If (1,2), (0,-1) and (2,-1) are the middle points of the sides of the triangle, find the coordinates of its centroid.

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(1, 2), Q(0, -1) and R(2, -1) are the midpoints of AB, BC and CA. Then,

…(i)

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

Adding (i), (iii) and (v), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 2 + 0 + 4

⇒ 2(x₁ + x₂ + x₃) = 6

⇒ x₁ + x₂ + x₃ = 3 …(vii)

From (i) and (vii), we get

x₃ = 3 – 2 = 1

From (iii) and (vii), we get

x₁ = 3 – 0 = 3

From (v) and (vii), we get

x₂ = 3 – 4 = -1

Now adding (ii), (iv) and (vi), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4 + (-2) + (-2)

⇒ 2(y₁ + y₂ + y₃) = 0

⇒ y₁ + y₂ + y₃ = 0 …(viii)

From (ii) and (viii), we get

y₃ = 0 – 4 = -4

From (iv) and (vii), we get

y₁ = 0 – (-2) = 2

From (vi) and (vii), we get

y₂ = 0 – (-2) = 2

Hence, the vertices of ΔABC are A(3, 2), B(-1, 2) and C(1, -4)

Now, we have to find the centroid of a triangle

The vertices of a triangle are A(3, 2), B(-1, 2) and C(1, -4)

Here, x₁ = 3, x₂ = -1, x₃ = 1

and y₁ = 2, y₂ = 2, y₃ = -4

Let the coordinates of the centroid be(x,y)

So,

= (1,0)

Hence, the centroid of a triangle is (1, 0)