Question

Find the lengths of the medians of a \(\triangle \mathrm{ABC}\) whose vertices are \(\mathrm{A}(-1, 3), \mathrm{B}(1, -1)\) and \(\mathrm{C}(5, 1)\). Also, find the co-ordinates of the centroid of \(\Delta \mathrm{ABC}\).

Answer 1

Let D, E, F be the mid-points of BC, CA and AB.

$$D = \left(\frac{1 + 5}{2}, \frac{-1 + 1}{2}\right) = (3, 0)$$

$$E = \left(\frac{5 + (-1)}{2}, \frac{1 + 3}{2}\right) = (2, 2)$$

$$F = \left(\frac{-1 + 1}{2}, \frac{3 - 1}{2}\right) = (0, 1)$$

Median AD:

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

$$= \sqrt{4^2 + (-3)^2}$$

$$= \sqrt{16 + 9}$$

$$= \sqrt{25}$$

$$= 5\text{ units}$$

Median BE:

$$BE = \sqrt{(2 - 1)^2 + (2-(-1))^2}$$

$$= \sqrt{1^2 + 3^2}$$

$$= \sqrt{10}\text{ units}$$

Median CF:

$$CF = \sqrt{(0-5)^2 + (1-1)^2}$$

$$= \sqrt{(-5)^2 + 0^2}$$

$$= 5\text{ units}$$

Centroid G:

$$G = \left(\frac{-1 + 1 + 5}{3}, \frac{3 - 1 + 1}{3}\right)$$

$$= \left(\frac{5}{3}, \frac{3}{3}\right)$$

$$= \left(\frac{5}{3}, 1\right)$$