[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-1}{-3-2}\implies \cfrac{4}{-5}\implies -\cfrac{4}{5} \\\\\\ \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-1=-\cfrac{4}{5}(x-2)\implies y-1=-\cfrac{4}{5}x+\cfrac{8}{5}[/tex]
[tex]\bf y=-\cfrac{4}{5}x+\cfrac{8}{5}+1\implies y=-\cfrac{4}{5}x+\cfrac{13}{5}\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} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{y-intercept}{\left(0~,~\frac{13}{5} \right)}~\hfill[/tex]
the Answer is 13/5
you first need to write the coordinates in a Y=MX+b formula then substitute 0 for x. The y-axis is a vertical line along x = 0
