Understand Slope of a Line

As we’ve been graphing linear equations, we’ve seen that some lines slant up as they go from left to right and some lines slant down. Some lines are very steep and some lines are flatter. What determines whether a line slants up or down, and if its slant is steep or flat?

The steepness of the slant of a line is called the slope of the line. The concept of slope has many applications in the real world. The pitch of a roof and the grade of a highway or wheelchair ramp are just some examples in which you literally see slopes. And when you ride a bicycle, you feel the slope as you pump uphill or coast downhill.

Use Geoboards to Model Slope

In this section, we will explore the concepts of slope.

Using rubber bands on a geoboard gives a concrete way to model lines on a coordinate grid. By stretching a rubber band between two pegs on a geoboard, we can discover how to find the slope of a line.

We’ll start by stretching a rubber band between two pegs to make a line as shown in Figure 1.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 4 row 2.

Does it look like a line?

Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure 2. We carefully make a $90 °$ angle around the third peg, so that one side is vertical and the other is horizontal.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 4,and column 4 row 2.

To find the slope of the line, we measure the distance along the vertical and horizontal legs of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in Figure 3.

This figure shows two arrows. The first arrow is vertical and is labeled “rise”. The second arrow begins at the end of the first arrow extending to the right and is labeled “run”.

To help remember the terms, it may help to think of the images shown in Figure 4.

The figure shows an image of a hot air balloon signifying rise as the balloon rises straight up, similar to a y-axis. The second image is of a person jogging, signifying run as the person runs as if they are on an x-axis

On our geoboard, the rise is $2$ units because the rubber band goes up $2$ spaces on the vertical leg. See Figure 5.

What is the run? Be sure to count the spaces between the pegs rather than the pegs themselves! The rubber band goes across $3$ spaces on the horizontal leg, so the run is $3$ units.

The slope of a line is the ratio of the rise to the run. So the slope of our line is $\frac{2}{3} .$ In mathematics, the slope is always represented by the letter $m .$

What is the slope of the line on the geoboard in Figure 5?

m = \frac{rise}{run}

m = \frac{2}{3}

The line has slope \frac{2}{3} .

When we work with geoboards, it is a good idea to get in the habit of starting at a peg on the left and connecting to a peg to the right. Then we stretch the rubber band to form a right triangle.

If we start by going up the rise is positive, and if we stretch it down the rise is negative. We will count the run from left to right, just like you read this paragraph, so the run will be positive.

Since the slope formula has rise over run, it may be easier to always count out the rise first and then the run.

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 5 row 2.

Solution

Use the definition of slope.

$m = \frac{rise}{run}$

Start at the left peg and make a right triangle by stretching the rubber band up and to the right to reach the second peg.

Count the rise and the run as shown.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 5,and column 5 row 2.

$\begin{matrix} The rise is 3 units . & m = \frac{3}{run} \\ The run is 4 units . & m = \frac{3}{4} \\ The slope is \frac{3}{4} . \end{matrix}$

What is the slope of the line on the geoboard shown?

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 3 and the point in column 4 row 4.

Solution

Use the definition of slope.

$m = \frac{rise}{run}$

Start at the left peg and make a right triangle by stretching the rubber band to the peg on the right. This time we need to stretch the rubber band down to make the vertical leg, so the rise is negative.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 3, column 1 row 4,and column 4 row 4.

$\begin{matrix} The rise is −1 . & m = \frac{−1}{run} \\ The run is 3 . & m = \frac{−1}{3} \\ m = - \frac{1}{3} \\ The slope is - \frac{1}{3} . \end{matrix}$

Notice that in the first example, the slope is positive and in the second example the slope is negative. Do you notice any difference in the two lines shown in Figure 6.

The figure shows two grids of evenly spaced dots. There are 5 rows and 5 columns in each. In the left grid A, a rubberband loops to connect the point in column 1, row 1 and the point in column 5, row 4. In the right grid B, a rubber band loops to connect the point in column 1, row 4 and the point in column 4, row 2.

As you read from left to right, the line in Figure A, is going up; it has positive slope. The line Figure B is going down; it has negative slope.

This image shows two arrows: the left arrow is labeled positive slope and points upward towards the right. The right arrow is labeled negative slope and points downward towards the right.

Use a geoboard to model a line with slope $\frac{1}{2} .$

Solution

To model a line with a specific slope on a geoboard, we need to know the rise and the run.

This table illustrates the steps and corresponding mathematical expressions for using and manipulating the slope formula.
Use the slope formula.	$m = \frac{rise}{run}$
Replace $m$ with $\frac{1}{2}$ .	$\frac{1}{2} = \frac{rise}{run}$

So, the rise is $1$ unit and the run is $2$ units.

Start at a peg in the lower left of the geoboard. Stretch the rubber band up $1$ unit, and then right $2$ units.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 3, column 1 row 4,and column 3 row 3.

