Learning Goal
By the end of this lesson I will be able to calculate the rate of change using rise over run of various lines.
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.
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CK12 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*}
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*}
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*}
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*}
Slope between \begin{align*}(x_1, y_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*}
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*}
Let’s pick 2 points on line \begin{align*}A\end{align*}
\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*}
Horizontal lines (\begin{align*}y\end{align*}
Now let’s consider line \begin{align*}B\end{align*}
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*}
Watch this video for help with the Examples above.
CK12 Foundation: The Slope of a Line
Vocabulary

Slope is a measure of change in the vertical direction for each step in the horizontal direction. Slope is often represented as “\begin{align*}m\end{align*}
m ”.  Slope can be expressed as \begin{align*}\frac{\text{rise}}{\text{run}}\end{align*}
riserun , or \begin{align*}\frac{\Delta y}{\Delta x}\end{align*}ΔyΔx .  The slope between two points \begin{align*}(x_1, y_1)\end{align*}
(x1,y1) and \begin{align*}(x_2, y_2)\end{align*}(x2,y2) is equal to \begin{align*}\frac{y_2  y_1}{x_2 x_1}\end{align*}y2−y1x2−x1 . 
Horizontal lines (where \begin{align*}y = a\end{align*}
y=a constant) all have a slope of 0. 
Vertical lines (where \begin{align*}x = a\end{align*}
x=a constant) all have an infinite (or undefined) slope.  The slope (or rate of change) of a distancetime graph is a velocity.
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_2y_1}{x_2x_1} = \frac{(1)(3)}{(5)(6)} = \frac{4}{11} \approx 0.364 && m = \frac{y_2y_1}{x_2x_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_2y_1}{x_2x_1} = \frac{(5)(6)}{(4)(4)} = \frac{11}{8} = 1.375 && m = \frac{y_2y_1}{x_2x_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.
Practice
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.