Find the lengths of the medians of a ΔABC whose vertices are A(-1, 3), B(1, -1) and C(5, 1). Also, find the co-ordinates of the centroid of ΔABC.

Using the Midpoint Formula,

$(x, y) = \Big( \dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2} \Big).

Let D be the midpoint of BC.

$\Rightarrow D = \Big( \dfrac{1 + 5}{2}, \dfrac{-1 + 1}{2} \Big) \\[1em] \Rightarrow D = \Big( \dfrac{6}{2}, \dfrac{0}{2} \Big) \\[1em] \Rightarrow D = (3, 0).$

Let E be the midpoint of AC.

$\Rightarrow E = \Big( \dfrac{-1 + 5}{2}, \dfrac{3 + 1}{2} \Big) \\[1em] \Rightarrow E = \Big( \dfrac{4}{2}, \dfrac{4}{2} \Big) \\[1em] \Rightarrow E = (2, 2).$

Let F be the midpoint of AB.

$\Rightarrow F = \Big( \dfrac{-1 + 1}{2}, \dfrac{3 + (-1)}{2} \Big) \\[1em] \Rightarrow F = \Big( \dfrac{0}{2}, \dfrac{2}{2} \Big) \\[1em] \Rightarrow F = (0, 1).$

Length of Median AD : A(-1, 3) and D(3, 0)

Using Distance Formula,

$D = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2}.

Substituting values we get:

$AD = \sqrt{(3 - (-1))^2 + (0 - 3)^2} \\[1em] = \sqrt{(4)^2 + (-3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Length of Median BE:

B(1, -1) and E(2, 2)

$BE = \sqrt{(2 - 1)^2 + (2 - (-1))^2} \\[1em]= \sqrt{(1)^2 + (3)^2} \\[1em] = \sqrt{1 + 9} \\[1em] = \sqrt{10} \text{ units}.$

Length of Median CF:

C(5, 1) and F(0, 1)

$CF = \sqrt{(0 - 5)^2 + (1 - 1)^2} \\[1em] = \sqrt{(-5)^2 + (0)^2} \\[1em] = \sqrt{25 + 0} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Centroid of triangle (G) $= \Big( \dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3} \Big)

Substituting values we get :

$\Rightarrow \text{Centroid} = \Big(\dfrac{-1 + 1 + 5}{3}, \dfrac{3 + (-1) + 1}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, \dfrac{3}{3}\Big) \\[1em] = \Big(\dfrac{5}{3}, 1\Big).$

Hence, AD = 5 units , $BE = \sqrt{10}$ units, CF = 5 units, coordinates of centroid are $\Big(\dfrac{5}{3}, 1\Big)$.