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Slope of a Line

Find the slope of a line

Find the Slope of a Line

When you graph linear equations, you may notice that some lines tilt up as they go from left to right and some lines tilt down. Some lines are very steep and some lines are flatter.

In mathematics, the measure of the steepness of a line is called the slope of the line.

The concept of slope has many applications in the real world. In construction the pitch of a roof, the slant of the plumbing pipes, and the steepness of the stairs are all applications of slope. and as you ski or jog down a hill, you definitely experience slope.

We can assign a numerical value to the slope of a line by finding the ratio of the rise and run. The rise is the amount the vertical distance changes while the run measures the horizontal change, as shown in this illustration. Slope is a rate of change. See .

To find the slope of a line, we locate two points on the line whose coordinates are integers. Then we sketch a right triangle where the two points are vertices and one side is horizontal and one side is vertical.

To find the slope of the line, we measure the distance along the vertical and horizontal sides of the triangle. The vertical distance is called the rise and the horizontal distance is called the run,

Example

Try it.

Find the slope of the line shown.

Solution

Locate two points on the graph whose coordinates are integers.	$(0,5)$ and $(3,3)$
Starting at $(0,5),$ sketch a right triangle to $(3,3)$ as shown in this graph.
Count the rise— since it goes down, it is negative.	The rise is $-2.$
Count the run.	The run is 3.
Use the slope formula.	$m=\frac{\text{rise}}{\text{run}}$
Substitute the values of the rise and run.	$m=\frac{-2}{3}$
Simplify.	$m=-\frac{2}{3}$
	The slope of the line is $-\frac{2}{3}.$
	So y decreases by 2 units as x increases by 3 units.

How do we find the slope of horizontal and vertical lines? To find the slope of the horizontal line, $y=4,$ we could graph the line, find two points on it, and count the rise and the run. Let’s see what happens when we do this, as shown in the graph below.

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

Graph a Line Given a Point and the Slope

Up to now, in this chapter, we have graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

We can also graph a line when we know one point and the slope of the line. We will start by plotting the point and then use the definition of slope to draw the graph of the line.

How to graph a Line Given a Point and the Slope

Try it.

Graph the line passing through the point $(1,-1)$ whose slope is $m=\frac{3}{4}.$

Solution

You can check your work by finding a third point. Since the slope is $m=\frac{3}{4},$ it can also be written as $m=\frac{-3}{-4}$ (negative divided by negative is positive!). Go back to $(1,-1)$ and count out the rise, $-3,$ and the run, $-4.$

Graph a Line Using its Slope and Intercept

We have graphed linear equations by plotting points, using intercepts, recognizing horizontal and vertical lines, and using one point and the slope of the line. 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.

See . Let’s look at the graph of the equation $y=\frac{1}{2}x+3$ and find its slope and y-intercept.

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

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

The y-intercept is $(0,3).$

Look at the equation of this line.

Look at the slope and y-intercept.

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.

Identify the slope and y-intercept of the line from the equation:

ⓐ $y=-\frac{4}{7}x-2$ ⓑ $x+3y=9$

Solution

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

Write the slope–intercept form of the equation of the line.
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.

Solve for y.	$x+3y=9$
Subtract x from each side.
Divide both sides by 3.
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.

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 the graph.
Identify the rise over the run.	$m=\frac{-1}{1}$
Count out the rise and run to mark the second point.	rise $-1$, run 1

Draw the line as shown in the graph.

Condensed — the full section is in OpenStax Intermediate 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. Let’s look for some patterns to help determine the most convenient method to graph a line.

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

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

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.

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

Condensed — the full section is in OpenStax Intermediate 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 uses real-world data, different letters are used for the variables, instead of using only x and y. The variable names remind us of what quantities are being measured.

Also, we often will need to extend the axes in our rectangular coordinate system to bigger positive and negative numbers to accommodate the data in the application.

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.

ⓓ Graph the equation.

