What if you were given two points that a line passes through like (-1, 0) and (2, 2)? How could you find the slope of that line? After completing this Concept, you'll be able to find the slope of any line.

### Watch This

CK-12 Foundation: 0406S Slope of a Line (H264)

### Guidance

Wheelchair ramps at building entrances must have a slope between \begin{align*}\frac{1}{16}\end{align*}

We come across many examples of slope in everyday life. For example, a slope is in the pitch of a roof, the grade or incline of a road, or the slant of a ladder leaning on a wall. In math, we use the word **slope** to define steepness in a particular way.

\begin{align*}\text{Slope} = \frac{\text{distance moved vertically}}{\text{distance moved horizontally}}\end{align*}

To make it easier to remember, we often word it like this:

\begin{align*}\text{Slope} = \frac{\text{rise}}{\text{run}}\end{align*}

In the picture above, the slope would be the ratio of the **height** of the hill to the horizontal **length** of the hill. In other words, it would be \begin{align*}\frac{3}{4}\end{align*}, or 0.75.

If the car were driving to the **right** it would **climb** the hill - we say this is a positive slope. Any time you see the graph of a line that goes up as you move to the right, the slope is **positive**.

If the car kept driving after it reached the top of the hill, it might go down the other side. If the car is driving to the **right** and **descending**, then we would say that the slope is **negative**.

Here’s where it gets tricky: If the car turned around instead and drove back down the left side of the hill, the slope of that side would still be positive. This is because the rise would be -3, but the run would be -4 (think of the \begin{align*}x-\end{align*}axis - if you move from right to left you are moving in the **negative** \begin{align*}x-\end{align*}direction). That means our slope ratio would be \begin{align*}\frac{-3}{-4}\end{align*}, and the negatives cancel out to leave 0.75, the same slope as before. In other words, the slope of a line is the same no matter which direction you travel along it.

**Find the Slope of a Line**

A simple way to find a value for the slope of a line is to draw a right triangle whose hypotenuse runs along the line. Then we just need to measure the distances on the triangle that correspond to the rise (the vertical dimension) and the run (the horizontal dimension).

#### Example A

*Find the slopes for the three graphs shown.*

**Solution**

There are already right triangles drawn for each of the lines - in future problems you’ll do this part yourself. Note that it is easiest to make triangles whose vertices are **lattice points** (i.e. points whose coordinates are all integers).

a) The rise shown in this triangle is 4 units; the run is 2 units. The slope is \begin{align*}\frac{4}{2}= 2\end{align*}.

b) The rise shown in this triangle is 4 units, and the run is also 4 units. The slope is \begin{align*}\frac{4}{4}= 1\end{align*}.

c) The rise shown in this triangle is 2 units, and the run is 4 units. The slope is \begin{align*}\frac{2}{4}= \frac{1}{2}\end{align*}.

#### Example B

*Find the slope of the line that passes through the points (1, 2) and (4, 7).*

**Solution**

We already know how to graph a line if we’re given two points: we simply plot the points and connect them with a line. Here’s the graph:

Since we already have coordinates for the vertices of our right triangle, we can quickly work out that the rise is \begin{align*}7 - 2 = 5\end{align*} and the run is \begin{align*}4 - 1 = 3\end{align*} (see diagram). So the slope is \begin{align*}\frac{7-2}{4-1} = \frac{5}{3}\end{align*}.

If you look again at the calculations for the slope, you’ll notice that the 7 and 2 are the \begin{align*}y-\end{align*}coordinates of the two points and the 4 and 1 are the \begin{align*}x-\end{align*}coordinates. This suggests a pattern we can follow to get a general formula for the slope between two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*}:

Slope between \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2) = \frac{y_2 - y_1} { x_2- x_1} \end{align*}

**or** \begin{align*}m= \frac{\Delta y}{\Delta x}.\end{align*}

In the second equation the letter \begin{align*}m\end{align*} denotes the slope (this is a mathematical convention you’ll see often) and the Greek letter delta \begin{align*}(\Delta)\end{align*} means ** change**. So another way to express slope is

*change in*\begin{align*}y\end{align*} divided by

*change in*\begin{align*}x\end{align*}. In the next section, you’ll see that it doesn’t matter which point you choose as point 1 and which you choose as point 2.

