Question

Write the equation of the line perpendicular to y= 5/6x + 1/6 that passes through the point (-1,5)

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{5}{6}}x+\cfrac{1}{6}\qquad \impliedby \begin{array}c \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{5}{6}} ~\hfill \stackrel{reciprocal}{\cfrac{6}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{6}{5} }}`

so we're really looking for the equation of a line whose slope is -6/5 and it passes through (-1 , 5)

`(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{6}{5} \\\\\\ \begin{array}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}{5}=\stackrel{m}{-\cfrac{6}{5}}(x-\stackrel{x_1}{(-1)}) \implies y -5 = -\cfrac{6}{5} ( x +1) \\\\\\ y-5=-\cfrac{6}{5}x-\cfrac{6}{5}\implies y=-\cfrac{6}{5}x-\cfrac{6}{5}+5\implies {\Large \begin{array}{llll} y=-\cfrac{6}{5}x+\cfrac{19}{5} \end{array}}`