[tex]\bf 5y=x-10\implies y=\cfrac{x-10}{5}\implies \stackrel{slope}{y=\stackrel{\downarrow }{\cfrac{1}{5}}x-2} \\\\\\ \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 we're really looking for the equation of a line whose slope is -5 and runs through 3, -3.
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-3})~\hspace{10em} slope = m\implies -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-(-3)=-5(x-3) \\\\\\ y+3=-5x+15\implies \stackrel{y}{f(x)}=-5x+12[/tex]
