to get the equation of any straight line, we simply need two points off of it, let's use those two points in the picture below.
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}}\implies \cfrac{3}{1}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_2=m(x-x_2) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_2}{5}=\stackrel{m}{3}(x-\stackrel{x_2}{1})[/tex]
keeping in mind that for the point-slope form, either point will do, in this case we used the second one, but the first one would have worked just the same.
