A(3, 1), B(y, 4) and C(1, x) are vertices of triangle ABC and G(3, 4) is its centroid. Find the values of x and y. Also, find the length of side BC.

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

Substituting values we get,

$\Rightarrow G_1 = \Big(\dfrac{3 + y + 1}{3}, \dfrac{1 + 4 + x}{3}\Big) \\[1em] \Rightarrow (3, 4) = \Big(\dfrac{y + 4}{3}, \dfrac{x + 5}{3}\Big) \\[1em] \Rightarrow 3 = \dfrac{y + 4}{3} \text{ and } 4 = \dfrac{x + 5}{3} \\[1em] \Rightarrow y + 4 = 9 \text{ and } 12 = x + 5 \\[1em] \Rightarrow y = 5 \text{ and } x = 7.$

B = (y, 4) = (5, 4) and C = (1, x) = (1, 7).

Distance between two points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

$BC = \sqrt{(1 - 5)^2 + (7 - 4)^2} \\[1em] = \sqrt{(-4)^2 + (3)^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} = 5 \text{ units}. \\[1em]$

Hence, x = 7, y = 5 and BC = 5 units.