Question

Write an equation in slope intercept from perpendicular to y = 1/5x - 9 through (7, -3)?​

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}{5}}x-9\qquad \impliedby \qquad \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}{ \cfrac{1}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{1} \implies -5}}`

so we're really looking for the equation of a line whose slope is -5 and it passes through (7 , -3)

`(\stackrel{x_1}{7}~,~\stackrel{y_1}{-3})\hspace{10em} \stackrel{slope}{m} ~=~ - 5 \\\\\\ \begin{array} \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}{(-3)}=\stackrel{m}{- 5}(x-\stackrel{x_1}{7}) \implies y +3= -5 (x -7) \\\\\\ y+3=-5x+35\implies {\Large \begin{array}{llll} y=-5x+32 \end{array}}`