Question

If the coordinates of the mid-points of the sides of a triangle are (3, 4), (4, 6) and (5, 7), find its vertices.

Answer 1

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

Let D(3, 4), E(4, 6) and F(5, 7) be the midpoints of sides BC, CA and AB respectively.

By midpoint formula.

x =\(\frac{x_1+x_2}2\), y = \(\frac{y_1+y_2}2\)

For midpoint D(3, 4) of side BC,

 3 =\(\frac{x_2+x_3}2\), 4 = \(\frac{y_2+y_3}2\)

∴ x₂ + x₃ = 6 and y₂ + y₃ = 8 …(1) 

For midpoint E(4, 6) of side CA,

  4 =\(\frac{x_1+x_3}2\), 4 = \(\frac{y_1+y_3}2\)

∴ x₁ + x₃ = 8 and y₁ + y₃ = 12 …(2) 

For midpoint F(5, 7) of side AB,

   5 =\(\frac{x_1+x_2}2\), 7 = \(\frac{y_1+y_2}2\)

∴ x₁ + x₂ = 10 and y₁ + y₂ = 14 …(3) 

Adding 1,2 and 3, we get,

x₂ + x₃ + x₁ + x₃ + x₁ + x₂ = 6 + 8 + 10

And y₂ + y₃ + y₁ + y₃ + y₁ + y₂ = 8 + 12 + 14

∴ 2(x₁ + x₂ + x₃) = 24 and 2(y₁ + y₂ + y₃) = 34

∴ x₁ + x₂ + x₃ = 12 and y₁ + y₂ + y₃ = 17

x₁ + 6 = 12 and y₁ + 8 = 17 …from 1

∴ x₁ = 6 and y₁ = 9

Substituting above values in 3, 

6 + x₂ and 9 + y₂ = 14 

∴ x₂ = 4 and y₂ = 5 

Similarly for equation 2, 

6 + x₃ = 8 and 9 + y₃ = 12 

∴ x₃ = 2 and y₃ = 3 

Hence the vertices of triangle are A(6 , 9), B(4 ,5) and C(2 ,3)