The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.

Let co-ordinates of P = (x, y).

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,

$\Rightarrow (2, -5) = \Big(\dfrac{x + (-6) + 11}{3}, \dfrac{y + 5 + 8}{3}\Big) \\[1em] \Rightarrow 2 = \dfrac{x - 6 + 11}{3} \text{ and } -5 = \dfrac{y + 13}{3} \\[1em] \Rightarrow 6 = x + 5 \text{ and } -15 = y + 13 \\[1em] \Rightarrow x = 1 \text{ and } y = -28.$

P = (x, y) = (1, -28).

Hence, co-ordinates of P = (1, -28).