Question

If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3, 4), find the vertices of the triangle.

Answer 1

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

Let D(1, 1), E(2, -3) and F(3, 4) be the midpoints of sides BC, CA and AB respectively. 

By midpoint formula.

∴ x₁ + x₂ = 6 and y₁ + y₂ = 8....(3)

Adding 1,2 and 3, we get,

x₂ +  x₃ + x₁ + x₃ + x₁ + x₂ = 2 +4 + 6

And y₂ + y₃ + y₁ + y₃ + y₁ + y₂ = 2 - 6 + 8

∴ 2(x₁ + x₂ + x3) = 12 and 2(y₁ + y₂ + y₃) = 2

∴x₁ + x₂ + x₃ = 6 and y₁ + y₂ + y₃ = 1

x₁+ 2 = 6 and y₁ + 2 = 2 …from 1

∴  x₁ = 4 and y₁ = 0

Substituting above values in 3, 

4 + x₂ = 6 and 0 + y₂ = 8 

∴ x₂ = 2 and y₂ = 89 

Similarly for equation 2, 

4 + x₃ = 6 and 0 + y₃ = -6 

∴ x₃ = 2 and y₃ = -6 

Hence the vertices of triangle are A(4 , 0), B(2 ,8) and C(0 ,-6)