The vertices of a triangle are A(-5, 3), B(p, -1) and C(6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, -1).

The vertices of the triangle ABC are A(-5, 3), B(p, -1), C(6, q) and the centroid of the △ABC will be

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

Putting the values in above formula,

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

Given, the centroid of the triangle is (1, -1).

Comparing we get,

$\dfrac{1 + p}{3} = 1 \text{ and } \dfrac{q + 2}{3} = -1$
⇒ 1 + p = 3 and 2 + q = -3
⇒ p = 3 - 1 and q = -3 - 2
⇒ p = 2 and q = -5.

Hence, the values are p = 2 and q = -5.