The hypotenuse of the right triangle formed by the rubber band represents a line with a slope of $\frac{1}{2} .$

Use a geoboard to model a line with slope $\frac{−1}{4},$

Solution

This table illustrates the steps involved in using the slope formula and replacing the slope variable with a specific numerical value.
Use the slope formula.	$m = \frac{rise}{run}$
Replace $m$ with $− \frac{1}{4}$ .	$− \frac{1}{4} = \frac{rise}{run}$

So, the rise is $−1$ and the run is $4 .$

Since the rise is negative, we choose a starting peg on the upper left that will give us room to count down. We stretch the rubber band down $1$ unit, then to the right $4$ units.
The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style triangle connecting three of the three points at column 1 row 2, column 1 row 3,and column 5 row 3.

The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is $- \frac{1}{4} .$

Find the Slope of a Line from its Graph

Now we’ll look at some graphs on a coordinate grid to find their slopes. The method will be very similar to what we just modeled on our geoboards.

To find the slope, we must count out the rise and the run. But where do we start?

We locate any two points on the line. We try to choose points with coordinates that are integers to make our calculations easier. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.

Find the slope of the line shown:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair 5, 1” and “ordered pair 0, -3”.

Solution

Locate two points on the graph, choosing points whose coordinates are integers. We will use $(0, −3)$ and $(5, 1) .$

Starting with the point on the left, $(0, −3),$ sketch a right triangle, going from the first point to the second point, $(5, 1) .$

Steps and calculations to determine the slope of a line using the rise over run method, with an illustrative graph.

Count the rise on the vertical leg of the triangle.	The rise is 4 units.
Count the run on the horizontal leg.	The run is 5 units.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{4}{5}$
	The slope of the line is $\frac{4}{5}$ .

Notice that the slope is positive since the line slants upward from left to right.

Find the slope of the line shown:

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 4, 2” and “ordered pair 3, 3”.

Solution

Locate two points on the graph. Look for points with coordinates that are integers. We can choose any points, but we will use (0, 5) and (3, 3). Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point.

Illustrates the step-by-step calculation of a line's slope using rise and run, featuring a graph, the formula, and numerical example.

Count the rise – it is negative.	The rise is −2.
Count the run.	The run is 3.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{−2}{3}$
Simplify.	$m = - \frac{2}{3}$
	The slope of the line is $- \frac{2}{3} .$

Notice that the slope is negative since the line slants downward from left to right.

What if we had chosen different points? Let’s find the slope of the line again, this time using different points. We will use the points $(−3, 7)$ and $(6, 1) .$
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 9. The y-axis runs from -1 to 7. A line passes through the points “ordered pair 0, 5” and “ordered pair 3, 3”. .

Starting at $(−3, 7),$ sketch a right triangle to $(6, 1) .$

Step-by-step calculation of the slope of a line using the rise and run method, illustrated by a graph.

Count the rise.	The rise is −6.
Count the run.	The run is 9.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{−6}{9}$
Simplify the fraction.	$m = - \frac{2}{3}$
	The slope of the line is $- \frac{2}{3} .$

It does not matter which points you use—the slope of the line is always the same. The slope of a line is constant!

The lines in the previous examples had $y$ -intercepts with integer values, so it was convenient to use the y-intercept as one of the points we used to find the slope. In the next example, the $y$ -intercept is a fraction. The calculations are easier if we use two points with integer coordinates.

Find the slope of the line shown: The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 8. A line passes through the points “ordered pair 2, 3” and “ordered pair 7, 6”.

Solution

This table outlines instructions and provides examples for working with points and sketching on a graph.
Locate two points on the graph whose coordinates are integers.	$(2, 3)$ and $(7, 6)$
Which point is on the left?	$(2, 3)$
Starting at $(2, 3)$ , sketch a right angle to $(7, 6)$ as shown below.

Steps to calculate the slope of a line from a graph using the rise and run method, including an illustrative image and formula.

Count the rise.	The rise is 3.
Count the run.	The run is 5.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values of the rise and run.	$m = \frac{3}{5}$
	The slope of the line is $\frac{3}{5} .$

Find the Slope of Horizontal and Vertical Lines

Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.

horizontal line $y = b;$ all the $y$ -coordinates are the same.
vertical line $x = a;$ all the $x$ -coordinates are the same.

So how do we find the slope of the horizontal line $y = 4 ?$ One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens in Figure 8. We’ll use the two points $(0, 4)$ and $(3, 4)$ to count the rise and run.

The graph shows the x y-coordinate plane. The x-axis runs from -1 to 5. The y-axis runs from -1 to 7. A horizontal line passes through the labeled points “ordered pair 0, 4” and “ordered pair 3, 4”.

This table illustrates the calculation of slope, demonstrating that a zero rise over a given run results in a zero slope.
What is the rise?	The rise is 0.
What is the run?	The run is 3.
What is the slope?	$m = \frac{rise}{run}$
	$m = \frac{0}{3}$
	$m = 0$

The slope of the horizontal line $y = 4$ is $0 .$

All horizontal lines have slope $0$ . When the $y$ -coordinates are the same, the rise is $0$ .