Solution

ⓐ

Find the Fahrenheit temperature for a Celsius temperature of 0.	$F=\frac{9}{5}C+32$
Find $F$ when $C=0.$	$F=\frac{9}{5}(0)+32$
Simplify.	$F=32$

ⓑ

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

ⓒ
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 $(0,32)$, and then count out the rise of 9 and the run of 5 to get a second point as shown in the graph.

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

Use Slopes to Identify Parallel and Perpendicular Lines

Two lines that have the same slope are called parallel lines. Parallel lines have the same steepness and 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 .

Verify that both lines have the same slope, $m=\frac{2}{5},$ 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, like the lines shown in this graph.

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

Try it.

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ $y=-4$ and $y=3$ ⓑ $x=-2$ and $x=-5.$

Solution

ⓐ $y=-4$ and $y=3$
We recognize right away from the equations that these are horizontal lines, and so we 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.
ⓑ $x=-2$ and $x=-5$
We recognize right away from the equations that these are vertical lines, and so we know their slopes are undefined.
Since the vertical lines cross the x-axis at $x=-2$ and $x=-5,$ we know the y-intercepts are $(-2,0)$ and $(-5,0).$
The lines are vertical and have different x-intercepts and so they are parallel.

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

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

Key Concepts

Slope of a Line
- The slope of a line is $m=\frac{\text{rise}}{\text{run}}.$
- The rise measures the vertical change and the run measures the horizontal change.
How to find the slope of a line from its graph using $m=\frac{\text{rise}}{\text{run}}.$
1. Locate two points on the line whose coordinates are integers.
2. Starting with one point, sketch a right triangle, going from the first point to the second point.
3. Count the rise and the run on the legs of the triangle.
4. Take the ratio of rise to run to find the slope: $m=\frac{\text{rise}}{\text{run}}.$
Slope of a line between two points.
- The slope of the line between two points $({x}_{1},{y}_{1})$ and $({x}_{2},{y}_{2})$ is:
  \[m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}.\]
How to graph a line given a point and the slope.
1. Plot the given point.
2. Use the slope formula $m=\frac{\text{rise}}{\text{run}}$ to identify the rise and the run.
3. Starting at the given point, count out the rise and run to mark the second point.
4. Connect the points with a line.
Slope Intercept Form of an Equation of a Line
- The slope–intercept form of an equation of a line with slope m and y-intercept, $(0,b)$ is $y=mx+b$
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 ${m}_{1}$ and ${m}_{2}$ are the slopes of two parallel lines then ${m}_{1}={m}_{2}.$
  Parallel vertical lines have different x-intercepts.
Perpendicular Lines
- Perpendicular lines are lines in the same plane that form a right angle.
- If ${m}_{1}$ and ${m}_{2}$ are the slopes of two perpendicular lines, then:
  their slopes are negative reciprocals of each other, ${m}_{1}=-\frac{1}{{m}_{2}}.$
  the product of their slopes is $-1,$${m}_{1}\cdot {m}_{2}=-1.$
- A vertical line and a horizontal line are always perpendicular to each other.

Slope of a Line

Find the Slope of a Line

In the following exercises, find the slope of each line shown.

Try it.

Solution

$\frac{2}{5}$

Try it.

Solution

$\frac{5}{4}$

Try it.

Solution

$-\frac{1}{3}$

Try it.

Solution

$-\frac{5}{2}$

Try it.

In the following exercises, find the slope of each line.

Try it.

$y=3$

Solution

Try it.

$y=-2$

Try it.

$x=-5$

Solution

undefined

Try it.

$x=4$

In the following exercises, use the slope formula to find the slope of the line between each pair of points.

Try it.

$(2,5),(4,0)$

Solution

$-\frac{5}{2}$

Try it.

$(3,6),(8,0)$

Try it.

$(-3,3),(4,-5)$

Solution

$-\frac{8}{7}$

Try it.

$(-2,4),(3,-1)$

Try it.

$(-1,-2),(2,5)$

Solution

$\frac{7}{3}$

Try it.

$(-2,-1),(6,5)$

Try it.

$(4,-5),(1,-2)$

Solution

$-1$

Try it.

$(3,-6),(2,-2)$

Graph a Line Given a Point and the Slope

In the following exercises, graph each line with the given point and slope.

