce is reimbursed on a day when he drives 0 miles.
  2. ⓑ Find the amount Bruce is reimbursed on a day when he drives 220 miles.
  3. ⓒ Interpret the slope and R-intercept of the equation.
  4. ⓓ Graph the equation.

Try it.

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

  1. ⓐ Find the cost if Janelle drives the car 0 miles one day.
  2. ⓑ Find the cost on a day when Janelle drives the car 400 miles.
  3. ⓒ Interpret the slope and C–intercept of the equation.
  4. ⓓ Graph the equation.
Solution
  1. ⓐ $15
  2. ⓑ $143
  3. ⓒ The slope, 0.32, means that the cost, C, increases by $0.32 when the number of miles driven, m, increases by 1. The C-intercept means that if Janelle drives 0 miles one day, the cost would be $15.
  4. ⓓ

Try it.

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

  1. ⓐ Find Cherie’s salary for a week when her sales were 0.
  2. ⓑ Find Cherie’s salary for a week when her sales were 3600.
  3. ⓒ Interpret the slope and S–intercept of the equation.
  4. ⓓ Graph the equation.

Try it.

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

  1. ⓐ Find Patel’s salary for a week when his sales were 0.
  2. ⓑ Find Patel’s salary for a week when his sales were 18,540.
  3. ⓒ Interpret the slope and S-intercept of the equation.
  4. ⓓ Graph the equation.
Solution
  1. ⓐ $750
  2. ⓑ $2418.60
  3. ⓒ The slope, 0.09, means that Patel’s salary, S, increases by $0.09 for every $1 increase in his sales. The S-intercept means that when his sales are $0, his salary is $750.
  4. ⓓ

Try it.

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

  1. ⓐ Find the cost if the number of guests is 40.
  2. ⓑ Find the cost if the number of guests is 80.
  3. ⓒ Interpret the slope and C-intercept of the equation.
  4. ⓓ Graph the equation.

Try it.

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

  1. ⓐ Find the cost if the number of guests is 50.
  2. ⓑ Find the cost if the number of guests is 100.
  3. ⓒ Interpret the slope and C–intercept of the equation.
  4. ⓓ Graph the equation.
Solution
  1. ⓐ $2850
  2. ⓑ $4950
  3. ⓒ The slope, 42, means that the cost, C, increases by $42 for when the number of guests increases by 1. The C-intercept means that when the number of guests is 0, the cost would be $750.
  4. ⓓ

Use Slopes to Identify Parallel Lines

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

Try it.

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

Try it.

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

Solution

parallel

Try it.

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

Try it.

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

Solution

parallel

Try it.

\(2x-4y=6;\ x-2y=3\)

Try it.

\(6x-3y=9;\ 2x-y=3\)

Solution

not parallel

Try it.

\(4x+2y=6;\ 6x+3y=3\)

Try it.

\(8x+6y=6;\ 12x+9y=12\)

Solution

parallel

Try it.

\(x=5;\ x=-6\)

Try it.

\(x=7;\ x=-8\)

Solution

parallel

Try it.

\(x=-4;\ x=-1\)

Try it.

\(x=-3;\ x=-2\)

Solution

parallel

Try it.

\(y=2;\ y=6\)

Try it.

\(y=5;\ y=1\)

Solution

parallel

Try it.

\(y=-4;\ y=3\)

Try it.

\(y=-1;\ y=2\)

Solution

parallel

Try it.

\(x-y=2;\ 2x-2y=4\)

Try it.

\(4x+4y=8;\ x+y=2\)

Solution

not parallel

Try it.

\(x-3y=6;\ 2x-6y=12\)

Try it.

\(5x-2y=11;\ 5x-y=7\)

Solution

not parallel

Try it.

\(3x-6y=12;\ 6x-3y=3\)

Try it.

\(4x-8y=16;\ x-2y=4\)

Solution

not parallel

Try it.

\(9x-3y=6;\ 3x-y=2\)

Try it.

\(x-5y=10;\ 5x-y=-10\)

Solution

not parallel

Try it.

\(7x-4y=8;\ 4x+7y=14\)

Try it.

\(9x-5y=4;\ 5x+9y=-1\)

Solution

not parallel

Use Slopes to Identify Perpendicular Lines

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

Try it.

\(3x-2y=8;2x+3y=6\)

Try it.

\(x-4y=8;4x+y=2\)

Solution

perpendicular

Try it.

\(2x+5y=3;5x-2y=6\)

Try it.

\(2x+3y=5;3x-2y=7\)

Solution

perpendicular

Try it.

\(3x-2y=1;2x-3y=2\)

Try it.

