K
Find the equation of a line that is perpendicular to the line y=
y= (Type your answer in slope-intercept form.)

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above[tex]y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{5}}x+5\qquad \impliedby \qquad \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} \\\\[-0.35em] ~\dotfill[/tex][tex]\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{1}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{1}\implies -5}}[/tex]so we're really looking for the equation of a line whose slope is -5 and that it passes through (-4 , 0)[tex](\stackrel{x_1}{-4}~,~\stackrel{y_1}{0})\hspace{10em} \stackrel{slope}{m} ~=~ - 5 \\\\\\ \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}{0}=\stackrel{m}{- 5}(x-\stackrel{x_1}{(-4)}) \\\\\\ y -0= -5 (x +4) \implies {\Large \begin{array}{llll} y=-5x-20 \end{array}}[/tex]