[tex](\stackrel{x_1}{-5}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{4}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{-3}-\underset{x_1}{(-5)}}}\implies \cfrac{-6}{-3+5}\implies \cfrac{-6}{2}\implies -3[/tex]
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-3}(x-\stackrel{x_1}{(-5)})\implies y-10=-3(x+5) \\\\\\ y-10=-3x-15\implies y=-3x-5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
as you can see, the "b" part is -5, or namely the y-intercept is at (0 , -5).
