Precalculus Examples

What is the slope of the line through $(- 9, 6)$ and $(- 3, 9)$ ?

Step 1

Slope is equal to the change in $y$ over the change in $x$ , or rise over run.

$m = \frac{change in y}{change in x}$

Step 2

The change in $x$ is equal to the difference in x-coordinates (also called run), and the change in $y$ is equal to the difference in y-coordinates (also called rise).

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Step 3

Substitute in the values of $x$ and $y$ into the equation to find the slope.

$m = \frac{9 - (6)}{- 3 - (- 9)}$

Step 4

Simplify.

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Step 4.1

Reduce the expression by cancelling the common factors.

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Step 4.1.1

Cancel the common factor of $9 - (6)$ and $- 3 - (- 9)$ .

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Step 4.1.1.1

Rewrite $9$ as $- 1 (- 9)$ .

$m = \frac{- 1 \cdot - 9 - (6)}{- 3 - (- 9)}$

Step 4.1.1.2

Factor $- 1$ out of $- 1 (- 9) - (6)$ .

$m = \frac{- 1 (- 9 + 6)}{- 3 - (- 9)}$

Step 4.1.1.3

Reorder terms.

$m = \frac{- 1 (- 9 + 6)}{- 3 - 9 \cdot - 1}$

Step 4.1.1.4

Factor $3$ out of $- 1 (- 9 + 6)$ .

$m = \frac{3 (- 1 (- 3 + 2))}{- 3 - 9 \cdot - 1}$

Step 4.1.1.5

Cancel the common factors.

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Step 4.1.1.5.1

Factor $3$ out of $- 3$ .

$m = \frac{3 (- 1 (- 3 + 2))}{3 (- 1) - 9 \cdot - 1}$

Step 4.1.1.5.2

Factor $3$ out of $- 9 \cdot - 1$ .

$m = \frac{3 (- 1 (- 3 + 2))}{3 (- 1) + 3 (- 3 \cdot - 1)}$

Step 4.1.1.5.3

Factor $3$ out of $3 (- 1) + 3 (- 3 \cdot - 1)$ .

$m = \frac{3 (- 1 (- 3 + 2))}{3 (- 1 - 3 \cdot - 1)}$

Step 4.1.1.5.4

Cancel the common factor.

$m = \frac{3 (- 1 (- 3 + 2))}{3 (- 1 - 3 \cdot - 1)}$

Step 4.1.1.5.5

Rewrite the expression.

$m = \frac{- 1 (- 3 + 2)}{- 1 - 3 \cdot - 1}$

Step 4.1.2

Add $- 3$ and $2$ .

$m = \frac{- 1 \cdot - 1}{- 1 - 3 \cdot - 1}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $- 3$ by $- 1$ .

$m = \frac{- 1 \cdot - 1}{- 1 + 3}$

Step 4.2.2

Add $- 1$ and $3$ .

$m = \frac{- 1 \cdot - 1}{2}$

Step 4.3

Multiply $- 1$ by $- 1$ .

$m = \frac{1}{2}$

Step 5