ABC is a triangle and G(4, 3) is its centroid. If A(1, 3), B(4, b) and C(a, 1) be the vertices, find the values of a and b and hence find the length of side BC.

By using the centroid formula,

$(x, y) = \Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

Given,

A(1, 3), B(4, b) and C(a, 1).

G(4, 3) is the centroid of triangle ABC.

Solving for x-coordinate :

$\Rightarrow 4 = \dfrac{1 + 4 + a}{3} \\[1em] \Rightarrow 12 = 5 + a \\[1em] \Rightarrow a = 12 - 5 \\[1em] \Rightarrow a = 7.$

C = (a, 1) = (7, 1)

Solving for y-coordinate :

$\Rightarrow 3 = \dfrac{3 + b + 1}{3} \\[1em] \Rightarrow 9 = 4 + b \\[1em] \Rightarrow b = 9 - 4 \\[1em] \Rightarrow b = 5.$

B = (4, b) = (4, 5)

By using distance formula,

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

Substituting values we get:

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

Hence, a = 7 and b = 5, length of BC = 5 units.