Now we’ll consider a vertical line, such as the line $x = 3$ , shown in Figure 9. We’ll use the two points $(3, 0)$ and $(3, 2)$ to count the rise and run.

The graph shows the x y-coordinate plane. Both axes run from -5 to 5. A vertical line passes through the labeled points “ordered pair 3, 2” and “ordered pair 3, 0”.

Slope calculation example defining rise and run, demonstrating an undefined slope with a run of zero.
What is the rise?	The rise is 2.
What is the run?	The run is 0.
What is the slope?	$m = \frac{rise}{run}$
	$m = \frac{2}{0}$

But we can’t divide by $0 .$ Division by $0$ is undefined. So we say that the slope of the vertical line $x = 3$ is undefined. The slope of all vertical lines is undefined, because the run is $0 .$

Find the slope of each line:

ⓐ $x = 8$
ⓑ $y = −5$

Solution

ⓐ $x = 8$

This is a vertical line, so its slope is undefined.

ⓑ $y = −5$

This is a horizontal line, so its slope is $0 .$

Use the Slope Formula to find the Slope of a Line between Two Points

Sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing.

Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair $(x, y)$ gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol $(x, y)$ be used to represent two different points?

Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable.

$(x_{1}, y_{1}) read x sub 1, y sub 1$
$(x_{2}, y_{2}) read x sub 2, y sub 2$

We will use $(x_{1}, y_{1})$ to identify the first point and $(x_{2}, y_{2})$ to identify the second point. If we had more than two points, we could use $(x_{3}, y_{3}), (x_{4}, y_{4}),$ and so on.

To see how the rise and run relate to the coordinates of the two points, let’s take another look at the slope of the line between the points $(2, 3)$ and $(7, 6)$ in Figure 10.

The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6. It os labeled x sub 2 minus x sub 1, 7 minus 2, 5.

Since we have two points, we will use subscript notation.

\overset{x_{1}, y_{1}}{(2, 3)} \overset{x_{2}, y_{2}}{(7, 6)}

On the graph, we counted the rise of $3 .$ The rise can also be found by subtracting the $y -coordinates$ of the points.

\begin{matrix} y_{2} - y_{1} \\ 6 - 3 \\ 3 \end{matrix}

We counted a run of $5 .$ The run can also be found by subtracting the $x -coordinates .$

\begin{matrix} x_{2} - x_{1} \\ 7 - 2 \\ 5 \end{matrix}

Step-by-step derivation of the slope formula, progressing from the basic rise/run definition to the coordinate-based formula.
We know	$m = \frac{rise}{run}$
So	$m = \frac{3}{5}$
We rewrite the rise and run by putting in the coordinates.	$m = \frac{6 - 3}{7 - 2}$
But 6 is the $y$ -coordinate of the second point, $y_{2}$ and 3 is the $y$ -coordinate of the first point $y_{1}$ . So we can rewrite the rise using subscript notation.	$m = \frac{y_{2} - y_{1}}{7 - 2}$
Also 7 is the $x$ -coordinate of the second point, $x_{2}$ and 2 is the $x$ -coordinate of the first point $x_{2}$ . So we rewrite the run using subscript notation.	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

We’ve shown that $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is really another version of $m = \frac{rise}{run} .$ We can use this formula to find the slope of a line when we have two points on the line.

Say the formula to yourself to help you remember it:

Slope is y of the second point minus y of the first point

over

x of the second point minus x of the first point.

Find the slope of the line between the points $(1, 2)$ and $(4, 5) .$

Solution

This table illustrates the step-by-step process of calculating the slope between two points, (1,2) and (4,5), using the slope formula.
We’ll call $(1, 2)$ point #1 and $(4, 5)$ point #2.	$\overset{x_{1}, y_{1}}{(1, 2)} and \overset{x_{2}, y_{2}}{(4, 5)}$
Use the slope formula.	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Substitute the values in the slope formula:
$y$ of the second point minus $y$ of the first point	$m = \frac{5 - 2}{x_{2} - x_{1}}$
$x$ of the second point minus $x$ of the first point	$m = \frac{5 - 2}{4 - 1}$
Simplify the numerator and the denominator.	$m = \frac{3}{3}$
	$m = 1$

Let’s confirm this by counting out the slope on the graph.

The rise is $3$ and the run is $3,$ so

$\begin{matrix} m = \frac{rise}{run} \\ m = \frac{3}{3} \\ m = 1 \end{matrix}$

How do we know which point to call #1 and which to call #2? Let’s find the slope again, this time switching the names of the points to see what happens. Since we will now be counting the run from right to left, it will be negative.

