Question

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

Answer 1

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

$y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{5}}x-1\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-2}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{-2}\implies \cfrac{5}{2}}}$

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

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{14})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{14}=\stackrel{m}{ \cfrac{5}{2}}(x-\stackrel{x_1}{(-2)}) \implies y -14= \cfrac{5}{2} (x +2) \\\\\\ y-14=\cfrac{5}{2}x+5\implies {\Large \begin{array}{llll} y=\cfrac{5}{2}x+19 \end{array}}$