## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$y-1=\dfrac{1}{3}(x+1)$
The equation of a line in the point-slope form is the following: $y-y_1=m(x-x_1)$, where $m$ is the slope and the point $(x_1,y_1)$ is on the graph. In order to determine the equation of a line, we have to calculate the slope. For this, we can use the formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$, where the two points on the line are $(x_1, y_1)$ and $(x_2,y_2)$. Here, if we plug in the given points we get: $m=\dfrac{2-1}{2-(-1)}=\dfrac{1}{3}$ Therefore, using the point $(-1, 1)$ and the slope $\frac{1}{3}$, the equation can be written as: $y-1=\dfrac{1}{3}[x-(-1)]$ $y-1=\dfrac{1}{3}(x+1)$