Question

Two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.

Answer 1

Given that A(- 1, 2) and C(3, 2) are the opposite vertices of a square.

Let us assume the other two vertices be B(x₁, y₁) and D(x₂, y₂) and the midpoint be M

We know that midpoint of AC = Midpoint of BD = M

⇒ x₁ + x₂ = 2 ..... (1)

⇒ y₁ + y₂ = 4 ..... (2)

We know that lengths of the sides of the square are equal.

AB = BC = CD = DA.

We know that distance between two points (x₁, y₁) and (x₂, y₂) is

⇒ AB = BC

⇒ AB² = BC²

⇒ (x₁ - (- 1))² + (y₁ - 2)² = (x₁ - 3)² + (y₁ - 2)²

⇒ x₁² + 1 + 2x₁ + y₁² + 4 - 4y₁ = x₁² - 6x₁ + 9 + y₁² + 4 - 4y₁

⇒ 8x₁ = 8

⇒ x₁ = 8/8

⇒ x₁ = 1 ..... (3)

From (1)

⇒ x₂ = 2 - 1 = 1 ..... (4)

We know that points ABC form right angled isosceles triangle.

We have AB² + BC² = AC²

⇒ 2AB² = (- 1 - 3)² + (2 - 2)²

⇒ 2((1 - (- 1))² + (y₁ - 2)²) = (- 4)² + (0)²

⇒ 2(2² + (y₁ - 2)²) = 8

⇒ 4 + (y₁ - 2)² = 8

⇒ (y₁ - 2)² = 4

⇒ y₁ - 2 = ±2

⇒ y₁ = 2 - 2 (or) y₁ = 2 + 2

⇒ y₁ = 0 (or) y₁ = 4

From (2)

⇒ y₂ = 4 - 0

⇒ y₂ = 4

⇒ y₂ = 4 - 4

⇒ y₂ = 0

It is clear that the other two points are (1, 0) and (1, 4).

∴ The other two points are (1, 0) and (1, 4).