Step-by-step calculation of the slope between two points (4,5) and (1,2) using the slope formula.
We’ll call $(4, 5)$ point #1 and $(1, 2)$ point #2.	$\overset{x_{1}, y_{1}}{(4, 5)} and \overset{x_{2}, y_{2}}{(1, 2)}$
Use the slope formula.	$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Substitute the values in the slope formula:
$y$ of the second point minus $y$ of the first point	$m = \frac{2 - 5}{x_{2} - x_{1}}$
$x$ of the second point minus $x$ of the first point	$m = \frac{2 - 5}{1 - 4}$
Simplify the numerator and the denominator.	$m = \frac{−3}{−3}$
	$m = 1$

The slope is the same no matter which order we use the points.

Find the slope of the line through the points $(−2, −3)$ and $(−7, 4) .$

Solution

Step-by-step calculation of the slope between two points, (-2,-3) and (-7,4), illustrating the application of the slope formula.
We’ll call $(−2, −3)$ point #1 and $(−7, 4)$ point #2.	$\overset{x_{1}, y_{1}}{(−2, −3)} and \overset{x_{2}, y_{2}}{(−7, 4)}$
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	$m = \frac{4 - (−3)}{x_{2} - x_{1}}$
$x$ of the second point minus $x$ of the first point	$m = \frac{4 - (−3)}{−7 - (−2)}$
Simplify.	$m = \frac{7}{−5}$
	$m = - \frac{7}{5}$

Let’s confirm this on the graph shown.

$\begin{matrix} m = \frac{rise}{run} \\ m = \frac{−7}{5} \\ m = - \frac{7}{5} \end{matrix}$

Graph a Line Given a Point and the Slope

In this chapter, we graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.

Another method we can use to graph lines is the point-slope method. Sometimes, we will be given one point and the slope of the line, instead of its equation. When this happens, we use the definition of slope to draw the graph of the line.

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

Solution

Plot the given point, $(1, −1) .$
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -3 to 4. A labeled point is drawn at “ordered pair 1, -1”.

Use the slope formula $m = \frac{rise}{run}$ to identify the rise and the run.

\begin{matrix} m = \frac{3}{4} \\ \frac{rise}{run} = \frac{3}{4} \\ rise = 3 \\ run = 4 \end{matrix}

Starting at the point we plotted, count out the rise and run to mark the second point. We count $3$ units up and $4$ units right.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. Two line segments are drawn. A vertical line segment connects the points “ordered pair 1, -1” and “order pair “1, 2”. It is labeled “3”. A horizontal line segment starts at the top of the vertical line segment and goes to the right, connecting the points “ordered pair 1, 2” and “ordered pair 5, 2”. It is labeled “4”.

Then we connect the points with a line and draw arrows at the ends to show it continues.

We can check our line by starting at any point and counting up $3$ and to the right $4 .$ We should get to another point on the line.

Graph the line with $y$ -intercept $(0, 2)$ and slope $m = - \frac{2}{3} .$

Solution

Plot the given point, the $y$ -intercept $(0, 2) .$
The graph shows the x y-coordinate plane. The x-axis runs from -1 to 4. The y-axis runs from -1 to 3. The point “ordered pair 0, 2” is labeled.

Use the slope formula $m = \frac{rise}{run}$ to identify the rise and the run.

\begin{matrix} m = - \frac{2}{3} \\ \frac{rise}{run} = \frac{−2}{3} \\ rise = –2 \\ run = 3 \end{matrix}

Starting at $(0, 2),$ count the rise and the run and mark the second point.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. A vertical line segment connects points at “ordered pair 0, 2” and “ordered pair 0, 0” and is labeled “down 2”. A horizontal line segment connects “ordered pair 0, 0” and “ordered pair 0, 3” and is labeled “right 3”.

Connect the points with a line.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. Two labeled points are drawn at “ordered pair 0, 2” and “ordered pair 3, 0”. A line passes through the points. Two line segments form a triangle with the line. A vertical line connects “ordered pair 0, 2” and “ordered pair 0, 0 ”. A horizontal line segment connects “ordered pair 0, 0” and “ordered pair 3, 0”.

Graph the line passing through the point $(−1, −3)$ whose slope is $m = 4 .$

Solution

Plot the given point.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. The point “ordered pair -1, -3” is labeled.

This table illustrates the steps to identify the rise and run from a given slope value, showing the mathematical representation at each stage.
Identify the rise and the run.	$m = 4$
Write 4 as a fraction.	$\frac{rise}{run} = \frac{4}{1}$
	$rise = 4 run = 1$

Count the rise and run.
The graph shows the x y-coordinate plane. Both axes run from -5 to 5. The y-axis runs from -4 to 2. A vertical line segment connects points at “ordered pair -1, -3” and “ordered pair -1, 1” and is labeled “up 4”. A horizontal line segment connects “ordered pair -1, 1” and “ordered pair 0, 1” and is labeled “over 1”.

Mark the second point. Connect the two points with a line.

Solve Slope Applications

At the beginning of this section, we said there are many applications of slope in the real world. Let’s look at a few now.

