Prove that the points $(2a, 4a), (2a, 6a)$ and $(2a + \sqrt3 a , 5a)$ are the vertices of an equilateral triangle.

Given:

Given points are $(2a, 4a), (2a, 6a)$ and $(2a + \sqrt3 a , 5a)$.

To do:

We have to prove that the given points are the vertices of an equilateral triangle.

Solution:

Let the vertices of the triangle be \( \mathrm{A}(2a,4a), \mathrm{B}(2a,6a) \) and \( C(2a+\sqrt3a,5a) \).

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

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

\( =\sqrt{(2 a-2 a)^{2}+(6 a-4 a)^{2}} \) \( =\sqrt{(0)^{2}+(2 a)^{2}} \)

\( =\sqrt{0+4 a^{2}} \)

\( =2 a \)

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

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

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

\( =\sqrt{4 a^{2}}=2 a \)

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

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

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

\( =\sqrt{4 a^{2}}=2 a \)

Here,

\( \mathrm{AB}=\mathrm{BC}=\mathrm{CA}=2 a \)

We know that an equilateral triangle has three equal sides.

Therefore, $(2a, 4a), (2a, 6a)$ and $(2a + \sqrt3 a , 5a)$ are the vertices of an equilateral triangle.

Hence proved.

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Updated on: 10-Oct-2022

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