## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{y=-7x+15}$.
Recall: (1) The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $$m=\dfrac{y_2-y_1}{x_2-x_1}$$ (2) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the $y$-intercept. Solve for the slope of the line using the formula above and the points $(2, 1)$ and $(4, -13)$: \begin{align*} m&=\frac{y_2-y_1}{x_2-x_1}\\ m&=\frac{-13-1}{4-2}\\\\ m&=\frac{-14}{2}\\\\ m&=-7 \end{align*} Hence, the tentative equation fo the line that contains the given points is: $$y=-7x+b$$ Solve for the $b$ by substituting the $x$ and $y$ values of the point $(2, 1)$ into the tentative equation above to obtain: \begin{align*} y&=-7x+b\\ 1&=-7(2)+b\\ 1&=-14+b\\ 1+14&=b\\ 15&=b\\ \end{align*} Therefore, the equation of the line that contains the given points in the table is $\color{blue}{y=-7x+15}$.