Question

Find the lengths of the medians of a $△ABC$ whose vertices are A(0,-1), B(2,1) and C(0, 3)

Answer 1

Let D be the midpoint of BC. So, the coordinates of D are

$D=\frac{(2+0)}{2},\frac{(1+3)}{2}.$

$\Rightarrow D=\frac{2}{2},\frac{4}{2}$

$D=(1,2)$

Let E be the midpoint of AC. So, the coordinates of E are

$E=\frac{(0+0)}{2},\frac{(-1+3)}{2}$

$E=(\frac{0}{2},\frac{2}{2}$

= E(0,1)

Let F be the midpoint of AB. So, the coordinates of F are

$F=\frac{(0+2)}{2},\frac{(-1+1)}{2}$

$F=\frac{2}{2},\frac{0}{2}$

= F(1,0)

$AD=\sqrt{(1-0{)}^2+(2-(-1){)}^2}$

$AD=(1{)}^2+(3{)}^2$

$AD=\sqrt{1+9}$

$AD=\sqrt{1}0units$

$BE=\sqrt{(0-2{)}^2+(1-1{)}^2}$

$BE=\sqrt{(-2{)}^2+(0{)}^2}$

$BE=\sqrt{4+0}$

$BE=2units$

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

$CF=\sqrt{(1{)}^2+(-3{)}^2}$

$CF=\sqrt{1+9}$

$CF=\sqrt{10}units$

Therefore, the lengths of the medians:$AD=\sqrt{10}units,BE=2units,CF=\sqrt{10}units$