Try it.

$(2,5);$$m=-\frac{1}{3}$

Solution

Try it.

$(1,4)$; $m=-\frac{1}{2}$

Try it.

$(-1,-4)$; $m=\frac{4}{3}$

Solution

Try it.

$(-3,-5)$; $m=\frac{3}{2}$

Try it.

y-intercept 3; $m=-\frac{2}{5}$

Solution

Try it.

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

Try it.

$(-4,2)$; $m=4$

Solution

Try it.

$(1,5)$; $m=-3$

Graph a Line Using Its Slope and Intercept

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

Try it.

$y=-7x+3$

Solution

$m=-7;(0,3)$

Try it.

$y=4x-10$

Try it.

$3x+y=5$

Solution

$m=-3;(0,5)$

Try it.

$4x+y=8$

Try it.

$6x+4y=12$

Solution

$m=-\frac{3}{2};(0,3)$

Try it.

$8x+3y=12$

Try it.

$5x-2y=6$

Solution

$m=\frac{5}{2};(0,-3)$

Try it.

$7x-3y=9$

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

Try it.

$y=3x-1$

Solution

Try it.

$y=2x-3$

Try it.

$y=\text{-}x+3$

Solution

Try it.

$y=\text{-}x-4$

Try it.

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

Solution

Try it.

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

Try it.

$3x-2y=4$

Solution

Try it.

$3x-4y=8$

Choose the Most Convenient Method to Graph a Line

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

Try it.

$x=2$

Solution

vertical line

Try it.

$y=5$

Try it.

$y=-3x+4$

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.

$3x-2y=-12$

Solution

intercepts

Try it.

$2x-5y=-10$

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.

ⓐ 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.

Solution

ⓐ $31
ⓑ $52
ⓒ The slope, $1.75,$ means that the payment, P, increases by $\text{\$}1.75$ when the number of units of water used, w, increases by 1. The P-intercept means that when the number units of water Tuyet used is 0, the payment is $31.
ⓓ

Try it.

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

ⓐ Find the payment for a month when R and y used 0 units of water.

ⓑ Find the payment for a month when R and y used 15 units of water.

ⓓ Graph the equation.

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.

ⓐ 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.

Solution

ⓐ $42
ⓑ $168.50
ⓒ The slope, 0.575 means that the amount he is reimbursed, R, increases by $0.575 when the number of miles driven, m, increases by 1. The R-intercept means that when the number miles driven is 0, the amount reimbursed is $42.
ⓓ

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.

ⓐ 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.

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.

ⓐ Find Cherie’s salary for a week when her sales were $0.

ⓑ Find Cherie’s salary for a week when her sales were $3,600.

ⓓ Graph the equation.

Solution

ⓐ $400
ⓑ $940
ⓒ The slope, $0.15,$ means that Cherie’s salary, S, increases by $0.15 for every $1 increase in her sales. The S-intercept means that when her sales are $0, her salary is $400.
ⓓ

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.

ⓐ 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.

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.

ⓐ Find the cost if the number of guests is 40.

ⓑ Find the cost if the number of guests is 80.

ⓓ Graph the equation.

Solution

ⓐ $1570
ⓑ $2690
ⓒ The slope gives the cost per guest. The slope, 28, means that the cost, C, increases by $28 when the number of guests increases by 1. The C-intercept means that if the number of guests was 0, the cost would be $450.
ⓓ

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.

ⓐ 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 and Perpendicular Lines

In the following exercises, use slopes and y-intercepts to determine if the lines are parallel, perpendicular, or neither.

Try it.

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

Solution

parallel

Try it.

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

Try it.

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

Solution

neither

Try it.

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

Try it.

$x=5;\ x=-6$

Solution

parallel

Try it.

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

Try it.

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

Solution

perpendicular

Try it.

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

Try it.

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

Solution

neither

Try it.

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

Try it.

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

Solution

perpendicular

Try it.

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

Try it.

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

Solution

perpendicular

Try it.

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

Try it.

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

Solution

neither

Try it.

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

Try it.

$y=2;\ y=6$

Solution

parallel

Try it.