The pitch of a building’s roof is the slope of the roof. Knowing the pitch is important in climates where there is heavy snowfall. If the roof is too flat, the weight of the snow may cause it to collapse. What is the slope of the roof shown?

This figure shows a house with a sloped roof. The roof on one half of the building is labeled “pitch of the roof”. There is a line segment with arrows at each end measuring the vertical length of the roof and is labeled “rise = 9 feet”. There is a line segment with arrows at each end measuring the horizontal length of the root and is labeled “run = 18 feet”.

Solution

Illustrates the step-by-step calculation of a roof's slope using the rise and run formula.
Use the slope formula.	$m = \frac{rise}{run}$
Substitute the values for rise and run.	$m = \frac{9 ft}{18 ft}$
Simplify.	$m = \frac{1}{2}$
	The slope of the roof is $\frac{1}{2}$ .

Have you ever thought about the sewage pipes going from your house to the street? Their slope is an important factor in how they take waste away from your house.

Sewage pipes must slope down $\frac{1}{4}$ inch per foot in order to drain properly. What is the required slope?

This figure shows a right triangle. The short leg is vertical and is labeled “1 over 4 inch”. The long leg labeled “1 foot”.

Solution

Step-by-step calculation of a pipe's slope, illustrating the application of the slope formula, unit conversion, and final simplification to determine the result.
Use the slope formula.	$m = \frac{rise}{run}$
	$m = \frac{- \frac{1}{4} in .}{1 ft}$
	$m = \frac{- \frac{1}{4} in .}{1 ft}$
Convert 1 foot to 12 inches.	$m = \frac{- \frac{1}{4} in .}{12 in.}$
Simplify.	$m = - \frac{1}{48}$
	The slope of the pipe is $- \frac{1}{48} .$

Key Concepts

Find the slope from a graph
1. Locate two points on the line whose coordinates are integers.
2. Starting with the point on the left, sketch a right triangle, with the hypotenuse 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, $m = \frac{rise}{run}$
Slope of a Horizontal Line
- The slope of a horizontal line, $y = b$ , is 0.
Slope of a Vertical Line
- The slope of a vertical line, $x = a$ , is undefined.
Slope Formula
- The slope of the line between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$
Graph a line given a point and a slope.
1. Plot the given point.
2. Use the slope formula 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.

Section Exercises

Practice Makes Perfect

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

Solution

$\frac{1}{4}$

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 4 and the point in column 5 row 2.

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 1 and the point in column 4 row 4.

Solution

$- \frac{3}{2}$

In the following exercises, model each slope. Draw a picture to show your results.

$\frac{2}{3}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 2 row 5 and the point in column 5 row 3.

$\frac{3}{4}$

$\frac{1}{4}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 5 row 3.

$\frac{4}{3}$

$- \frac{1}{2}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 4 and the point in column 3 row 5.

$- \frac{3}{4}$

$- \frac{2}{3}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4.

$- \frac{3}{2}$

Find the Slope of a Line from its Graph

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

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, -4” and “ordered pair 10, 0”.

Solution

$\frac{2}{5}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, -5” and “ordered pair 3, 0”.

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from -12 to 12. A line passes through the points “ordered pair 0, -1” and “ordered pair 1, 0”.

Solution

$\frac{5}{4}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 3” and “ordered pair 6, 0”.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 2” and “ordered pair 6, 0”.

Solution

$- \frac{1}{3}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 6” and “ordered pair 8, 0”.

Solution

$- \frac{3}{4}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -1, 0” and “ordered pair 0, -1”.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -4, 0” and “ordered pair -4, 6”.

Solution

$\frac{3}{4}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. The y-axis runs from -10 to 10. A line passes through the points “ordered pair -2, 0” and “ordered pair 4, 4”.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 4” and “ordered pair 4, -6”.

Solution

$- \frac{5}{2}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair -8, 8” and “ordered pair 8, -4”.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 1, 4” and “ordered pair 7, 0”.

Solution

$- \frac{2}{3}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 0, 3” and “ordered pair 7, 0”.

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 2, 0” and “ordered pair 10, 4”.

Solution

$\frac{1}{4}$

The graph shows the x y-coordinate plane. The x-axis runs from -10 to 10. A line passes through the points “ordered pair 6, 2” and “ordered pair 0, -3”.

