If A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of a triangle ABC, find the length of the median through A and the coordinates of the centroid of triangle ABC.

A(5, -1), B(-3, -2) and C(-1, 8) are the vertices of △ABC. Below figure shows the triangle:

Let D be the midpoint of BC. By midpoint formula, coordinates of D are,

$= \Big(\dfrac{-3 + (-1)}{2}, \dfrac{-2 + 8}{2}\Big) \\[1em] = \Big(\dfrac{-4}{2}, \dfrac{6}{2}\Big) \\[1em] = (-2, 3).$

By distance formula we get,

$AD = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] = \sqrt{(-2 - 5)^2 + (3 - (-1))^2} \\[1em] = \sqrt{(-7)^2 + (3 + 1)^2} \\[1em] = \sqrt{49 + 16} \\[1em] = \sqrt{65} \text{ units.}

Coordinates of centroid is given by $\Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

$= \Big(\dfrac{5 + (-3) + (-1)}{3}, \dfrac{-1 + (-2) + 8}{3}\Big) \\[1em] = \Big(\dfrac{1}{3}, \dfrac{5}{3}\Big).$

Hence, the coordinates of the centroid of triangle is $\Big(\dfrac{1}{3}, \dfrac{5}{3}\Big)$ and the length of the median through A is $\sqrt{65}$ units.