Question

Find the equation of a line perpendicular to 3x+4y = 24 that passes through the point (-3,-7)

Answer 1

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

`3x+4y=24\implies 4y=-3x+24\implies y=\cfrac{-3x+24}{4} \\\\\\ y=\stackrel{\stackrel{\textit{\\ormalsize m}}{\downarrow }}{-\cfrac{3}{4}}x+6\qquad \impliedby \qquad \begin{array}c \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} \implies \cfrac{4}{ 3 }}}`

so we are really looking for the equation of a line whose slope is 4/3 and it passes through (-3 , -7)

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{-7})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{4}{3} \\\\\\ \begin{array}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}{(-7)}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{(-3)}) \implies y +7 = \cfrac{4}{3} ( x +3) \\\\\\ y+7=\cfrac{4}{3}x+4\implies {\Large \begin{array}{llll} y=\cfrac{4}{3}x-3 \end{array}}`