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Sagot :
to get the equation of any straight line all we need is two points from it, let's use those ones in the picture.
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{0}}}\implies \cfrac{2}{3}\implies \cfrac{2}{3} \\\\\\ \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}{1}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{0})\implies y=\cfrac{2}{3}x+1[/tex]
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