\(3x-4y=8;4x-3y=6\)

Solution

not perpendicular

Try it.

\(5x+2y=6;2x+5y=8\)

Try it.

\(2x+4y=3;6x+3y=2\)

Solution

not perpendicular

Try it.

\(4x-2y=5;3x+6y=8\)

Try it.

\(2x-6y=4;12x+4y=9\)

Solution

perpendicular

Try it.

\(6x-4y=5;8x+12y=3\)

Try it.

\(8x-2y=7;3x+12y=9\)

Solution

perpendicular

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Add: \(\frac{x}{4}+\frac{1}{4}.\)
    If you missed this problem, review .

    Bonisa impendulo

    \(\frac{x+1}{4}\)

  2. Find the reciprocal of \(\frac{3}{7}.\)
    If you missed this problem, review .

    Bonisa impendulo

    \(\frac{7}{3}\)

  3. Solve \(2x-3y=12\ \text{for}\ y\).
    If you missed this problem, review .

    Bonisa impendulo

    \(y=\frac{12-2x}{-3}\)

  4. Use the graph to find the slope and y-intercept of the line, \(y=2x+1\).

    Compare these values to the equation\(y=mx+b\).

    Bonisa impendulo

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

    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 \(x\) and the y-coordinate of the y-intercept is the same as the constant term.

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

    Bonisa impendulo

    slope \(m=\frac{2}{3}\) and y-intercept \((0,-1)\)

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

    Bonisa impendulo

    slope \(m=\frac{1}{2}\) and y-intercept \((0,3)\)

  7. Identify the slope and y-intercept of the line with equation \(y=-3x+5\).

    Bonisa impendulo

    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.

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

    Bonisa impendulo

    \(\frac{2}{5};(0,-1)\)

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

    Bonisa impendulo

    \(-\frac{4}{3};(0,1)\)

  10. Identify the slope and y-intercept of the line with equation \(x+2y=6\).

    Bonisa impendulo

    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\(y\).

    Solve for y.\(x+2y=6\)
    Subtract x from each side.
    Divide both sides by 2.
    Simplify.
    \((\text{Remember:}\ \frac{a+b}{c}=\frac{a}{c}+\frac{b}{c})\)
    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.

  11. Identify the slope and y-intercept of the line \(x+4y=8\).

    Bonisa impendulo

    \(-\frac{1}{4};(0,2)\)

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

    Bonisa impendulo

    \(-\frac{3}{2};(0,6)\)

  13. Graph the line of the equation \(y=4x-2\) using its slope and y-intercept.

  14. Graph the line of the equation \(y=4x+1\) using its slope and y-intercept.

    Bonisa impendulo

  15. Graph the line of the equation \(y=2x-3\) using its slope and y-intercept.

    Bonisa impendulo

  16. Graph the line of the equation \(y=\text{-}x+4\) using its slope and y-intercept.

    Bonisa impendulo

    \(y=mx+b\)
    The equation is in slope–intercept form.\(y=\text{-}x+4\)
    Identify the slope and y-intercept.\(m=-1\)
    y-intercept is (0, 4)
    Plot the y-intercept.See graph below.
    Identify the rise and the run.\(m=\frac{-1}{1}\)
    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.
    \(\begin{array}{lll} \\ y & = & \text{-}x+4 \\ 0 & \overset{?}{=} & -4+4 \\ 0 & = & 0✓\end{array}\)

  17. Graph the line of the equation \(y=\text{-}x-3\) using its slope and y-intercept.

    Bonisa impendulo

  18. Graph the line of the equation \(y=\text{-}x-1\) using its slope and y-intercept.

    Bonisa impendulo

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

    Bonisa impendulo

    \(y=mx+b\)
    The equation is in slope–intercept form.\(y=-\frac{2}{3}x-3\)
    Identify the slope and y-intercept.\(m=-\frac{2}{3}\); 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.

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

    Bonisa impendulo

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

    Bonisa impendulo

  22. Graph the line of the equation \(4x-3y=12\) using its slope and y-intercept.

    Bonisa impendulo

    \(4x-3y=12\)
    Find the slope–intercept form of the equation.\(\ -3y=-4x+12\)
    \(\ -\frac{3y}{3}=\frac{-4x+12}{-3}\)
    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.

  23. Graph the line of the equation \(2x-y=6\) using its slope and y-intercept.

    Bonisa impendulo

  24. Graph the line of the equation \(3x-2y=8\) using its slope and y-intercept.

    Bonisa impendulo

  25. Graph the line of the equation \(y=0.2x+45\) using its slope and y-intercept.

    Bonisa impendulo

    We’ll use a grid with the axes going from about \(-80\) to 80.

