Question

Which equation represents the line that is perpendicular to

the graph
of 4x+3y=9 and passes through (-2, 3)?

Answer 1

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

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

so a perpendicular line to that will have a slope of

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

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

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{3})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{3}{4} \\\\\\ \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}{3}=\stackrel{m}{ \cfrac{3}{4}}(x-\stackrel{x_1}{(-2)}) \implies y -3= \cfrac{3}{4} (x +2) \\\\\\ y-3=\cfrac{3}{4}x+\cfrac{3}{2}\implies y=\cfrac{3}{4}x+\cfrac{3}{2}+3\implies {\Large \begin{array}{llll} y=\cfrac{3}{4}x+\cfrac{9}{2} \end{array}}`