Question

Find the lengths of the medians of a triangle ABC whose vertices are A(7, –3), B(5, 3) and C(3, –1). [4 MARKS]

Answer 1

Formula: 1 Mark
Concept: 1 Mark
Answer: 2 Marks

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

Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore AD=\sqrt{(7-4{)}^2+(-3-1{)}^2}$

$\Rightarrow AD=\sqrt{9+16}=5units$

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

$\Rightarrow BE=\sqrt{0+25}=5units$

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

$\Rightarrow CF=\sqrt{9+1}=\sqrt{10}units$