Show that the following points form an equilateral triangle.

(a, 0), (−a, 0) and (0, a√3)

Formula used:

(a, 0), (–a, 0) and (0, a√3)

Let the points be A (a, 0), B (–a, 0) and C (0, a√3)

Distance of AB

⇒ AB = √ ((–a – a)² + (0 – 0)²)

⇒ AB = √ ((–2a)² + (0)²)

⇒ AB = √ (4a² + 0)

⇒ AB = √4a²

⇒ AB = 2a

Distance of BC

⇒ B C= √ ((0 – a)² + (a√3 – 0)²)

⇒ BC = √ ((–a)² + (a√3)²)

⇒ BC = √ (a² + 3a²)

⇒ BC = √4a²

⇒ BC = 2a

Distance of AC

⇒ AC = √ ((0 – a)² + (a√3 – 0))²)

⇒ AC = √ ((–a)² + (a√3)²)

⇒ AC = √ (a² + 3a²)

⇒ AC = √ 4a²

⇒ AC = 2a

∴ AB = BC = AC = 2a

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