Question

What is the equation of the line that is perpendicular to the line y =1/2x+5 and passes through the point (-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 }}{\cfrac{1}{2}}x+5\impliedby \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}} {\stackrel{slope}{\cfrac{1}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{1}\implies -2}}`

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

`(\stackrel{x_1}{-2}~,~\stackrel{y_1}{3})\hspace{10em} \stackrel{slope}{m} ~=~ - 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}{- 2}(x-\stackrel{x_1}{(-2)})\implies y-3=-2(x+2) \\\\\\ y-3=-2x-4\implies y=-2x-1`