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

$y=-\dfrac{2}{3}x-\dfrac{1}{3}$
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. Here, our line's slope is: $m=-\frac{2}{3}$ Therefore, using the point $(4, -3)$ and the slope $-\frac{2}{3}$, the equation can be written as: $y-(-1)=-\dfrac{2}{3}(x-1)$ $y+1=-\dfrac{2}{3}(x-1)\\ y+1=-\dfrac{2}{3}x+\dfrac{2}{3}\\ y=-\dfrac{2}{3}x+\dfrac{2}{3}-1\\ y=-\dfrac{2}{3}x-\dfrac{1}{3}$