## Precalculus (6th Edition)

$y=-\frac{1}{3}x+\frac{1}{3}$ Refer to the graph below.
Solve for the slope $m$ using the formula $m=\dfrac{y_2-y_1}{x_2-x_1}$ where $(x_1, y_1)$ and $(x-2, y_2)$ are points on the line. $m=\dfrac{-1-1}{4-(-2)} \\m=\dfrac{-2}{4+2} \\m=\dfrac{-2}{6} \\m=-\dfrac{1}{3}$ The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. The line's slope is $-\frac{1}{3}$ so its tentative equation in slope-intercept form is $y=-\frac{1}{3}x+b$. To find the value of $b$, substitute the x and y coordinates of $(-2, 1)$ to obtain: $y=-\frac{1}{3}x+b \\1=-\frac{1}{3}(-2)+b \\1=\frac{2}{3}+b \\1-\frac{2}{3}=b \\\frac{1}{3}=b$ Thus, the equation of the line is $y=-\frac{1}{3}x+\frac{1}{3}$. Refer to the graph in the answer part above.