first off, what's the slope of "r" anyway, [tex]\bf y=\stackrel{slope}{-\cfrac{1}{2}}x-4[/tex].
low and behold, since "r" is in slope-intercept form, notice, it has a slope of -1/2.
now, any line perpendicular to "r", will have a negative reciprocal slope to it, that is,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{2}\\\\
negative\implies +\cfrac{1}{{{ 2}}}\qquad reciprocal\implies + \cfrac{{{ 2}}}{1}\implies 2[/tex]
so we're really looking for a line whose slope is 2, and runs through 4,-3,
[tex]\bf \begin{array}{ccccccccc}
&&x_1&&y_1\\
% (a,b)
&&(~{{ 4}} &,&{{ -3}}~)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}\implies 2
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-(-3)=2(x-4)\implies y+3=2x-8
\\\\\\
y=2x-11[/tex]
