Find the length of the medians of a ΔABC having vertices at A(0, −1), B(2, 1) and C(0, 3).
We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (0,−1); B (2, 1) and C (0, 3).
So we should find the mid-points of the sides of the triangle.
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 P of side AB can be written as,
$P(x, y)=\left(\frac{2+0}{2}, \frac{1-1}{2}\right)$
Now equate the individual terms to get,
$x=1$
$y=0$
So co-ordinates of P is (1, 0)
Similarly mid-point Q of side BC can be written as,
$Q(x, y)=\left(\frac{2+0}{2}, \frac{3+1}{2}\right)$
Now equate the individual terms to get,
$x=1$
$y=2$
So co-ordinates of Q is (1, 2)
Similarly mid-point R of side AC can be written as,
$R(x, y)=\left(\frac{0+0}{2}, \frac{3-1}{2}\right)$
Now equate the individual terms to get,
$x=0$
$y=1$
So co-ordinates of R is (0, 1)
Therefore length of median from A to the side BC is,
$\mathrm{AQ}=\sqrt{(0-1)^{2}+(-1-2)^{2}}$
$=\sqrt{1+9}$
$=\sqrt{10}$
Similarly length of median from B to the side AC is,
BR $=\sqrt{(2-0)^{2}+(1-1)^{2}}$
$=\sqrt{4}$
$=2$
Similarly length of median from C to the side AB is
$\mathrm{CP}=\sqrt{(0-1)^{2}+(3-0)^{2}}$
$=\sqrt{1+9}$
$=\sqrt{10}$