$y=-1;\ y=2$

Condensed — the full section is in OpenStax Intermediate 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.

Simplify: $\frac{(1-4)}{(8-2)}.$
If you missed this problem, review .

Reveal the answer

$-\frac{1}{2}$
Divide: $\frac{0}{4},\frac{4}{0}.$
If you missed this problem, review .

Reveal the answer

$0$; undefined
Simplify: $\frac{15}{-3},\frac{-15}{3},\frac{-15}{-3}.$
If you missed this problem, review .

Reveal the answer

$-5;\ -5;\ 5$

Find the slope of the line shown.

Reveal the answer

Locate two points on the graph whose coordinates are integers.	$(0,5)$ and $(3,3)$
Starting at $(0,5),$ sketch a right triangle to $(3,3)$ as shown in this graph.
Count the rise— since it goes down, it is negative.	The rise is $-2.$
Count the run.	The run is 3.
Use the slope formula.	$m=\frac{\text{rise}}{\text{run}}$
Substitute the values of the rise and run.	$m=\frac{-2}{3}$
Simplify.	$m=-\frac{2}{3}$
	The slope of the line is $-\frac{2}{3}.$
	So y decreases by 2 units as x increases by 3 units.

Find the slope of the line shown.

Reveal the answer

$-\frac{4}{3}$
Find the slope of the line shown.

Reveal the answer

$-\frac{3}{5}$
Find the slope of each line: ⓐ $x=8$ ⓑ $y=-5.$

Reveal the answer

ⓐ $x=8$
This is a vertical line. Its slope is undefined.
ⓑ $y=-5$
This is a horizontal line. It has slope 0.
Find the slope of the line: $x=-4.$

Reveal the answer

undefined
Find the slope of the line: $y=7.$

Reveal the answer

0

Use the slope formula to find the slope of the line through the points $(-2,-3)$ and $(-7,4).$

Reveal the answer

We’ll call $(-2,-3)$ point #1 and $(-7,4)$ point #2.	$(\begin{array}{ll}{x}_{1}, & {y}_{1} \\ -2, & -3\end{array})\ (\begin{array}{ll}{x}_{2}, & {y}_{2} \\ -7, & 4\end{array})$
Use the slope formula.	$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$
Substitute the values. $y$ of the second point minus $y$ of the first point
$x$ of the second point minus $x$ of the first point	$m=\frac{4-(-3)}{-7-(-2)}$
Simplify.	$\begin{array}{l} \\ \\ m=\frac{7}{-5} \\ m=-\frac{7}{5}\end{array}$
Let’s verify this slope on the graph shown.

\[\begin{array}{l} \\ \\ m=\frac{\text{rise}}{\text{run}} \\ m=\frac{7}{-5} \\ m=-\frac{7}{5}\end{array}\]

Use the slope formula to find the slope of the line through the pair of points: $(-3,4)$ and $(2,-1).$

Reveal the answer

$-1$
Use the slope formula to find the slope of the line through the pair of points: $(-2,6)$ and $(-3,-4).$

Reveal the answer

10
Graph the line passing through the point $(1,-1)$ whose slope is $m=\frac{3}{4}.$

Reveal the answer

You can check your work by finding a third point. Since the slope is $m=\frac{3}{4},$ it can also be written as $m=\frac{-3}{-4}$ (negative divided by negative is positive!). Go back to $(1,-1)$ and count out the rise, $-3,$ and the run, $-4.$
Graph the line passing through the point $(2,-2)$ with the slope$m=\frac{4}{3}.$

Reveal the answer
Graph the line passing through the point $(-2,3)$ with the slope $m=\frac{1}{4}.$

Reveal the answer

Identify the slope and y-intercept of the line from the equation:

ⓐ $y=-\frac{4}{7}x-2$ ⓑ $x+3y=9$

Reveal the answer

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

Write the slope–intercept form of the equation of the line.
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.

Solve for y.	$x+3y=9$
Subtract x from each side.
Divide both sides by 3.
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.

