Show that the following points form an equilateral triangle.

(−√3, 1), (2√3, −2) and (2√3, 4)

Formula used:

(–√3, 1), (2√3, –2) and (2√3, 4)

Let the points be A (–√3, 1), B (2√3, –2) and C (2√3, 4)

Distance of AB

⇒ AB = √ ((2√3 – (–√3))² + (–2 – 1)²)

⇒ AB = √ ((2√3 + √3))² + (–2 – 1)²)

⇒ AB = √ ((12 + 12 + 3)² + (–3)²)

⇒ AB = √ (27 + 9)

⇒ AB = √36

⇒ AB = 6

Distance of BC

⇒ B C= √ ((2√3 – 2√3)² + (4 – (–2))²)

⇒ B C= √ ((2√3 – 2√3)² + (4 + 2)²)

⇒ BC = √ ((0)² + (6)²)

⇒ BC = √ (0 + 36)

⇒ BC = √36

⇒ BC = 6

Distance of AC

⇒ AC = √ ((2√3 – (–√3))² + (4 – 1))²)

⇒ AC = √ ((2√3 + √3))² + (4 – 1))²)

⇒ AC = √ ((3√3)² + (3)²)

⇒ AC = √ (27 + 9)

⇒ AC = √ 36

⇒ AC = 6

∴ AB = BC = AC = 6

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