If A (−1, 3) , B(1, −1) and C (5, 1) are the vertices of a triangle ABC,

We have a triangle $\triangle \mathrm{ABC}$ in which the co-ordinates of the vertices are $\mathrm{A}(-1,3) \mathrm{B}(1,-1)$ and

$\mathrm{C}(5,1)$. In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point D of side BC can be written as,

$\mathrm{D}(x, y)=\left(\frac{5+1}{2}, \frac{1-1}{2}\right)$

Now equate the individual terms to get,

$x=3$

$y=0$

So co-ordinates of D is (3, 0)

So the length of median from A to the side BC,

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

$=\sqrt{16+9}$

$=5$