Question

Find the equation for the line running

to y = -5x + 5 and passes through the
perpendicular
point (10,7).

Answer 1

let's reword this a bit differently, since it looks jumbled

let's find the equation of a line that is perpendicular to y = -5x + 5, and it also passes through (10 , 7), well, 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 }}{-5}x+5\qquad \impliedby \begin{array} \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}{ -5 \implies \cfrac{-5}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{-5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{-5} \implies \cfrac{1}{ 5 }}}`

so we're really looking for the equation of a line with a slope of 1/5 and that it passes through (10 , 7)

`(\stackrel{x_1}{10}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{5} \\\\\\ \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}{7}=\stackrel{m}{ \cfrac{1}{5}}(x-\stackrel{x_1}{10}) \\\\\\ y-7=\cfrac{1}{5}x-2\implies {\Large \begin{array}{llll} y=\cfrac{1}{5}x+5 \end{array}}`