Prove that the points (2a, 4a), (2a, 6a) and

Question:

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

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In an equilateral triangle all the sides have equal length.

Here the three points are $A(2 a, 4 a), B(2 a, 6 a)$ and $C(2 a+a \sqrt{3}, 5 a)$.

Let us now find out the lengths of all the three sides of the given triangle.

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

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

$=\sqrt{0+4 a^{2}}$

$A B=2 a$

$B C=\sqrt{(2 a-2 a-a \sqrt{3})^{2}+(6 a-5 a)^{2}}$

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

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

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

$B C=2 a$

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

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

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

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

$A C=2 a$

Since all the three sides have equal lengths the triangle has to be an equilateral triangle.

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