Question

Find (a) pq to the nearest tenth and (b) the coordinates of the midpoint of pq p(2,2) q(-3,-6)

Answer 1

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P(\stackrel{x_1}{2}~,~\stackrel{y_1}{2})\qquad Q(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-6})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ PQ=√([-3 - 2]^2 + [-6 - 2]^2)\implies PQ=√((-5)^2+(-8)^2) \\\\\\ PQ=√(25+64)\implies PQ=√(89)\implies \boxed{PQ\approx 9.4}`

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{2}~,~\stackrel{y_1}{2})\qquad Q(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -3 + 2}{2}~~~ ,~~~ \cfrac{ -6 + 2}{2} \right)\implies \left(\cfrac{-1}{2}~~,~~\cfrac{-4}{2} \right)\implies \left( -\cfrac{1}{2}~~,~~-2 \right)`