Examine whether t

Formula used:

(0, –1), (2, 1), (0, 3) and (–2, 1)

Let the vertices be taken as A (0, –1), B (2, 1), C (0, 3) and D (–2, 1).

Distance of AB

⇒ AB = √ ((2 – 0)² + ((1 – (–1))²)

⇒ AB = √ ((2 – 0))² + (1 + 1)²)

⇒ AB = √ ((2)² + (2)²)

⇒ AB = √ (4 + 4)

⇒ AB = √ 8

Distance of BC

⇒ BC= √ ((0 – 2)² + (3 – 1)²)

⇒ BC = √ ((–2)² + (2)²)

⇒ BC = √ (4 + 4)

⇒ BC = √ 8

Distance of CD

⇒ CD = √ ((–2 – 0)² + (1 – 3)²)

⇒ CD = √ ((–2)² + (–2)²)

⇒ CD = √ (4 + 4)

⇒ CD = √8

Distance of AD

⇒ AD = √ ((–2 – 0)² + (1 – (–1))²)

⇒ AD = √ ((–2 – 0)² + (1 + 1)²)

⇒ AD = √ ((–2)² + (2)²)

⇒ AD = √ (4 + 4)

⇒ AD = √ 8

Distance of AC

⇒ AC = √ ((0 – 0)² + (3 – (–1))²)

⇒ AC = √ ((0 – 0)² + (3 + 1)²)

⇒ AC = √ ((0)² + (4)²)

⇒ AC = √ (0 + 16)

⇒ AC = √ 16

⇒ AC = 4

Distance of BD

⇒ AC = √ ((–2 – 2)² + (1 – 1)²)

⇒ AC = √ ((–4)² + (0)²)

⇒ AC = √ (16 + 0)

⇒ AC = √ 16

⇒ AC = 4

AB = BC = CD = DA = √8 (That is, all the sides are equal.)

AC = BD = 4. (That is, the diagonals are equal.)

Hence the points A, B, C and D form a square.