Question

Prove that the points (2a,4a),(2a,6a) and (2a+√3a,5a) are the vertices of an equilateral triangle whose side is 2a​

Answer 1

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See the attachment mate :)

Answer 2

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AB=√ 0+ (2a)²

= 2a

BC = √ (2a-2a+√3a)² + (6a-5a)²

= √ 3a² +a²

= 2a

AC=√ (2a-2a-√3a)² + (5a-4a)²

AC= √3a²+a²

= 2a

Here all the sides are 2a hence proved