    \(y=mx+b\)
    The equation is in slope–intercept form.\(y=0.2x+45\)
    Identify the slope and y-intercept.\(m=0.2\)
    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 \(m=0.2\); in fraction form this means \(m=\frac{2}{10}\). Given the scale of our graph, it would be easier to use the equivalent fraction \(m=\frac{10}{50}\).
    Draw the line.

  26. Graph the line of the equation \(y=0.5x+25\) using its slope and y-intercept.

    Bonisa impendulo

  27. Graph the line of the equation \(y=0.1x-30\) using its slope and y-intercept.

    Bonisa impendulo

  28. Determine the most convenient method to graph each line.

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

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

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

    Bonisa impendulo

    ⓐ intercepts ⓑ horizontal line ⓒ slope–intercept ⓓ vertical line

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

    Bonisa impendulo

    ⓐ vertical line ⓑ slope–intercept ⓒ horizontal line ⓓ intercepts

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

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

    ⓒ Interpret the slope and F-intercept of the equation.

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

    The slope, \(\frac{9}{5}\), 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 \(0\text{^{\circ}}\) on the Celsius scale, it is \(32\text{^{\circ}}\) on the Fahrenheit scale.

    ⓓ Graph the equation.

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

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

    1. ⓐ Estimate the height of a child who wears women’s shoe size 0.
    2. ⓑ Estimate the height of a woman with shoe size 8.
    3. ⓒ Interpret the slope and h-intercept of the equation.
    4. ⓓ Graph the equation.
    Bonisa impendulo
    1. ⓐ 50 inches
    2. ⓑ 66 inches
    3. ⓒ The slope, 2, means that the height, h, increases by 2 inches when the shoe size, s, increases by 1. The h-intercept means that when the shoe size is 0, the height is 50 inches.
    4. ⓓ

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

    1. ⓐ Estimate the temperature when there are no chirps.
    2. ⓑ Estimate the temperature when the number of chirps in one minute is 100.
    3. ⓒ Interpret the slope and T-intercept of the equation.
    4. ⓓ Graph the equation.
    Bonisa impendulo
    1. ⓐ 40 degrees
    2. ⓑ 65 degrees
    3. ⓒ The slope, \(\frac{1}{4}\), means that the temperature Fahrenheit (F) increases 1 degree when the number of chirps, n, increases by 4. The T-intercept means that when the number of chirps is 0, the temperature is \(40\text{^{\circ}}\).
    4. ⓓ

  34. Stella has a home business selling gourmet pizzas. The equation \(C=4p+25\) 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.

    Bonisa impendulo

    ⓐ Find Stella's cost for a week when she sells no pizzas.
    Find C when \(p=0\).
    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 \(p=15\).
    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.

  35. Sam drives a delivery van. The equation \(C=0.5m+60\) 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.

    Bonisa impendulo
    1. ⓐ $60
    2. ⓑ $185
    3. ⓒ The slope, 0.5, means that the weekly cost, C, increases by $0.50 when the number of miles driven, n, increases by 1. The C-intercept means that when the number of miles driven is 0, the weekly cost is $60
    4. ⓓ

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

    1. ⓐ Find Loreen’s cost for a week when she writes no invitations.
    2. ⓑ Find the cost for a week when she writes 75 invitations.
    3. ⓒ Interpret the slope and C-intercept of the equation.
    4. ⓓ Graph the equation.
    Bonisa impendulo
    1. ⓐ $35
    2. ⓑ $170
    3. ⓒ The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n, increases by 1.
      The C-intercept means that when the number of invitations is 0, the weekly cost is $35.;
    4. ⓓ

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

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

    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.

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

    Bonisa impendulo

    parallel

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

    Bonisa impendulo

    parallel

  40. Use slopes and y-intercepts to determine if the lines \(y=-4\) and \(y=3\) are parallel.

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

    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 \(y=-4\) and at \(y=3\), we know the y-intercepts are \((0,-4)\) and \((0,3)\). The lines have the same slope and different y-intercepts and so they are parallel.

Symbols used here

^\circ
degrees
1/360 of a full turn. 180° = π radians.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Use the Slope-Intercept Form of an Equation of a Line

  1. Recognize the relation between the graph and the slope–intercept form of an equation of a line
  2. Identify the slope and y-intercept form of an equation of a line
  3. Graph a line using its slope and intercept
  4. Choose the most convenient method to graph a line
  5. Graph and interpret applications of slope–intercept
  6. Use slopes to identify parallel lines
  7. Use slopes to identify perpendicular lines
  8. Find the slope-intercept form of the equation of the line.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama ngokwakho

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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