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Question

Slope of the line $x=\sqrt{3} y$ is $\frac{1}{\sqrt{3}}$ which is $30^{\circ}$ with the positive $x$ -axis. As it is an equilateral triangle the other

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Slope of the line $x=\sqrt{3} y$ is $\frac{1}{\sqrt{3}}$ which is $30^{\circ}$ with the positive $x$ -axis. As it is an equilateral triangle the other vertex must be on $x=-\sqrt{3} y$

Let the side be $(x, y)$. Hence, $x^{2}+y^{2}=a^{2}$

$4 y^{2}=a^{2}$

$y=\pm \frac{a}{2}$

similarily, $x=\pm \frac{\sqrt{3} a}{2}$

Hence, if the triangle lies in the 1 st quadrant and 4 th quadrant vertex is $\left(\frac{\sqrt{3} a}{2},-\frac{a}{2}\right)$

If the triangle lies in the 2 nd and 3 rd quadrant the vertex is $\left(-\frac{\sqrt{3} a}{2},-\frac{a}{2}\right)$

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