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

(5, 2), (1, 5), (−2, 1) and (2, −2)

Formula used:

(5, 2), (1, 5), (–2, 1) and (2, –2)

Let the vertices be taken as A (5, 2), B (1, 5), C (–2, 1) and D (2, –2).

Distance of AB

⇒ AB = √ ((1 – 5)² + ((5 – 2)²)

⇒ AB = √ ((–4)² + (3)²)

⇒ AB = √ (16 + 9)

⇒ AB = √25

⇒ AB = 5

Distance of BC

⇒ BC= √ ((–2 – 1)² + (1 – 5)²)

⇒ BC = √ ((–3)² + (–4)²)

⇒ BC = √ (9 + 16)

⇒ BC = √ 25

⇒ BC = 5

Distance of CD

⇒ CD = √ ((2 – (–2))² + (–2 – 1)²)

⇒ CD = √ ((2 + 2)² + (–2 – 1)²)

⇒ CD = √ ((4)² + (–3)²)

⇒ CD = √ (16 + 9)

⇒ CD = √25

⇒ CD = 5

Distance of AD

⇒ AD = √ ((2 – 5)² + (–2 – 2)²)

⇒ AD = √ ((–3)² + (–4)²)

⇒ AD = √ (9 + 16)

⇒ AD = √ 25

⇒ AD = 5

Distance of AC

⇒ AC = √ ((–2 – 5)² + (1 – 2)²)

⇒ AC = √ ((–7)² + (–1)²)

⇒ AC = √ (49 + 1)

⇒ AC = √50

⇒ AC = 5√2

Distance of BD

⇒ BD = √ ((2 – 1)² + (–2 – 5)²)

⇒ BD = √ ((1)² + (–7)²)

⇒ BD = √ (1 + 49)

⇒ BD = √ 50

⇒ BD = 5√2

AB = BC = CD = DA = 5 (That is, all the sides are equal.)

AC = BD = 5√2. (That is, the diagonals are equal.)

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