The co-ordinates of mid-points of the sides of a triangle ABC are (0, 1), (1, 1) and (1, 0); let us find the co-ordinates of its centroid.

Let the vertices of ΔABC be

A = (x₁, y₁) and

B = (x₂, y₂) and

C = (x₃, y₃)

Centroid of a triangle is given by

G =

Let (0, 1), (1, 1) and (1, 0) be midpoints of AB, BC and AC respectively

Point (0, 1) is the midpoint of AB

⇒ (0, 1) =

Equating x and y coordinates

⇒ = 0 and = 1

⇒ x₁ + x₂ = 0 … (a) and y₁ + y₂ = 2 … (i)

Point (1, 1) is the midpoint of BC

⇒ (1, 1) =

Equating x and y coordinates

⇒ = 1 and = 1

⇒ x₂ + x₃ = 2 … (b) and y₂ + y₃ = 2 … (ii)

Point (1, 0) is the midpoint of AC

⇒ (1, 0) =

Equating x and y coordinates

⇒ = 1 and = 0

⇒ x₁ + x₃ = 2 … (c) and y₁ + y₃ = 0 … (iii)

Adding equations (a), (b) and (c) & adding equations (i), (ii) and (iii) we get

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 4 and y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4

⇒ 2 (x₁ + x₂ + x₃) = 4 and 2 (y₁ + y₂ + y₃) = 4

⇒ x₁ + x₂ + x₃ = 2 and y₁ + y₂ + y₃ = 2

Divide by 3

⇒ = and =

Hence centroid is G =