# Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).

Question:

Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).

Solution:

The vertices of ∆ABC are A(0, −1), B(2, 1) and C(0, 3).
Let ADBE and CF be the medians of ∆ABC.

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

$D\left(\frac{2+0}{2}, \frac{1+3}{2}\right)$ i. e. $D\left(\frac{2}{2}, \frac{4}{2}\right)$ i. e. $D(1,2)$

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

$E\left(\frac{0+0}{2}, \frac{-1+3}{2}\right)$ i. e. $E\left(\frac{0}{2}, \frac{2}{2}\right)$ i. e. $E(0,1)$

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

$F\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)$ i. e. $F\left(\frac{2}{2}, \frac{0}{2}\right)$ i. e. $F(1,0)$

$A D=\sqrt{(1-0)^{2}+(2-(-1))^{2}}=\sqrt{(1)^{2}+(3)^{2}}=\sqrt{1+9}=\sqrt{10}$ units.

$B E=\sqrt{(0-2)^{2}+(1-1)^{2}}=\sqrt{(-2)^{2}+(0)^{2}}=\sqrt{4+0}=\sqrt{4}=2$ units.

$C F=\sqrt{(1-0)^{2}+(0-3)^{2}}=\sqrt{(1)^{2}+(-3)^{2}}=\sqrt{1+9}=\sqrt{10}$ units.

Therefore, the lengths of the medians: $A D=\sqrt{10}$ units, $B E=2$ units and $C F=\sqrt{10}$ units