bearing in mind that perpendicular lines have negative reciprocal slopes, let's find the slope of 5x - y = 9 then.
[tex]\bf 5x-y=9\implies -y=-5x+9\implies y=\stackrel{\stackrel{m}{\downarrow }}{5}x-9\leftarrow \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]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{5\implies \cfrac{5}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{5}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{5}}}[/tex]
so then, we're really looking for the equation of a line whose slope is -1/5 and runs through (3,-10).
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-10})~\hspace{10em} slope = m\implies -\cfrac{1}{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-(-10)=-\cfrac{1}{5}(x-3)\implies y+10=-\cfrac{1}{5}x+\cfrac{3}{5} \\\\\\ y=-\cfrac{1}{5}x+\cfrac{3}{5}-10\implies y=-\cfrac{1}{5}x+\cfrac{53}{5}[/tex]
and it looks like the one in the picture below.
