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Sagot :
to get the equation of any straight line, all we need is two points from it, Check the picture below, let's use those two.
[tex](\stackrel{x_1}{-5}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-5)}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{1+5}{0+5}\implies \cfrac{6}{5}[/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}{(-5)}=\stackrel{m}{\cfrac{6}{5}}(x-\stackrel{x_1}{(-5)}) \\\\\\ y+5=\cfrac{6}{5}(x+5)\implies y+5=\cfrac{6}{5}x+6\implies y=\cfrac{6}{5}x+1[/tex]
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