168k views
3 votes
What is the equation of the line that passes through the point (-2,14) and is perpendicular to the line with the following equation? y= -2/5 x-1

1 Answer

6 votes

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


y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}x-1\qquad \impliedby \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{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}

so we're really looking for the equation of a line whose slope is 5/2 and it passes through (-2 , 14)


(\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \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}{14}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -14= \cfrac{5}{2} (x +2) \\\\\\ y-14=\cfrac{5}{2}x+5\implies {\Large \begin{array}{llll} y=\cfrac{5}{2}x+19 \end{array}}

User Joeellis
by
7.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories