Question

Point A (3,-2) and Point B (5,q) are two points on line M. The equation of Line P is 2x+3y-5=0. Line P is perpendicular to line M. Determine the value of q Responses

Answer 1

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation of line P

`2x+3y-5=0\implies 2x+3y=5\implies 3y=-2x+5 \\\\\\ y=\cfrac{-2x+5}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+\cfrac{5}{3}\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}`

and we know that line M is perpendicular to P, so

`\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{ \cfrac{-2}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-2}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-2} \implies \cfrac{3}{ 2 }}}`

so then

`A(\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{q}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{q}-\stackrel{y1}{(-2)}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{3}}} ~~ = ~~\stackrel{\stackrel{\textit{\small slope}}{\downarrow }}{ \cfrac{ 3 }{ 2 }}\implies \cfrac{q+2}{2}=\cfrac{3}{2}\implies q+2=3\implies \boxed{q=1}`