Finite Math Examples

$(1, 13)$ , $(0, 9)$

Find the slope of the line between $(1, 13)$ and $(0, 9)$ using $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , which is the change of $y$ over the change of $x$ .

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

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

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).

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

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

$m = \frac{9 - (13)}{0 - (1)}$

Simplify.

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Simplify the numerator.

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Multiply $- 1$ by $13$ .

$m = \frac{9 - 13}{0 - (1)}$

Subtract $13$ from $9$ .

$m = \frac{- 4}{0 - (1)}$

Simplify the denominator.

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Multiply $- 1$ by $1$ .

$m = \frac{- 4}{0 - 1}$

Subtract $1$ from $0$ .

$m = \frac{- 4}{- 1}$

Divide $- 4$ by $- 1$ .

$m = 4$

Use the slope $m = 4$ and one of the given points such as $(1, 13)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m \cdot (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (13) = 4 \cdot (x - (1))$

Solve for $y$ to get the equation.

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Multiply $- 1$ by $13$ .

$y - 13 = 4 \cdot (x - (1))$

Simplify $4 \cdot (x - (1))$ .

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Multiply $- 1$ by $1$ .

$y - 13 = 4 \cdot (x - 1)$

Apply the distributive property.

$y - 13 = 4 x + 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$y - 13 = 4 x - 4$

Move all terms not containing $y$ to the right side of the equation.

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Add $13$ to both sides of the equation.

$y = 4 x - 4 + 13$

Add $- 4$ and $13$ .

$y = 4 x + 9$

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