Question

If the vertices of triangle pqr are p(2,2) q(8,4) r(3,7), find the equation of the perpendicular bisector of segment pq

Answer 1

First off, let's find the slope of PQ

$\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &P&(~ 2 &,& 2~) % (c,d) &Q&(~ 8 &,& 4~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-2}{4-2}\implies \cfrac{6}{2}\implies 3$

now, a perpendicular line to PQ, will have a negative reciprocal slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad 3\implies \cfrac{3}{1}\\\\ negative\implies -\cfrac{3}{ 1}\qquad reciprocal\implies - \cfrac{ 1}{3}$

now, we also know that, the line is a bisector of PQ, therefore, it cuts PQ in two equal halves, so it passes half-way through PQ, what is that point anyway?

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &P&(~ 2 &,& 2~) % (c,d) &Q&(~ 8 &,& 4~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+2}{2}~~,~~\cfrac{4+2}{2} \right)\implies (5~,~3)$

so then, what's the equation of a line whose slope is -1/3 and runs through 5,3?

$\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ % (a,b) &&(~ 5 &,& 3~) \end{array} \\\\\\ % slope = m slope = m\implies -\cfrac{1}{3} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-3=-\cfrac{1}{3}(x-5) \implies y-3=-\cfrac{1}{3}x+\cfrac{5}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{5}{3}+3\implies y=-\cfrac{1}{3}x+\cfrac{14}{3}$