Question

Given a triangle with vertices P(-9,7), Q(-3,7), and R(-3,1), and S is the midpoint of PQ and T is the midpoint of QR, what is the length of ST?

Answer 1

Check the picture below.

so we know S is the midpoint of PQ, and T is the midpoint of QR, thus

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &P&(~ -9 &,& 7~) % (c,d) &Q&(~ -3 &,& 7~) \end{array}\quad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ S=\left( \cfrac{-3-9}{2}~,~\cfrac{7+7}{2} \right)\implies S=(-6,7)\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &Q&(~ -3 &,& 7~) % (c,d) &R&(~ -3 &,& 1~) \end{array} \\\\\\ T=\left(\cfrac{-3-3}{2}~,~\cfrac{1+7}{2} \right)\implies T=(-3,4)$

so, what's the distance from S to T?

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &S&(~ -6 &,& 7~) % (c,d) &T&(~ -3 &,& 4~) \end{array}\\\\\\ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ST=\sqrt{[-3-(-6)]^2+[4-7]^2}\implies ST=\sqrt{(-3+6)^2+(4-7)^2} \\\\\\ ST=\sqrt{3^2+(-3)^2}\implies ST=\sqrt{18}\implies ST=\sqrt{9\cdot 2} \\\\\\ ST=\sqrt{3^2\cdot 2}\implies ST=3\sqrt{2}$