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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]5x-y=-7\implies 5x-y+7=0 \\\\\\ \stackrel{\stackrel{m}{\downarrow }}{5} x+7=y\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 a line perpendicular to that one above will have a slope of
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{5\implies \cfrac{5}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{5}}}[/tex]
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