keeping in mind that a perpendicular line to another, will have a negative reciprocal slope, hmmmmm so what's the slope of the equation above anyway?
[tex]\bf y=16-\cfrac{x}{3}\implies \stackrel{\textit{slope-intercept form}}{y = -\cfrac{x}{3}+16}\implies y=\stackrel{slope}{-\cfrac{1}{3}}x+16
\\\\\\
\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]