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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]y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{6}}x+\cfrac{7}{6}\qquad \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]
so we can say that
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{5}{6}} ~\hfill \stackrel{reciprocal}{\cfrac{6}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{6}{5}}}[/tex]
so we're really looking for the equation of a line with a slope of -6/5 and that passes through (-8 , 9)
[tex](\stackrel{x_1}{-8}~,~\stackrel{y_1}{9})\qquad\qquad \stackrel{slope}{m}\implies -\cfrac{6}{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}{9}=\stackrel{m}{-\cfrac{6}{5}}(x-\stackrel{x_1}{(-8)})\implies y-9=-\cfrac{6}{5}(x+8) \\\\\\ y-9=-\cfrac{6}{5}x-\cfrac{48}{5}\implies y=-\cfrac{6}{5}x-\cfrac{48}{5}+9\implies y=-\cfrac{6}{5}x-\cfrac{3}{5}[/tex]
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