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Sagot :
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]\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}\impliedby y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{3}{5}}x-8[/tex]
so we're really looking for the line of a slope of 3/5 and passing through (0,-3)
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{-3})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{3}{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-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{3}{5}}(x-\stackrel{x_1}{0}) \\\\\\ y+3=\cfrac{3}{5}x\implies y=\cfrac{3}{5}x-3[/tex]
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