Question

Prove that the points $\left(2a,4a\right),\left(2a,6a\right)$ and $\left(2a+\sqrt{3}a,5a\right)$ are the vertices of an equilateral triangle whose side is $2a$.

Answer 1

Consider the given point

$A(2a,4a),B(2a,6a)andC(2a+\sqrt{3}a,5a)$

Distance between $AB,BCandCA$ respectively.

$AB=\sqrt{{(2a-2a)}^2+{(4a-6a)}^2}$

$=\sqrt{0+{(-2a)}^2}$

$=\sqrt{4a^2}=2a$

$BC=\sqrt{{(2a-(2a+\sqrt{3}a\left)\right)}^2+{(6a-5a)}^2}$

$=\sqrt{3a^2+a^2}$

$=\sqrt{4a^2}=2a$

$CA=\sqrt{{(2a+\sqrt{3}a-2a)}^2+{(5a-4a)}^2}$

$=\sqrt{3a^2+a^2}$

$=\sqrt{4a^2}=2a$

Hence, $AB=BC=CA$

It is an equilateral triangle.