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

(12, 9), (20, −6), (5, −14) and (−3, 1)

Formula used:

(12, 9), (20, –6), (5, –14) and (–3, 1)

Let the vertices be taken as A (12, 9), B (20, –6), C (5, –14) and D (–3, 1).

Distance of AB

⇒ AB = √ ((20 – 12)² + ((–6 – 9)²)

⇒ AB = √ ((8)² + (–15)²)

⇒ AB = √ (64 + 225)

⇒ AB = √289

Distance of BC

⇒ BC= √ ((5 – 20)² + (–14 – (–6))²)

⇒ BC= √ ((5 – 20)² + (–14 + 6)²)

⇒ BC = √ ((–15)² + (–8)²)

⇒ BC = √ (225 + 64)

⇒ BC = √ 289

Distance of CD

⇒ CD = √ ((–3 – 5)² + (1 – (–14))²)

⇒ CD = √ ((–3 – 5)² + (1 + 14)²)

⇒ CD = √ ((–8)² + (15)²)

⇒ CD = √ (64 + 225)

⇒ CD = √289

Distance of AD

⇒ AD = √ ((–3 – 12)² + (1 – 9)²)

⇒ AD = √ ((–15)² + (–8)²)

⇒ AD = √ (225 + 64)

⇒ AD = √ 289

Distance of AC

⇒ AC = √ ((5 – 12)² + (–14 – 9)²)

⇒ AC = √ ((–7)² + (–23)²)

⇒ AC = √ (49 + 529)

⇒ AC = √578

Distance of BD

⇒ BD = √ ((–3 – 20)² + (1 – (–6))²)

⇒ BD = √ ((–3 – 20)² + (1 + 6)²)

⇒ BD = √ ((–23)² + (7)²)

⇒ BD = √ (529 + 49)

⇒ BD = √ 578

AB = BC = CD = DA = √ 289 (That is, all the sides are equal.)

AC = BD = √578. (That is, the diagonals are equal.)

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