well, keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of MN anyway, well, let's use those two points of 2,3 and -3, 2 to get the slope of it.
[tex] \bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{2})\\\\\\slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-3}{-3-2}\implies \cfrac{-1}{-5}\implies \cfrac{1}{5}\\\\[-0.35em]\rule{34em}{0.25pt} [/tex]
[tex] \bf \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 K(3,-3)
[tex] \bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{-3})~\hspace{10em}
slope = m\implies -5
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-3)=-5(x-3)
\\\\\\
y+3=-5x+15\implies y=-5x+12
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\stackrel{x=2}{y=-5(2)+12}\implies y=-10+12\implies y=2~\hfill \boxed{(2,2)} [/tex]
