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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]\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}\qquad \impliedby y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{5}}x+7 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{1}{5}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{5}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{5}{1}\implies -5}}[/tex]
so then line K has a slope of -5 and pass through (-1, -3)
[tex](\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad \qquad slope=m=-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}{-5}[x-\stackrel{x_1}{(-1)}] \\\\\\ y+3=-5(x+1)\implies y+3=-5x-5\implies y=-5x-8[/tex]
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