If the points (10,5),(8,4) and (6,6) are the mid-points of the sides of a triangle, find its vertices.

Consider a ΔABC with A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). If P(10, 5), Q(8, 4) and R(6, 6) 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 = 20 + 16 + 12

⇒ 2(x₁ + x₂ + x₃) = 48

⇒ x₁ + x₂ + x₃ = 24 …(vii)

From (i) and (vii), we get

x₃ = 24 – 20 = 4

From (iii) and (vii), we get

x₁ = 24 – 16 = 8

From (v) and (vii), we get

x₂ = 24 – 12 = 12

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

y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 10 + 8 + 12

⇒ 2(y₁ + y₂ + y₃) = 30

⇒ y₁ + y₂ + y₃ = 15 …(viii)

From (ii) and (viii), we get

y₃ = 15 – 10 = 5

From (iv) and (vii), we get

y₁ = 15 – 8 = 7

From (vi) and (vii), we get

y₂ = 15 – 12 = 3

Hence, the vertices of ΔABC are A(8, 7), B(12, 3) and C(4, 5)