Question

What is an equation of the line that is perpendicular to y=4/5x+15 and passes through the point (-3, 4) ?

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{4}{5}}x+15\impliedby \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{4}{5}} ~\hfill \stackrel{reciprocal}{\cfrac{5}{4}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{5}{4} }}`

so we are really looking for the equation of a line whose slope is -5/4 and it passes through (-3 , 4)

`(\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{5}{4} \\\\\\ \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}{4}=\stackrel{m}{-\cfrac{5}{4}}(x-\stackrel{x_1}{(-3)}) \implies y -4 = -\cfrac{5}{4} ( x +3) \\\\\\ y-4=-\cfrac{5}{4}x-\cfrac{15}{4}\implies y=-\cfrac{5}{4}x-\cfrac{15}{4}+4\implies {\Large \begin{array}{llll} y=-\cfrac{5}{4}x+\cfrac{1}{4} \end{array}}`