Question

Write the equation of a line perpendicular to 3x-4y=-9 that passes through the point ​(-3.2)

Answer 1

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

`3x-4y=-9\implies 3x=4y-9\implies 3x+9=4y\implies \cfrac{3x+9}{4}=y \\\\\\ \stackrel{\stackrel{m}{\downarrow }}{\cfrac{3}{4}}x+\cfrac{9}{4}=y\qquad \impliedby \qquad \begin{array}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`

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{3}{4}} ~\hfill \stackrel{reciprocal}{\cfrac{4}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{4}{3}}}`

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

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{4}{3} \\\\\\ \begin{array} \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}{2}=\stackrel{m}{- \cfrac{4}{3}}(x-\stackrel{x_1}{(-3)}) \implies y -2= -\cfrac{4}{3} (x +3) \\\\\\ y-2=- \cfrac{4}{3}x-4\implies {\Large \begin{array}{llll} y=- \cfrac{4}{3}x-2 \end{array}}`