Question

Christopher wants to build a road named Comet Parkway that connects Sun Bank to Moonbeam Drive. He wants this road to be perpendicular to Moonbeam Drive. Determine the equation of the line that represents Comet Parkway.

Answer 1

Check the picture below.

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

`(\stackrel{x_1}{15}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{20}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{20}-\underset{x_1}{15}}} \implies \cfrac{-4}{5}`

that means that the slope of Comet Parkway will be

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

so we're really looking for the equation of a line whose slope is 5/4 and that it passes through (2 , 6)

`(\stackrel{x_1}{2}~,~\stackrel{y_1}{6})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{4} \\\\\\ \begin{array}c \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}{6}=\stackrel{m}{ \cfrac{5}{4}}(x-\stackrel{x_1}{2}) \\\\\\ y-6=\cfrac{5}{4}x-\cfrac{5}{2}\implies y=\cfrac{5}{4}x-\cfrac{5}{2}+6\implies {\LARGE \begin{array}{llll} y=\cfrac{5}{4}x+\cfrac{7}{2} \end{array}}`