Question

PLEASE HELPPPP

Write an equation for the line perpendicular to the given line through the given point.

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{1}{3}}x-\cfrac{2}{5}\qquad \impliedby \qquad \begin{array}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{1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{1} \implies -3}}`

so we are really looking for the equation of a line whose slope is -3 and it passes through (5 , -1)

`(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\hspace{10em} \stackrel{slope}{m} ~=~ -3 \\\\\\ \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}{(-1)}=\stackrel{m}{-3}(x-\stackrel{x_1}{5}) \implies y +1 = -3 ( x -5) \\\\\\ y+1=-3x+15\implies {\Large \begin{array}{llll} y=-3x+14 \end{array}}`