Identify the slope and y-intercept from the equation of the line.
ⓐ $y=\frac{2}{5}x-1$ ⓑ $x+4y=8$

Reveal the answer

ⓐ $m=\frac{2}{5};(0,-1)$
ⓑ $m=-\frac{1}{4};(0,2)$
Identify the slope and y-intercept from the equation of the line.
ⓐ $y=-\frac{4}{3}x+1$ ⓑ $3x+2y=12$

Reveal the answer

ⓐ $m=-\frac{4}{3};(0,1)$
ⓑ $m=-\frac{3}{2};(0,6)$

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

Reveal the answer

	$y=mx+b$
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 the graph.
Identify the rise over the run.	$m=\frac{-1}{1}$
Count out the rise and run to mark the second point.	rise $-1$, run 1

Draw the line as shown in the graph.

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

Reveal the answer
Graph the line of the equation$y=\text{-}x-1$ using its slope and y-intercept.

Reveal the answer
Determine the most convenient method to graph each line:
ⓐ $y=5$ ⓑ $4x-5y=20$ ⓒ $x=-3$ ⓓ $y=-\frac{5}{9}x+8$

Reveal the answer

ⓐ $y=5$
This equation has only one variable, y. Its graph is a horizontal line crossing the y-axis at $5$.
ⓑ $4x-5y=20$
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.
ⓒ $x=-3$
There is only one variable, x. The graph is a vertical line crossing the x-axis at$-3.$
ⓓ $y=-\frac{5}{9}x+8$
Since this equation is in $y=mx+b$ form, it will be easiest to graph this line by using the slope and y-intercepts.
Determine the most convenient method to graph each line:
ⓐ $3x+2y=12$ ⓑ $y=4$ ⓒ $y=\frac{1}{5}x-4$ ⓓ $x=-7.$

Reveal the answer

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

Reveal the answer

ⓐ vertical line ⓑ slope-intercept ⓒ horizontal line
ⓓ intercepts
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.

Reveal the answer

ⓐ
Find the Fahrenheit temperature for a Celsius temperature of 0. $F=\frac{9}{5}C+32$
Find $F$ when $C=0.$ $F=\frac{9}{5}(0)+32$
Simplify. $F=32$
ⓑ
Find the Fahrenheit temperature for a Celsius temperature of 20. $F=\frac{9}{5}C+32$
Find $F$ when $C=20.$ $F=\frac{9}{5}(20)+32$
Simplify. $F=36+32$
Simplify. $F=68$
ⓒ
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 $(0,32)$, and then count out the rise of 9 and the run of 5 to get a second point as shown in the graph.
The equation $h=2s+50$ 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.

Reveal the answer

ⓐ 50 inches
ⓑ 66 inches
ⓒ 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.
ⓓ
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.
ⓐ 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.

Reveal the answer

ⓐ 40 degrees
ⓑ 65 degrees
ⓒ 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°.
ⓓ
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.

Reveal the answer

ⓐ
Find Sam’s cost for a week when he drives 0 miles. $C=0.5m+60$
Find $C$ when $m=0.$ $C=0.5(0)+60$
Simplify. $C=60$
Sam’s costs are $60 when he drives 0 miles.
ⓑ
Find the cost for a week when he drives 250 miles. $C=0.5m+60$
Find $C$ when $m=250.$ $C=0.5(250)+60$
Simplify. $C=185$
Sam’s costs are $185 when he drives 250 miles.
ⓒ Interpret the slope and C-intercept of the equation.

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.
ⓓ Graph the equation.
We’ll need to use a larger scale than our usual. Start at the C-intercept $(0,60)$.
To count out the slope m = 0.5, we rewrite it as an equivalent fraction that will make our graphing easier.
$m=0.5$
Rewrite as a fraction. $m=\frac{0.5}{1}$
Multiply numerator and denominator by 100. $m=\frac{0.5(100)}{1(100)}$
Simplify. $m=\frac{50}{100}$
So to graph the next point go up 50 from the intercept of 60 and then to the right 100. The second point will be (100, 110).
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.

