Examine whether the following points taken in order form a square.

(−1, 2), (1, 0), (3, 2) and (1, 4)

Formula used:

(–1, 2), (1, 0), (3, 2) and (1, 4)

Let the vertices be taken as A (–1, 2), B (1, 0), C (3, 2) and D (1, 4).

Distance of AB

⇒ AB = √ ((1 – (–1))² + ((0 – 2)²)

⇒ AB = √ ((1 + 1)² + (0 – 2)²)

⇒ AB = √ ((2)² + (–2)²)

⇒ AB = √ (4 + 4)

⇒ AB = √8

Distance of BC

⇒ BC= √ ((3 – 1)² + (2 – 0)²)

⇒ BC = √ ((2)² + (2)²)

⇒ BC = √ (4 + 4)

⇒ BC = √ 8

Distance of CD

⇒ CD = √ ((1 – 3)² + (4 – 2))²)

⇒ CD = √ ((–2)² + (2)²)

⇒ CD = √ (4 + 4)

⇒ CD = √8

Distance of AD

⇒ AD = √ ((1 – (–1))² + (4 – 2)²)

⇒ AD = √ ((1 + 1)² + (4 – 2)²)

⇒ AD = √ ((2)² + (2)²)

⇒ AD = √ (4 + 4)

⇒ AD = √ 8

Distance of AC

⇒ AC = √ ((3 – (–1))² + (2 – 2)²)

⇒ AC = √ ((3 + 1) + (2 – 2)²)

⇒ AC = √ ((4)² + (0)²)

⇒ AC = √ (16 + 0)

⇒ AC = √16

⇒ AC = 4

Distance of BD

⇒ BD = √ ((1 – 1)² + (4 – 0)²)

⇒ BD = √ ((0)² + (4)²)

⇒ BD = √ (0 + 16)

⇒ BD = √16

⇒ BD = 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.