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{3}{4}}x+6 ~~ \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 then
[tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{-3}\implies \cfrac{4}{3}}}[/tex]
so we're really looking for the equation of a line whose slope is 4/3 and passes through (3 , 9)
[tex](\stackrel{x_1}{3}~,~\stackrel{y_1}{9})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{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}{9}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3}) \\\\\\ y-9=\cfrac{4}{3}x-4\implies y=\cfrac{4}{3}x+5[/tex]
