The vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(0,-1), \mathrm{B}(2,1)$ and $\mathrm{C}(0,3)$.

Let $A D, B E$ and $C F$ be the medians of $\triangle A B C$.

Let $\mathrm{D}$ be the midpoint of $\mathrm{BC}$. So, the coordinates of $\mathrm{D}$ are

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

Let $\mathrm{E}$ be the midpoint of $\mathrm{AC}$. So the coordinate of $\mathrm{E}$ are

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

Let $F$ be the midpoint of $A B$. So, the coordinates of $F$ are

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

$\mathrm{AD}=\sqrt{(1-0)^{2}+(2-(-1))^{2}}=\sqrt{(1)^{2}+(3)^{2}}=\sqrt{1+9}=\sqrt{10} \text { units. }$

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

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

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