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

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

KnowledgeBoat · Accepted Answer

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). ⇒ x₁ + x₂ = 4 ........(1) ⇒ y₁ + y₂ = -2 ........(2) Mid-point of BC, i.e. E(-1, 4). ⇒ x₂ + x₃ = -2 ........(3) ⇒ y₂ + y₃ = 8 ........(4) Mid-point of AC, i.e. F(-2, 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).

Mathematics

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

Section Formula

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