Reveal the answer

ⓐ $25
ⓑ $85
ⓒ The slope, 4, means that the weekly cost, C, increases by $4 when the number of pizzas sold, p, increases by 1. The C-intercept means that when the number of pizzas sold is 0, the weekly cost is $25.
ⓓ
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.
ⓐ 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.

Reveal the answer

ⓐ $35
ⓑ $170
ⓒ The slope, $1.8,$ means that the weekly cost, C, increases by $\text{\$}1.80$ 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.
ⓓ

Use slopes and y-intercepts to determine if the lines are parallel:

ⓐ $3x-2y=6$ and $y=\frac{3}{2}x+1$ ⓑ $y=2x-3$ and $-6x+3y=-9.$

Reveal the answer

ⓐ

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

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.

ⓑ

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

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

Use slopes and y-intercepts to determine if the lines are parallel:
ⓐ $2x+5y=5$ and $y=-\frac{2}{5}x-4$ ⓑ $y=-\frac{1}{2}x-1$ and $x+2y=-2.$

Reveal the answer

ⓐ parallel ⓑ not parallel; same line
Use slopes and y-intercepts to determine if the lines are parallel:
ⓐ $4x-3y=6$ and $y=\frac{4}{3}x-1$ ⓑ $y=\frac{3}{4}x-3$ and $3x-4y=12.$

Reveal the answer

ⓐ parallel ⓑ not parallel; same line
Use slopes and y-intercepts to determine if the lines are parallel:
ⓐ $y=-4$ and $y=3$ ⓑ $x=-2$ and $x=-5.$

Reveal the answer

ⓐ $y=-4$ and $y=3$
We recognize right away from the equations that these are horizontal lines, and so we 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.
ⓑ $x=-2$ and $x=-5$
We recognize right away from the equations that these are vertical lines, and so we know their slopes are undefined.
Since the vertical lines cross the x-axis at $x=-2$ and $x=-5,$ we know the y-intercepts are $(-2,0)$ and $(-5,0).$
The lines are vertical and have different x-intercepts and so they are parallel.
Use slopes and y-intercepts to determine if the lines are parallel:
ⓐ $y=8$ and $y=-6$ ⓑ $x=1$ and $x=-5.$

Reveal the answer

ⓐ parallel ⓑ parallel
Use slopes and y-intercepts to determine if the lines are parallel:
ⓐ $y=1$ and $y=-5$ ⓑ $x=8$ and $x=-6.$

Reveal the answer

ⓐ parallel ⓑ parallel

Use slopes to determine if the lines are perpendicular:

ⓐ $y=-5x-4$ and $x-5y=5$ ⓑ $7x+2y=3$ and $2x+7y=5$

Reveal the answer

ⓐ

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

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes, Since $-5(\frac{1}{5})=-1,$ it checks.

ⓑ

Solve the equations for $y\text{.}$	$\begin{array}{ll}7x+2y & =3 \\ 2y & =-7x+3 \\ \frac{2y}{2} & =\frac{-7x+3}{2} \\ y & =-\frac{7}{2}x+\frac{3}{2}\end{array}$	$\ \begin{array}{ll}2x+7y & =5 \\ 7y & =-2x+5 \\ \frac{7y}{7} & =\frac{-2x+5}{7} \\ y & =-\frac{2}{7}x+\frac{5}{7}\end{array}$
Identify the slope of each line.	$\ \begin{array}{ll}y & =mx+b \\ {m}_{1} & =-\frac{7}{2}\end{array}$	$\ \begin{array}{ll}y & =mx+b \\ {m}_{1} & =-\frac{2}{7}\end{array}$

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

Use slopes to determine if the lines are perpendicular:
ⓐ $y=-3x+2$ and $x-3y=4$ ⓑ $5x+4y=1$ and $4x+5y=3.$

Reveal the answer

ⓐ perpendicular ⓑ not perpendicular
Use slopes to determine if the lines are perpendicular:
ⓐ $y=2x-5$ and $x+2y=-6$ ⓑ $2x-9y=3$ and $9x-2y=1.$

Reveal the answer

ⓐ perpendicular ⓑ not perpendicular
$(2,5),(4,0)$

Reveal the answer

$-\frac{5}{2}$

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.

