Question

Find the equation of a line passing through (-4,-3) and perpendicular to 3 x + 2 y = 14.

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so therefore

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}$

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}$