Question

Show that the points A(2, 1), B(0, 3), C(–2, 1) and D(0, –1) are the vertices of a square.

Answer 1

Given: A(2, 1), B(0, 3), C(–2, 1), D(0, –1).

Step-wise calculation:

1. Compute squared side-lengths using the distance formula:

AB² = (0 – 2)² + (3 – 1)²

= (–2)² + 2²

= 4 + 4

= 8

⇒ AB = `2sqrt(2)`

BC² = (–2 – 0)² + (1 – 3)²

= (–2)² + (–2)²

= 4 + 4

= 8

⇒ BC = `2sqrt(2)`

CD² = (0 – (–2))² + (–1 – 1)² 

= 2² + (–2)²

= 4 + 4

= 8

⇒ CD = `2sqrt(2)`

DA² = (2 – 0)² + (1 – (–1))²

= 2² + 2²

= 4 + 4

= 8

⇒ DA = `2sqrt(2)` 

Thus, AB = BC = CD = DA, all four sides are equal.

2. Compute diagonals:

AC² = (–2 – 2)² + (1 – 1)² 

= (–4)² + 0²

= 16

⇒ AC = 4

BD² = (0 – 0)² + (–1 – 3)² 

= 0 + (–4)²

= 16

⇒ BD = 4 

Thus, AC = BD diagonals are equal.

Slope AB = `(3 - 1)/(0 - 2)` 

= `2/(-2)` 

= –1

Slope BC = `(1 - 3)/(-2 - 0)`

= `(-2)/(-2)` 

= 1

And (–1)·1 = –1 

So, AB ⟂ BC.

All four sides are equal and the diagonals are equal and adjacent sides are perpendicular, so the quadrilateral ABCD is a square.