The coordinates of the middle points of the sides of a triangle are (1,1), (2,3) and (4,1), find the coordinates of its vertices.

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(1, 1), Q(2, 3) and R(4, 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 + 4 + 8

⇒ 2(x₁ + x₂ + x₃) = 14

⇒ x₁ + x₂ + x₃ = 7 …(vii)

From (i) and (vii), we get

x₃ = 7 – 2 = 5

From (iii) and (vii), we get

x₁ = 7 – 4 = 3

From (v) and (vii), we get

x₂ = 7 – 8 = -1

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

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 2 + 6 + 2

⇒ 2(y₁ + y₂ + y₃) = 10

⇒ y₁ + y₂ + y₃ = 5 …(viii)

From (ii) and (viii), we get

y₃ = 5 – 2 = 3

From (iv) and (vii), we get

y₁ = 5 – 6 = -1

From (vi) and (vii), we get

y₂ = 5 – 2 = 3

Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)