Find the equation of the linePerpendicular to y=2/3x+2 and passes through the point (3,4)

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check firstly what's the slope of the equation above hmmmm[tex]\bf y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+2\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 (3,4)[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})~\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}{4}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{3}) \implies y-4=-\cfrac{3}{2}x-\cfrac{9}{2} \\\\\\ y=-\cfrac{3}{2}x-\cfrac{9}{2}+4\implies y = -\cfrac{3}{2}x-\cfrac{1}{2}[/tex]