## Algebra and Trigonometry 10th Edition

$(0,\frac{19}{3})$, $(7,4)$, $(10,3)$
Use the point-slope form: $y-y_1=m(x-x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. $y-5=-\frac{1}{3}(x-4)$ $y-5=-\frac{1}{3}x+\frac{4}{3}$ $y=-\frac{1}{3}x+\frac{19}{3}$ Now, choose three different values for $x$ to find three different points on the line. For example, $x=0$, $x=7$ and $x=10$: $x=0$: $y=-\frac{1}{3}(0)+\frac{19}{3}=\frac{19}{3}$ Point: $(0,\frac{19}{3})$ $x=7$: $y=-\frac{1}{3}(7)+\frac{19}{3}=\frac{12}{3}=4$ Point: $(7,4)$ $x=10$: $y=-\frac{1}{3}(10)+\frac{19}{3}=\frac{9}{3}=3$ Point: $(10,3)$