Question

Find the equation of a line perpendicular y=1+2x that passes through the point (4,-4)(4,−4).

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

`y=1+2x\implies y=\stackrel{\stackrel{m}{\downarrow }}{2}x+1\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}{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 that it passes through (4 , -4)

`(\stackrel{x_1}{4}~,~\stackrel{y_1}{-4})\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}{(-4)}=\stackrel{m}{- \cfrac{1}{2}}(x-\stackrel{x_1}{4}) \implies y +4= -\cfrac{1}{2} (x -4) \\\\\\ y+4=-\cfrac{1}{2}x+2\implies {\Large \begin{array}{llll} y=-\cfrac{1}{2}x-2 \end{array}}`