# Find the lengths of the medians of

Question:

Find the lengths of the medians of a ΔABC having vertices at A(5, 1), B(1, 5), and C(−3, −1).

Solution:

We have to find the lengths of the medians of a triangle whose co-ordinates of the vertices are A (5, 1); B (1, 5) and C (−3,−1).

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,

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

Now equate the individual terms to get,

$x=3$

$y=3$

So co-ordinates of P is (3, 3)

Similarly mid-point Q of side BC can be written as,

$Q(x, y)=\left(\frac{1-3}{2}, \frac{5-1}{2}\right)$

Now equate the individual terms to get,

$x=1$

$y=0$

So co-ordinates of R is (1, 0)

Therefore length of median from A to the side BC is,

$\mathrm{AQ}=\sqrt{(5+1)^{2}+(1-2)^{2}}$

$=\sqrt{25}$

$=5$

Similarly length of median from C to the side AB is

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

$=\sqrt{36+16}$

$=2 \sqrt{13}$