Question

Find the lengths of the medians of a triangle 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)

$\therefore AD$ = $\sqrt{(7-4{)}^2+(-3-1{)}^2}$ = $\sqrt{9+16}$ = 5 units

BE = $\sqrt{(5-5{)}^2+(-2-3{)}^2}$ = $\sqrt{0+25}$ = 5 units

and, CF = $\sqrt{(6-3{)}^2+(0+1{)}^2}$ = $\sqrt{9+1}$ = $\sqrt{10}$ units