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: Slope of a Line

Find the slope of a line
Graph a line given a point and the slope
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 and perpendicular lines
Locate two points on the line whose coordinates are integers.
Starting with one point, sketch a right triangle, going from the first point to the second point.

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.

Try your own

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

More in Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Find the Fahrenheit temperature for a Celsius temperature of 0.	\(F=\frac{9}{5}C+32\)
Find \(F\) when \(C=0.\)	\(F=\frac{9}{5}(0)+32\)
Simplify.	\(F=32\)

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

We’ll call \((-2,-3)\) point #1 and \((-7,4)\) point #2.	\((\begin{array}{ll}{x}_{1}, & {y}_{1} \\ -2, & -3\end{array})\ (\begin{array}{ll}{x}_{2}, & {y}_{2} \\ -7, & 4\end{array})\)
Use the slope formula.	\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)
Substitute the values. \(y\) of the second point minus \(y\) of the first point
\(x\) of the second point minus \(x\) of the first point	\(m=\frac{4-(-3)}{-7-(-2)}\)
Simplify.	\(\begin{array}{l} \\ \\ m=\frac{7}{-5} \\ m=-\frac{7}{5}\end{array}\)
Let’s verify this slope on the graph shown.

Find the Fahrenheit temperature for a Celsius temperature of 0.	\(F=\frac{9}{5}C+32\)
Find \(F\) when \(C=0.\)	\(F=\frac{9}{5}(0)+32\)
Simplify.	\(F=32\)

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

Locate two points on the graph whose coordinates are integers.	\((0,5)\) and \((3,3)\)
Starting at \((0,5),\) sketch a right triangle to \((3,3)\) as shown in this graph.
Count the rise— since it goes down, it is negative.	The rise is \(-2.\)
Count the run.	The run is 3.
Use the slope formula.	\(m=\frac{\text{rise}}{\text{run}}\)
Substitute the values of the rise and run.	\(m=\frac{-2}{3}\)
Simplify.	\(m=-\frac{2}{3}\)
	The slope of the line is \(-\frac{2}{3}.\)
	So y decreases by 2 units as x increases by 3 units.

Solve for y.	\(x+3y=9\)
Subtract x from each side.
Divide both sides by 3.
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.

	\(y=mx+b\)
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 the graph.
Identify the rise over the run.	\(m=\frac{-1}{1}\)
Count out the rise and run to mark the second point.	rise \(-1\), run 1

Find Sam’s cost for a week when he drives 0 miles.	\(C=0.5m+60\)
Find \(C\) when \(m=0.\)	\(C=0.5(0)+60\)
Simplify.	\(C=60\)
	Sam’s costs are $60 when he drives 0 miles.

Find the cost for a week when he drives 250 miles.	\(C=0.5m+60\)
Find \(C\) when \(m=250.\)	\(C=0.5(250)+60\)
Simplify.	\(C=185\)
	Sam’s costs are $185 when he drives 250 miles.

	\(m=0.5\)
Rewrite as a fraction.	\(m=\frac{0.5}{1}\)
Multiply numerator and denominator by 100.	\(m=\frac{0.5(100)}{1(100)}\)
Simplify.	\(m=\frac{50}{100}\)

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

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

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

Solve the equations for \(y\text{.}\)	\(\begin{array}{ll}7x+2y & =3 \\ 2y & =-7x+3 \\ \frac{2y}{2} & =\frac{-7x+3}{2} \\ y & =-\frac{7}{2}x+\frac{3}{2}\end{array}\)	\(\ \begin{array}{ll}2x+7y & =5 \\ 7y & =-2x+5 \\ \frac{7y}{7} & =\frac{-2x+5}{7} \\ y & =-\frac{2}{7}x+\frac{5}{7}\end{array}\)
Identify the slope of each line.	\(\ \begin{array}{ll}y & =mx+b \\ {m}_{1} & =-\frac{7}{2}\end{array}\)	\(\ \begin{array}{ll}y & =mx+b \\ {m}_{1} & =-\frac{2}{7}\end{array}\)