Question

Write the equation of the line that is perpendicular to the graph of y=1/2x +6, and whose y-intercept is (0,-2).

Answer 1

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

$y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so 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 we're really looking for the equation of a line whose slope is -2 and passes through (0 , -2)

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{-2})\qquad \qquad \stackrel{slope}{m}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-2)}=\stackrel{m}{-2}(x-\stackrel{x_1}{0}) \\\\\\ y+2=-2x\implies y=-2x-2$