Question

Determine an equation of a line that passes through point P(3,-3)that is perpendicular to line m y=7/6x+5

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{7}{6}}x+5\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{7}{6}} ~\hfill \stackrel{reciprocal}{\cfrac{6}{7}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{6}{7} }}`

so we are really looking at the equation of a line whose slope is -6/7 and it passes through (3 , -3)

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{-3})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{6}{7} \\\\\\ \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}{(-3)}=\stackrel{m}{-\cfrac{6}{7}}(x-\stackrel{x_1}{3}) \implies y +3 = -\cfrac{6}{7} ( x -3) \\\\\\ y+3=-\cfrac{6}{7}x+\cfrac{18}{7}\implies y=-\cfrac{6}{7}x+\cfrac{18}{7}-3\implies {\Large \begin{array}{llll} y=-\cfrac{6}{7}x-\cfrac{3}{7} \end{array}}`