Show that the points $ \mathrm{A}(a, a), \mathrm{B}(-a,-a) $ and $ C(-\sqrt{3} a, \sqrt{3} a) $ are the vertices of an equilateral triangle.

Given:

\( \mathrm{A}(a, a), \mathrm{B}(-a,-a) \) and \( C(-\sqrt{3} a, \sqrt{3} a) \)

To do:

We have to show that the points \( \mathrm{A}(a, a), \mathrm{B}(-a,-a) \) and \( C(-\sqrt{3} a, \sqrt{3} a) \) are the vertices of an equilateral triangle.

Solution:

Let the given points be $A(a, a), B(-a,-a)$ and $C(-\sqrt{3} a, \sqrt{3} a)$

Therefore,

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

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

$=2 \sqrt{2} a$

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

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

$=\sqrt{8 a^{2}}$

$=2 \sqrt{2} a$

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

$=\sqrt{a^{2}+3 a^{2}+2 \sqrt{3} a^{2}+a^{2}+3 a^{2}-2 \sqrt{3} a^{2}}$
$=\sqrt{8 a^{2}}$

$=2 \sqrt{2} a$

Here,

$AB = BC = AC$, this implies $ABC$ is an equilateral triangle.

Hence proved. 

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Updated on: 10-Oct-2022

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