Find the equation of the like that passes through the point (-4,13) and is perpendicular to the equation below:y = -2/3x + 1

bearing in mind that perpendicular lines have negative reciprocal slopes.[tex]\bf y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+1\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} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{\cfrac{3}{2}}}[/tex]so we're really looking for the equation of a line whose slope is 3/2 and runs through (-4,13)[tex]\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{13})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{3}{2} \\\\\\ \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-\stackrel{y_1}{13}=\stackrel{m}{\cfrac{3}{2}}[x-\stackrel{x_1}{(-4)}] \\\\\\ y-13=\cfrac{3}{2}(x+4)\implies y-13=\cfrac{3}{2}x+6\implies y=\cfrac{3}{2}x+19[/tex]