Question

Find an equation of a line through (2, 1) and perpendicular to - 2x + 4y = 8.

Answer 1

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

$-2x+4y=8\implies 4y=2x+8\implies y=\cfrac{2x+8}{4} \\\\\\ y=\cfrac{2x}{4}+\cfrac{8}{4}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+2\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

since we know that's its slope, then

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

so then we're really looking for the equation of a line whose slope is -2 and passes through (2 , 1)

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad \qquad \stackrel{slope}{m}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{-2}(x-\stackrel{x_1}{2}) \\\\\\ y-1=-2x-4\implies y=-2x-3$