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Sagot :
to get the equation of any straight line, we simply need two points off of it, let's use the ones provided in the picture below.
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{-7}{2}[/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}{4}=\stackrel{m}{-\cfrac{7}{2}}(x-\stackrel{x_1}{1}) \\\\\\ y-4=-\cfrac{7}{2}x+\cfrac{7}{2}\implies y=-\cfrac{7}{2}x+\cfrac{7}{2}+4\implies y=-\cfrac{7}{2}x+\cfrac{15}{2}[/tex]
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