First, use the slope formula to find the slope of this line: $$\begin{aligned} m=\frac{y_2-y_1}{x_2-x_1}\amp=\frac{23-(-108)}{-42-13}\\ \amp=\frac{131}{-55}\\ \amp=-\frac{131}{55}\text{.} \end{aligned}$$ The generic point-slope equation is $y=m(x-x_0)+y_0\text{.}$ We have found the slope, $m\text{,}$ and we may use $(13,-108)$ for $(x_0,y_0)\text{.}$ So an equation in point-slope form is ${y = \frac{-131}{55}\mathopen{}\left(x-13\right)-108}\text{.}$ To find a slope-intercept form equation, we can take the generic $y=mx+b$ and substitute in the value of $m$ we found. Also, we know that $(x,y)=(13,-108)$ should make the equation true. So we have $$\begin{aligned} y\amp=mx+b\\ -108\amp=-\frac{131}{55}(13)+b\amp\text{Now we may solve for }b\text{.}\\ -108\multiplyright{55}\amp=\left(-\frac{131}{55}(13)+b\right)\multiplyright{55}\\ -5940\amp=-131(13)+55b\\ -5940\amp=-1703+55b\\ -5940\addright{1703}\amp=-1703+55b\addright{1703}\\ -4237\amp=55b\\ \divideunder{-4237}{55}\amp=\divideunder{55b}{55}\\ b\amp=-\frac{4237}{55}\text{.} \end{aligned}$$ So the slope-intercept equation is ${y = \frac{-131}{55}x-\frac{4237}{55}}\text{.}$