Find the Slope of Horizontal and Vertical Lines

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

$y = 3$

Solution

$y = 1$

$x = 4$

Solution

undefined

$x = 2$

$y = −2$

Solution

$y = −3$

$x = −5$

Solution

undefined

$x = −4$

Use the Slope Formula to find the Slope of a Line between Two Points

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

$(1, 4), (3, 9)$

Solution

$\frac{5}{2}$

$(2, 3), (5, 7)$

$(0, 3), (4, 6)$

Solution

$\frac{3}{4}$

$(0, 1), (5, 4)$

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

Solution

$- \frac{5}{2}$

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

$(−3, 3), (2, −5)$

Solution

$- \frac{8}{5}$

$(−2, 4), (3, −1)$

$(−1, −2), (2, 5)$

Solution

$\frac{7}{3}$

$(−2, −1), (6, 5)$

$(4, −5), (1, −2)$

Solution

−1

$(3, −6), (2, −2)$

Graph a Line Given a Point and the Slope

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

$(1, −2); m = \frac{3}{4}$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 5, 1” and “ordered pair 1, -2”

$(1, −1); m = \frac{1}{2}$

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

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 2, 5” and “ordered pair 5, 4”.

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

$(−3, 4); m = - \frac{3}{2}$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -3, 4” and “ordered pair -1, 1”.

$(−2, 5); m = - \frac{5}{4}$

$(−1, −4); m = \frac{4}{3}$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 2,0” and “ordered pair -1, -4”.

$(−3, −5); m = \frac{3}{2}$

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

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 0, 3” and “ordered pair 5, 1”.

$(0, 5); m = - \frac{4}{3}$

$(−2, 0); m = \frac{3}{4}$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -2,0” and “ordered pair 2, 3”.

$(−1, 0); m = \frac{1}{5}$

$(−3, 3); m = 2$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair -3, 3” and “ordered pair -2, 5”.

$(−4, 2); m = 4$

$(1, 5); m = −3$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -12 to 12. The y-axis runs from 12 to -12. A line passes through the points “ordered pair 1, 5” and “ordered pair 2, 2”.

$(2, 3); m = −1$

Solve Slope Applications

In the following exercises, solve these slope applications.

Slope of a roof A fairly easy way to determine the slope is to take a $12-inch$ level and set it on one end on the roof surface. Then take a tape measure or ruler, and measure from the other end of the level down to the roof surface. You can use these measurements to calculate the slope of the roof. What is the slope of the roof in this picture?

The figure shows a wood board at a diagonal representing a side-view slice of a pitched roof. A vertical line segment with arrows on both ends measures the vertical change in height of the roof and is labeled “4 inches”. A level tool is in a horizontal position above the board and above it is a line segment with arrows on both ends labeled “12 inches”.

Solution

$\frac{1}{3}$

What is the slope of the roof shown?

The figure shows a diagonal side-view slice of a pitched roof. A ruler in vertical position is at the bottom of the roof segment and shows unit labels 1 through 8 and extends one further unit. A second ruler starts at the “7” label of the vertical ruler and extends horizontally until it hits the rising roof. The horizontal ruler has unit labels 1 through 11 and extends one further unit.

Road grade A local road has a grade of $6% .$ The grade of a road is its slope expressed as a percent.

ⓐ Find the slope of the road as a fraction and then simplify the fraction.
ⓑ What rise and run would reflect this slope or grade?

Solution

ⓐ $\frac{3}{50}$ ⓑ $rise = 3; run = 50$

Highway grade A local road rises $2$ feet for every $50$ feet of highway.

ⓐ What is the slope of the highway?
ⓑ The grade of a highway is its slope expressed as a percent. What is the grade of this highway?

Everyday Math

Wheelchair ramp The rules for wheelchair ramps require a maximum $1$ inch rise for a $12$ inch run.

ⓐ What run must the ramp have to accommodate a $24-inch$ rise to the door?
ⓑ Draw a model of this ramp.

Solution

ⓐ 288 inches (24 feet)
ⓑ Models will vary.

Wheelchair ramp A $1-inch$ rise for a $16-inch$ run makes it easier for the wheelchair rider to ascend the ramp.

ⓐ What run must the ramp have to easily accommodate a $24-inch$ rise to the door?
ⓑ Draw a model of this ramp.

Writing Exercises

What does the sign of the slope tell you about a line?

Solution

Answers will vary.

How does the graph of a line with slope $m = \frac{1}{2}$ differ from the graph of a line with slope $m = 2 ?$

Why is the slope of a vertical line undefined?

Solution

Answers will vary.

Explain how you can graph a line given a point and its slope.

Self Check

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

A self-assessment checklist for students to rate their understanding and confidence in various skills related to finding, graphing, and applying slope, with options: Confidently, With some help, or No-I don't get it!

ⓑ On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

Chapter Review Exercises

Use the Rectangular Coordinate System

Plot Points in a Rectangular Coordinate System

In the following exercises, plot each point in a rectangular coordinate system.

$(1, 3), (3, 1)$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from 6 to -6. The points “ordered pair 1,3” and “ordered pair 3,1” are plotted.

$(2, 5), (5, 2)$

In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.

ⓐ $(−1, −5)$
ⓑ $(−3, 4)$
ⓒ $(2, −3)$
ⓓ $(1, \frac{5}{2})$

Solution

ⓐ III
ⓑ II
ⓒ IV
ⓐ I

A Cartesian coordinate plane, also known as an xy-grid, displays four quadrants (I, II, III, IV) and four distinct points labeled a, b, c, and d, plotted within these quadrants.

