Question

Find the length of the medians of a $△$ ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

Answer 1

Let D,E,F be the mid-points of the sides BC, CA and AB respectively. Then, the coordinates of D, E and F are$D(\frac{5+3}{2},\frac{3+1}{2})=D(4,1),E(\frac{3+7}{2},\frac{-1-3}{2})=E(5,-2)$

$and,F(\frac{7+5}{2},\frac{-3+3}{2})=F(6,0)$

$\therefore AD=\sqrt{{(7-4)}^2+{(-3-1)}^2}=\sqrt{9+16}=5units$
$BE=\sqrt{{(5-5)}^2+{(-2-3)}^2}=\sqrt{0+25}=5units$
$and,CF=\sqrt{{(6-3)}^2+{(0+1)}^2}=\sqrt{9+1}=\sqrt{10}units.$