What if you were given the coordinates of two points? How would you determine the steepness of the line they form? After completing this Concept, you'll be able to find the slope of a line through two points.

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CK-12 Slope in the Coordinate Plane

### Guidance

Recall from Algebra I that **slope** is the measure of the steepness of a line. Two points \begin{align*}(x_1, y_1)\end{align*} and \begin{align*}(x_2, y_2)\end{align*} have a slope of \begin{align*}m = \frac{(y_2-y_1)}{(x_2-x_1)}\end{align*}. You might have also learned slope as \begin{align*}\frac{rise}{run}\end{align*}. This is a great way to remember the formula. Also remember that if an equation is written in slope-intercept form, \begin{align*}y=mx+b\end{align*}, then \begin{align*}m\end{align*} is always the slope of the line.

Slopes can be positive, negative, zero, or undefined as shown in the pictures below:

**Positive:**

**Negative:**

**Zero:**

**Undefined:**

#### Example A

What is the slope of the line through (2, 2) and (4, 6)?

Use (2, 2) as \begin{align*}(x_1, y_1)\end{align*} and (4, 6) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m=\frac{6-2}{4-2} = \frac{4}{2} = 2\end{align*}

#### Example B

Find the slope between (-8, 3) and (2, -2).

Use (-8, 3) as \begin{align*}(x_1, y_1)\end{align*} and (2, -2) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m = \frac{-2-3}{2-(-8)} = \frac{-5}{10} = -\frac{1}{2}\end{align*}

#### Example C

The picture shown is the California Incline, a short road that connects Highway 1 with Santa Monica. The length of the road is 1532 feet and has an elevation of 177 feet. *You may assume that the base of this incline is zero feet*. **Can you find the slope of the California Incline?**

In order to find the slope, we need to first find the horizontal distance in the triangle shown. This triangle represents the incline and the elevation. To find the horizontal distance, we need to use the Pythagorean Theorem (a concept you will be introduced to formally in a future lesson), \begin{align*}a^2+b^2 = c^2\end{align*}, where \begin{align*}c\end{align*} is the hypotenuse.

\begin{align*}177^2 + run^2 & = 1532^2\\ 31,329 + run^2 &= 2,347,024\\ run^2 & = 2,315,695\\ run & \approx 1521.75\end{align*}

The slope is then \begin{align*}\frac{177}{1521.75}\end{align*}, which is roughly \begin{align*}\frac{3}{25}\end{align*}.

CK-12 Slope in the Coordinate Plane

### Guided Practice

1. Find the slope between (-5, -1) and (3, -1).

2. What is the slope of the line through (3, 2) and (3, 6)?

3. Find the slope between (-5, 2) and (3, 4).

**Answers:**

1. Use (-5, -1) as \begin{align*}(x_1, y_1)\end{align*} and (3, -1) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m=\frac{-1-(-1)}{3-(-5)} = \frac{0}{8} = 0\end{align*}

The slope of this line is 0, or a horizontal line. Horizontal lines always pass through the \begin{align*}y-\end{align*}axis. The \begin{align*}y-\end{align*}coordinate for both points is -1. So, the equation of this line is \begin{align*}y = -1\end{align*}.

2. Use (3, 2) as \begin{align*}(x_1, y_1)\end{align*} and (3, 6) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m = \frac{6-2}{3-3} = \frac{4}{0} = undefined\end{align*}

The slope of this line is undefined, which means that it is a vertical line. Vertical lines always pass through the \begin{align*}x-\end{align*}axis. The \begin{align*}x-\end{align*}coordinate for both points is 3.

So, the equation of this line is \begin{align*}x = 3\end{align*}.

3. Use (-5, 2) as \begin{align*}(x_1, y_1)\end{align*} and (3, 4) as \begin{align*}(x_2, y_2)\end{align*}.

\begin{align*}m = \frac{4-2}{3-(-5)} = \frac{2}{8} = \frac{1}{4}\end{align*}

### Practice

Find the slope between the two given points.

- (4, -1) and (-2, -3)
- (-9, 5) and (-6, 2)
- (7, 2) and (-7, -2)
- (-6, 0) and (-1, -10)
- (1, -2) and (3, 6)
- (-4, 5) and (-4, -3)
- (-2, 3) and (-2, -3)
- (4, 1) and (7, 1)

For 9-10, determine if the statement is true or false.

- If you know the slope of a line you will know whether it is pointing up or down from left to right.
- Vertical lines have a slope of zero.