Question

Find the co-ordinates of centroid of a triangle if points D(–7, 6), E(8, 5) and F(2, –2) are the mid-points of the sides of that triangle.

Answer 1

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of MBC.

Given that D(–7, 6), E(8, 5) and F(2, –2) are the midpoints of sides AB, BC and AC respectively.

Let G(x, y) be the centroid.

D is now the midpoint of AB.

∴ By midpoint formula. 

–7 = `(x_1 + x_2)/2`

∴ x₁ + x₂ = –7 × 2

∴ x₁ + x₂ = –14  ......(i)

Similarly x₂ + x₃ = 16  ......(ii)

And x₁ + x₃ = 4  ......(iii)

Adding equations (i), (ii), and (iii), we get

x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = –14 + 16 + 4

∴ 3x₁ + 2x₂ + 2x₃ = 6

∴ 2(x₁ + x₂ + x₃) = 6

∴ x₁ + x₂ + x₃ = `6/2`

∴ x₁ + x₂ + x₃ = 3  ......(iv)

D is the midpoint of AB.

∴  By the midpoint formula,

6 = `("y"_1 + "y"_2)/2`

∴ y₁ + y₂ = 12  ......(v)

Similarly, y₂ + y₃ = 10  ......(vi)

y₁ + y₃ = –4  ......(vii)

Adding equations (v), (vi), and (vii), we get

y₁ + y₂ + y₂ + y₃ + y₁ + y₄ = 12 + 10 – 4

∴ 2y₁ + 2y₂ + 2y₃ = 18

∴ 2(y₁ + y₂ + y₃) = 18

∴ y₁ + y₂ + y₃ = `18/2`

∴ y₁ + y₂ + y₃ = 9  ......(viii)

G(x, y) is the centroid of ΔABC.

∴ By centroid formula,

a = `(x_1 + x_2 + x_3)/3`

∴ x = `3/3`  ......[From (iv)]

∴ x = 1

b = `("y"_1 + "y"_2 + "y"_3)/3`

∴ y = `9/3` ......[From (viii)]

∴ y = 3