Question

Write an equation of the Line that passes thru (5,-2);and is perpendicular to y=5/3x -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{5}{3}}x-3\qquad \impliedby \qquad \begin{array}c \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{5}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{5}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{5} }}`

so we're really looking for the equation of a line whose slope is -3/5 and it passes through (5 , -2)

`(\stackrel{x_1}{5}~,~\stackrel{y_1}{-2})\hspace{10em} \stackrel{slope}{m} ~=~ - \cfrac{3}{5} \\\\\\ \begin{array}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}{- \cfrac{3}{5}}(x-\stackrel{x_1}{5}) \implies y +2= -\cfrac{3}{5} (x -5) \\\\\\ y+2=-\cfrac{3}{5}x+3\implies {\Large \begin{array}{llll} y=-\cfrac{3}{5}x+1 \end{array}}`