Question

The straight line L has equation y = 2x − 5

Find an equation of the straight line perpendicular to L which passes through (−2, 3).

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 }}{2}x-5\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}`

so then

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{2\implies \cfrac{2}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{2}}}`

so we're really looking for the equation of a line with a slope of -1/2 and that passes through (-2 , 3)

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{3})\qquad\qquad \stackrel{slope}{m}\implies -\cfrac{1}{2} \\\\\\ \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}{3}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{(-2)}) \\\\\\ y-3=-\cfrac{1}{2}(x+2)\implies y-3=-\cfrac{1}{2}x-1\implies y=-\cfrac{1}{2}x+2`