Question

What is an equation in point-slope form for the line perpendicular to y =-5x - 14 that contains (10, -4)?

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 }}{-5}x-14\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}} {-5\implies \stackrel{slope}{\cfrac{-5}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-5}\implies \cfrac{1}{5}}}`

so we're really looking for the equation of a line whose slope is 1/5 and it passes through (10, -4)

`(\stackrel{x_1}{10}~,~\stackrel{y_1}{-4})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{5} \\\\\\ \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}{(-4)}=\stackrel{m}{ \cfrac{1}{5}}(x-\stackrel{x_1}{10}) \implies {\large \begin{array}{llll} y +4= \cfrac{1}{5} (x -10) \end{array}}`