Question

What is the equation of the line that is perpendicular to y=2x+3 and passes through the point (-4,8)

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+3\qquad \impliedby \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}{ 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 whose slope is -1/2 and it passes through (-4 , 8)

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{1}{2} \\\\\\ \begin{array}c \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}{8}=\stackrel{m}{- \cfrac{1}{2}}(x-\stackrel{x_1}{(-4)}) \implies y -8= -\cfrac{1}{2} (x +4) \\\\\\ y-8=-\cfrac{1}{2}x-2\implies {\Large \begin{array}{llll} y=-\cfrac{1}{2}x+6 \end{array}}`