Using distance formula, prove that the points 1 , - 1 + 3 i 2 and - 1 - 3 i 2 form the vertices of an equilateral triangle in complex plane.

Let A z 1 = 1 , B z 2 = - 1 + 3 i 2 and C z 3 = - 1 - 3 i 2 We know that distance between two complex numbers z 1 , z 2 is z 2 - z 1 ∴ A B = z 2 - z 1 = - 1 + 3 i 2 - 1 ⇒ A B = - 1 + 3 i - 2 2 ⇒ A B = - 3 + 3 i 2 ⇒ A B = 9 4 + 3 4 ⇒ A B = 3 → 1 Similarly, B C = - 1 - 3 i 2 - - 1 + 3 i 2 ⇒ B C = - 1 - 3 i 2 + 1 - 3 i 2 ⇒ B C = 2 3 i 2 ⇒ B C = 3 → 2 and C A = 3 → 3 From 1 , 2 and 3 , we have A B = B C = C A We know that, in an equilateral triangle length of three sides are equal. Hence,proved.

Using distance formula, prove that the points 1 , - 1 + 3 i 2 and - 1 - 3 i 2 form the vertices of an equilateral triangle in complex plane.

Let A z 1 = 1 , B z 2 = - 1 + 3 i 2 and C z 3 = - 1 - 3 i 2 We know that distance between two complex numbers z 1 , z 2 is z 2 - z 1 ∴ A B = z 2 - z 1 = - 1 + 3 i 2 - 1 ⇒ A B = - 1 + 3 i - 2 2 ⇒ A B = - 3 + 3 i 2 ⇒ A B = 9 4 + 3 4 ⇒ A B = 3 → 1 Similarly, B C = - 1 - 3 i 2 - - 1 + 3 i 2 ⇒ B C = - 1 - 3 i 2 + 1 - 3 i 2 ⇒ B C = 2 3 i 2 ⇒ B C = 3 → 2 and C A = 3 → 3 From 1 , 2 and 3 , we have A B = B C = C A We know that, in an equilateral triangle length of three sides are equal. Hence,proved.