Let A(1, 2), B(0, –1) and C(2, –1) be the mid-points of the sides PQ, QR and RP of the triangle PQR.
Let the co-ordinates of P, Q and R be (x1, y1), (x2, y2) and (x3 , y3) respectively. Then, by the mid-point formula.
\(\frac{x_1+x_2}{2}\) = 1 and \(\frac{y_1+y_2}{2}\) = 2 ⇒ x1 + x2 = 2 and y1 + y2 = 4 ...(i)
\(\frac{x_1+x_3}{2}\) = 2 and \(\frac{y_1+y_3}{2}\) = -1 ⇒ x1 + x3 = 2 and y1 + y3 = -2 ...(ii)
\(\frac{x_2+x_3}{2}\) = 0 and \(\frac{y_2+y_3}{2}\) = -1 ⇒ x2 + x3 = 0 and y2 + y3 = -2 ...(iii)
From (i), (ii) and (iii)
x1 + x2 + x1 + x3 +x2 + x3 = 2 + 4 + 0 and y1 + y2 + y1 + y3 + y2 + y3 = 4 + (–2) + (–2)
⇒ x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0
Now, (x1 + x2 + x3) – (x1 + x2) = 3 – 2 ⇒ x3 = 1 and
(y1 + y2 + y3) – (y1 + y2) = 0 – 4 ⇒ y3 = – 4
(x1 + x2 + x3) – (x1 + x3) = 3 – 4 ⇒ x2 = –1 and
(y1 + y2 + y3) – (y1 + y3) = 0 – (–2) ⇒ y2 = 2
(x1 + x2 + x3) – (x2 + x3) = 3 – 0 ⇒ x1 = 3 and
(y1 + y2 + y3) – (y2 + y3) = 0 – (–2) ⇒ y1 = 2
∴ The vertices of the ΔPQR are P(3, 2), Q(–1, 2) and R(1, –4)
Hence, co-ordinates of centroid of ΔPQR = \(\bigg(\frac{3+(-1)+1}{3},\frac{2+2+(-4)}{3}\bigg)\) = (1, 0).