The points (2, -1), (-1, 4) and (-2, 2) are mid-points of the sides of a triangle. Find its vertices.

Let D = (2, -1), E = (-1, 4) and F = (-2, 2).

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

Mid-point of AB, i.e. D(2, -1).

$\therefore 2 = \dfrac{x$1 + x2}{2} \text{ and } -1 = \dfrac{y1 + y2}{2}

⇒ x₁ + x₂ = 4 ……..(1)

⇒ y₁ + y₂ = -2 ……..(2)

Mid-point of BC, i.e. E(-1, 4).

$\therefore -1 = \dfrac{x$2 + x3}{2} \text{ and } 4 = \dfrac{y2 + y3}{2}

⇒ x₂ + x₃ = -2 ……..(3)

⇒ y₂ + y₃ = 8 ……..(4)

Mid-point of AC, i.e. F(-2, 2).

$\therefore -2 = \dfrac{x$1 + x3}{2} \text{ and } 2 = \dfrac{y1 + y3}{2}

⇒ x₁ + x₃ = -4 ……..(5)

⇒ y₁ + y₃ = 4 ……..(6)

Adding 1, 3 and 5 we get,

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 4 + (-2) + (-4)

⇒ 2(x₁ + x₂ + x₃) = -2

⇒ x₁ + x₂ + x₃ = -1.

From (1),

⇒ 4 + x₃ = -1

⇒ x₃ = -5.

Substituting value of x₃ in (5) we get,

⇒ x₁ + (-5) = -4

⇒ x₁ = 1.

Substituting value of x₃ in (3) we get,

⇒ x₂ + (-5) = -2

⇒ x₂ = 3.

Adding (2), (4) and (6) we get,

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = -2 + 8 + 4

⇒ 2(y₁ + y₂ + y₃) = 10

⇒ y₁ + y₂ + y₃ = 5

From (2)

⇒ -2 + y₃ = 5

⇒ y₃ = 7.

Substituting value of y₃ in (4) we get,

⇒ y₂ + 7 = 8

⇒ y₂ = 1.

Substituting value of y₃ in (6) we get,

⇒ y₁ + 7 = 4

⇒ y₁ = -3.

A = (x₁, y₁) = (1, -3), B = (x₂, y₂) = (3, 1), C = (x₃, y₃) = (-5, 7).

Hence, A = (1, -3), B = (3, 1) and C = (-5, 7).