Answer :
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 }}{-2}x+8\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} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-2\implies \cfrac{-2}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-2}\implies \cfrac{1}{2}}}[/tex]
so, we're really looking for the equation of a line whose slope is 1/2 and that it passes through (6 , -2)
[tex](\stackrel{x_1}{6}~,~\stackrel{y_1}{-2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \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}{(-2)}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{6}) \implies y +2= \cfrac{1}{2} (x -6) \\\\\\ y+2=\cfrac{1}{2}x-3\implies \LARGE \begin{array}{llll} y=\cfrac{1}{2}x-5 \end{array}[/tex]
