Question

What is an equation of the line that passes through the point (4,2)(4,2) and is perpendicular to the line 4x+3y=214x+3y=21?

Answer 1

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

`4x+3y=21\implies 3y=-4x+21\implies y=\cfrac{-4x+21}{3} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}}x+7\qquad \impliedby \qquad \begin{array} \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{-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 it passes through (4 , 2)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{3}{4} \\\\\\ \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{3}{4}}(x-\stackrel{x_1}{4}) \\\\\\ y-2=\cfrac{3}{4}x-3\implies {\Large \begin{array}{llll} y=\cfrac{3}{4}x-1 \end{array}}`