**Find the Slopes of Horizontal and Vertical lines**

#### Example C

*Determine the slopes of the two lines on the graph below.*

**Solution**

There are 2 lines on the graph: \begin{align*}A (y = 3)\end{align*} and \begin{align*}B (x = 5)\end{align*}.

Let’s pick 2 points on line \begin{align*}A\end{align*}—say, \begin{align*}(x_1, y_1) = (-4, 3)\end{align*} and \begin{align*}(x_2, y_2) = (5, 3)\end{align*}—and use our equation for slope:

\begin{align*}m= \frac{y_2 - y_1} {x_2 - x_1} = \frac{(3)-(3)}{(5)-(-4)} = \frac {0}{9} = 0.\end{align*}

If you think about it, this makes sense - if \begin{align*}y\end{align*} doesn’t change as \begin{align*}x\end{align*} increases then there is no slope, or rather, the slope is zero. You can see that this must be true for all horizontal lines.

Horizontal lines (\begin{align*}y\end{align*} = *constant*) all have a slope of 0.

Now let’s consider line \begin{align*}B\end{align*}. If we pick the points \begin{align*}(x_1 , y_1) = (5, -3)\end{align*} and \begin{align*}(x_2 , y_2) = (5, 4)\end{align*}, our slope equation is \begin{align*}m = \frac{y_2 -y_1}{x_2 - x_1} = \frac{(4)-(-3)}{(5)-(5)} = \frac{7}{0}\end{align*}. But dividing by zero isn’t allowed!

In math we often say that a term which involves division by zero is **undefined.** (Technically, the answer can also be said to be infinitely large—or infinitely small, depending on the problem.)

Vertical lines \begin{align*}(x =\end{align*} *constant*) all have an infinite (or undefined) slope.

Watch this video for help with the Examples above.

CK-12 Foundation: The Slope of a Line

### Guided Practice

*Find the slopes of the lines on the graph below.*

**Solution**

Look at the lines - they both slant down (or decrease) as we move from left to right. Both these lines have **negative slope.**

The lines don’t pass through very many convenient lattice points, but by looking carefully you can see a few points that look to have integer coordinates. These points have been circled on the graph, and we’ll use them to determine the slope. We’ll also do our calculations twice, to show that we get the same slope whichever way we choose point 1 and point 2.

For Line \begin{align*}A\end{align*}:

\begin{align*}&(x_1, y_1) = (-6, 3) \qquad (x_2, y_2) = (5, -1) && (x_1, y_1) = (5, -1) \qquad (x_2, y_2) = (-6, 3)\\ & m = \frac{y_2-y_1}{x_2-x_1} = \frac{(-1)-(3)}{(5)-(-6)} = \frac{-4}{11} \approx -0.364 && m = \frac{y_2-y_1}{x_2-x_1} = \frac{(3)-(-1)}{(-6)-(5)} = \frac{4}{-11} \approx -0.364\end{align*}

For Line \begin{align*}B\end{align*}

\begin{align*}&(x_1, y_1) = (-4, 6) \qquad (x_2, y_2) = (4, -5) && (x_1, y_1) = (4, -5) \qquad (x_2, y_2) = (-4, 6)\\ & m = \frac{y_2-y_1}{x_2-x_1} = \frac{(-5)-(6)}{(4)-(-4)} = \frac{-11}{8} = -1.375 && m = \frac{y_2-y_1}{x_2-x_1} = \frac{(6)-(-5)}{(-4)-(4)} = \frac{11}{-8} = -1.375\end{align*}

You can see that whichever way round you pick the points, the answers are the same. Either way, **Line \begin{align*}A\end{align*} has slope -0.364, and Line \begin{align*}B\end{align*} has slope -1.375.**

### Explore More

Use the slope formula to find the slope of the line that passes through each pair of points.

- (-5, 7) and (0, 0)
- (-3, -5) and (3, 11)
- (3, -5) and (-2, 9)
- (-5, 7) and (-5, 11)
- (9, 9) and (-9, -9)
- (3, 5) and (-2, 7)
- (2.5, 3) and (8, 3.5)

For each line in the graphs below, use the points indicated to determine the slope.

- For each line in the graphs above, imagine another line with the same slope that passes through the point (1, 1), and name one more point on that line.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 4.6.