A(-4, 2), B(0, 2) and C(-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB, respectively. Show that the centroid of △PQR is the same as the centroid of △ABC.

By formula,

Mid-point (M) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Given,

P is mid-point of BC.

$\therefore P = \Big(\dfrac{0 + (-2)}{2}, \dfrac{2 + (-4)}{2}\Big) \\[1em] = \Big(\dfrac{-2}{2}, \dfrac{-2}{2}\Big) \\[1em] = (-1, -1).$

Q is mid-point of CA.

$\therefore Q = \Big(\dfrac{-2 + (-4)}{2}, \dfrac{-4 + 2}{2}\Big) \\[1em] = \Big(\dfrac{-6}{2}, \dfrac{-2}{2}\Big) \\[1em] = (-3, -1).$

R is mid-point of AB.

$\therefore R = \Big(\dfrac{-4 + 0}{2}, \dfrac{2 + 2}{2}\Big) \\[1em] = \Big(\dfrac{-4}{2}, \dfrac{4}{2}\Big) \\[1em] = (-2, 2).$

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

Let G₁ and G₂ be centroid of △ABC and △PQR.

Substituting values we get,

$G$1 = \Big(\dfrac{-4 + 0 + (-2)}{3}, \dfrac{2 + 2 + (-4)}{3}\Big) \\[1em] = \Big(-\dfrac{6}{3}, \dfrac{0}{3}\Big) \\[1em] = (-2, 0). \\[1em] G2 = \Big(\dfrac{(-1) + (-3) + (-2)}{3}, \dfrac{(-1) + (-1) + 2}{3}\Big) \\[1em] = \Big(\dfrac{-6}{3}, \dfrac{0}{3}\Big) \\[1em] = (-2, 0).

Since, G₁ = G₂.

Hence, proved that the centroid of △PQR is the same as the centroid of △ABC.