What is perpendicular to y=-(1/3)x+5 but goes though the point (1,-10) in slope-intersect form

bearing in mind that perpendicular lines have negative reciprocal slopes hmmmmm wait a second, what's the slope of that line above anyway? [tex]\bf y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}} x+5\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}[/tex]therefore any perpendicular line to that [tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}[/tex]so, we're really looking for the equation of a line whose slope is 3 and runs through (1, -10)[tex]\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{-10})~\hspace{10em} slope = m\implies 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-(-10)=3(x-1) \\\\\\ y+10=3x-3\implies y=3x-13[/tex]