ⓐ $(3, −2)$
ⓑ $(−4, −1)$
ⓒ $(−5, 4)$
ⓓ $(2, \frac{10}{3})$

Identify Points on a Graph

In the following exercises, name the ordered pair of each point shown in the rectangular coordinate system.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 5, 3, “b” at 2, -1, “c” at -3,-2, and “d” at -1,4.

Solution

ⓐ (5,3)
ⓑ (2,−1)
ⓒ (−3,−2)
ⓓ (−1,4)

The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at -2, 2, “b” at 3, 5, “c” at 4,-1, and “d” at -1,3.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 2, 0, “b” at 0, -5, “c” at -4,0, and “d” at 0,3.

Solution

ⓐ (2,0)
ⓑ (0,−5)
ⓒ (−4,0)
ⓓ (0,3)

The graph shows the x y-coordinate plane. The axes run from -7 to 7. “a” is plotted at 0, 4, “b” at 5, 0, “c” at 0,-1, and “d” at -3,0.

Verify Solutions to an Equation in Two Variables

In the following exercises, find the ordered pairs that are solutions to the given equation.

$5 x + y = 10$

ⓐ $(5, 1)$
ⓑ $(2, 0)$
ⓒ $(4, −10)$

Solution

ⓑ (2, 0),ⓒ (4, –10)

$y = 6 x - 2$

ⓐ $(1, 4)$
ⓑ $(\frac{1}{3}, 0)$
ⓒ $(6, −2)$

Complete a Table of Solutions to a Linear Equation in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

$y = 4 x - 1$

$x$	$y$	$(x, y)$
$0$
$1$
$−2$

Solution

$x$	$y$	$(x, y)$
$0$	$−1$	$(0, −1)$
$1$	$3$	$(1, 3)$
$−2$	$−9$	$(−2, −9)$

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

$x$	$y$	$(x, y)$
$0$
$1$
$−2$

$x + 2 y = 5$

$x$	$y$	$(x, y)$
	$0$
$1$
$−1$

Solution

$x$	$y$	$(x, y)$
$5$	$0$	$(5, 0)$
$1$	$2$	$(1, 2)$
$−1$	$3$	$(−1, 3)$

$3 x - 2 y = 6$

$x$	$y$	$(x, y)$
$0$
	$0$
$−2$

Find Solutions to a Linear Equation in Two Variables

In the following exercises, find three solutions to each linear equation.

$x + y = 3$

Solution

Answers will vary.

$x + y = −4$

$y = 3 x + 1$

Solution

Answers will vary.

$y = - x - 1$

Graphing Linear Equations

Recognize the Relation Between the Solutions of an Equation and its Graph

In each of the following exercises, an equation and its graph is shown. For each ordered pair, decide

ⓐ if the ordered pair is a solution to the equation.
ⓑ if the point is on the line.

$y = - x + 4$
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0, 4” and “ordered pair 4, 0”.

$(0, 4)$
$(−1, 3)$
$(2, 2)$
$(−2, 6)$

Solution

ⓐ yes ⓑ yes
ⓐ no ⓑ no
ⓐ yes ⓑ yes
ⓐ yes ⓑ yes

$y = \frac{2}{3} x - 1$
The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0, -1” and “ordered pair 3, 1”.

$(0, −1)$
$(3, 1)$
$(−3, −3)$
$(6, 4)$

Graph a Linear Equation by Plotting Points

In the following exercises, graph by plotting points.

$y = 4 x - 3$

Solution

The graph shows the x y-coordinate plane. Each axis runs from -6 to 6. A line passes through the points “ordered pair 1, 1” and “ordered pair 0, -3”.

$y = −3 x$

$2 x + y = 7$

Solution

The graph shows the x y-coordinate plane. Each axis runs from -6 to 6. A line passes through the points “ordered pair 1, 5” and “ordered pair 0, 7”.

Graph Vertical and Horizontal lines

In the following exercises, graph the vertical or horizontal lines.

$y = −2$

$x = 3$

Solution

The graph shows the x y-coordinate plane. Each axis runs from -6 to 6. A vertical line passes through the point “ordered pair 0, 3”.

Graphing with Intercepts

Identify the Intercepts on a Graph

In the following exercises, find the $x -$ and $y -intercepts .$

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0, 4” and “ordered pair -4, 0”.

Solution

(0,3) (3,0)

Find the Intercepts from an Equation of a Line

In the following exercises, find the intercepts.

$x + y = 5$

$x - y = −1$

Solution

(−1,0) (0,1)

$y = \frac{3}{4} x - 12$

$y = 3 x$

Solution

(0,0)

Graph a Line Using the Intercepts

In the following exercises, graph using the intercepts.

