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Sagot :
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
[tex]3x+y=7\implies y=\stackrel{\stackrel{m}{\downarrow }}{-3} x+7\impliedby \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}[/tex]
therefore
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-3\implies \cfrac{-3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-3}\implies \cfrac{1}{3}}}[/tex]
so we're really looking for the equation of a line whose slope is 1/3 and passes through (3 , -3)
[tex](\stackrel{x_1}{3}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{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}{(-3)}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y+3=\cfrac{1}{3}x-1\implies y=\cfrac{1}{3}x-4[/tex]
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