) arc the vertices of ∆ABC.

Fig. 7.33.
Let D(3, 4), E(4. 1), F(2, 0) are the mid-points of sides BC, CA and AB respectively. Since D is the mid-point of BC.



Similarly, E and F are mid-points of CA and AB respectively.

⇒ x₁ + x₂ = 4 and y₁ + y₂ = 0    ...(iii)

From (i), (ii) and (iii), we get
2 (x₁ + x₂ + x₃) = 18 and 2(y₁ + y₂ + y₃) = 10
⇒ x₁ + x₂ + ₃, = 9 and y₁ + y₂ + y₃ = 5 ...(iv)
From (i) and (iv), we get
x₁ + 6 = 9 and y₁ + 8 = 5
⇒    x₁ = 3 and y₁ = -3
∴Co-ordinates of A are (3, -3)
From (ii) and (iv), we get
x₂ + 8 = 9 and y₂ + 2 = 5
⇒    x₂ = 1 and y₂ = 3
∴ Co-ordinates of B are (1, 3)
From (ii) and (iv), we get
4 + x₃ = 9 and 0 + y₃ = 5
⇒    x₃ = 5 and y₃ = 5
∴ Coordinates of C are (5, 5)
Hence, the vertices of ∆ABC are A(3, -3), B(1, 3), C(5, 5).