Show that the following points form an equilateral triangle.

(2, 2), (−2, −2) and (−2√3, 2√3)

Formula used:

(2, 2), (–2, –2) and (–2√3, 2√3)

Let the points be A (2, 2), B (–2, –2) and C (–2√3, 2√3)

Distance of AB

⇒ AB = √ ((–2 – 2)² + (–2 – 2)²)

⇒ AB = √ ((–4)² + (–4)²)

⇒ AB = √ (16 + 16)

⇒ AB = √32

⇒ AB = 4√2

Distance of BC

⇒ B C= √ ((–2√3 – (–2))² + (2√3 – (–2))²)

⇒ B C= √ ((–2√3 + 2))² + (2√3 + 2)²)

⇒ BC = √ (((–2√3)² + 2 (–2√3) (2) + (2)²) + ((2√3)² + 2 (2√3) (2) + (2)²))

⇒ BC = √ (12 – 8√3 + 4 + 12 + 8√3 + 4)

⇒ BC = √ (12 + 4 + 12 + 4

⇒ BC = √ 32

⇒ BC = 4√2

Distance of AC

⇒ AC= √ ((–2√3 – 2))² + (2√3 – 2)²)

⇒ AC = √ (((–2√3)² + 2 (–2√3) (–2) + (2)²) + ((2√3)² + 2 (2√3) (–2) + (–2)²))

⇒ AC = √ (12 + 8√3 + 4 + 12 – 8√3 + 4)

⇒ AC = √ (12 + 4 + 12 + 4)

⇒ AC = √ 32

⇒ AC = 4√2

∴ AB = BC = AC = 4√2

Since, all the sides are equal the points form an equilateral triangle.