$- x + 3 y = 3$

$x + y = −2$

Solution

This answer graph shows the x y-coordinate plane. The x and y-axis each run from -6 to 6. The equation x plus y equals -2 is shown. A line passes through the intercepts with coordinates 0, –2 and –2, 0.

Choose the Most Convenient Method to Graph a Line

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

$x = 5$

$y = −3$

Solution

horizontal line

$2 x + y = 5$

$x - y = 2$

Solution

intercepts

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

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

Solution

plotting points

Understand Slope of a Line

Use Geoboards to Model Slope

In the following exercises, find the slope modeled on each geoboard.

Solution

$\frac{4}{3}$

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 4 row 4.

Solution

$- \frac{2}{3}$

In the following exercises, model each slope. Draw a picture to show your results.

$\frac{1}{3}$

$\frac{3}{2}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 5 and the point in column 3 row 2.

$- \frac{2}{3}$

$- \frac{1}{2}$

Solution

The figure shows a grid of evenly spaced dots. There are 5 rows and 5 columns. There is a rubber band style loop connecting the point in column 1 row 2 and the point in column 3 row 3.

Find the Slope of a Line from its Graph

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

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 0, 0” and “ordered pair 2, -6”.

Solution

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair -4, -4” and “ordered pair 5, -1”.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair -3, 6” and “ordered pair 5, 2”.

Solution

$- \frac{1}{2}$

Find the Slope of Horizontal and Vertical Lines

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

$y = 2$

$x = 5$

Solution

undefined

$x = −3$

$y = −1$

Solution

Use the Slope Formula to find the Slope of a Line between Two Points

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

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

$(−1, −1), (0, −5)$

Solution

−4

$(3, 5), (4, −1)$

$(−5, −2), (3, 2)$

Solution

$\frac{1}{2}$

Graph a Line Given a Point and the Slope

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

$(2, −2); m = \frac{5}{2}$

$(−3, 4); m = - \frac{1}{3}$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from -4 to 2. A line passes through the points “ordered pair -3, 4” and “ordered pair 1, 3”.

Solve Slope Applications

In the following exercise, solve the slope application.

A roof has rise $10$ feet and run $15$ feet. What is its slope?

Chapter Practice Test

Plot and label these points:

ⓐ $(2, 5)$
ⓑ $(−1, −3)$
ⓒ $(−4, 0)$
ⓓ $(3, −5)$
ⓔ $(−2, 1)$

Solution

The graph shows the x y-coordinate plane. The axes extend from -6 to 6. a is plotted at 2, 5, b at -1, -3, c at -4, 0, d at 3, -5, and e at -2,1.

Name the ordered pair for each point shown.

The graph shows the x y-coordinate plane. The axes extend from -7 to 7. A is plotted at -4, 1, B at 3, 2, C at 0, -2, D at -1, -4, and E at 4,-3.

Find the $x -intercept$ and $y -intercept$ on the line shown.

The graph shows the x y-coordinate plane. The x-axis runs from -7 to 7. The y-axis runs from -7 to 7. A line passes through the points “ordered pair 4, 0” and “ordered pair 0, -2”.

Solution

(4,0), (0,−2)

Find the $x -intercept$ and $y -intercept$ of the equation $3 x - y = 6 .$

Is $(1, 3)$ a solution to the equation $x + 4 y = 12 ?$ How do you know?

Solution

no; 1 + 4 · 3 ≠ 12

Complete the table to find four solutions to the equation $y = - x + 1 .$

$x$	$y$	$(x, y)$
$0$
$1$
$3$
$−2$

Complete the table to find three solutions to the equation $4 x + y = 8$

$x$	$y$	$(x, y)$
$0$
	$0$
$3$

Solution

$x$	$y$	$(x, y)$
$0$	$8$	$(0, 8)$
$2$	$0$	$(2, 0)$
$3$	$−4$	$(3, −4)$

In the following exercises, find three solutions to each equation and then graph each line.

$y = −3 x$

$2 x + 3 y = −6$

Solution

The graph shows the x y-coordinate plane. The x-axis runs from -6 to 6. The y-axis runs from -6 to 6. A line passes through the points “ordered pair 0, -2” and “ordered pair -3, 0”.

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

The graph shows the x y-coordinate plane. The axes run from -7 to 7. The y-axis runs from -5 to -4. A line passes through the points “ordered pair 6, 4” and “ordered pair 0, -3”.

The graph shows the x y-coordinate plane. The axes run from -7 to 7. A line passes through the points “ordered pair 3, 0” and “ordered pair 1, 5”.

Solution

$- \frac{5}{2}$

Use the slope formula to find the slope of the line between $(0, −4)$ and $(5, 2) .$

Find the slope of the line $y = 2 .$

Solution

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

A bicycle route climbs $20$ feet for $1,000$ feet of horizontal distance. What is the slope of the route?

Solution

$\frac{1}{50}$

slope of a line

The slope of a line is

m = \frac{rise}{run}

. The rise measures the vertical change and the run measures the horizontal change.