If G(-2, 1) is the centroid of ΔABC, two of whose vertices are A(1, -6) and B(-5, 2), find the third vertex of the triangle.

Let coordinates of third vertex be C(x₃, y₃).

By using the centroid formula,

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

Given,

G(-2, 1) is the centroid.

Substituting values we get :

$\Rightarrow (-2, 1) = \Big( \dfrac{1 + (-5) + x$3}{3}, \dfrac{-6 + 2 + y3}{3}\Big) \\[1em] \Rightarrow -2 = \Big( \dfrac{1 + (-5) + x3}{3}\Big), 1 = \Big( \dfrac{-6 + 2 + y3}{3}\Big) \\[1em] \Rightarrow -6 = 1 + (-5) + x3, 3 = -6 + 2 + y3 \\[1em] \Rightarrow -6 = -4 + x3, 3 = -4 + y3 \\[1em] \Rightarrow -6 + 4 = x3, 3 + 4 = y3 \\[1em] \Rightarrow x3 = -2, y3 = 7.

Hence, coordinates of third vertex = (-2, 7).