# If One Vertex Of An Equilateral Triangle Of Side 'a' Lies At The Origin And The Other Lies On Line X− Sqrt3y = 0, Then The Co-ordinates Of The Third Vertex Are

Given:

Slope of the line x = $\sqrt{3}y = \frac{1}{\sqrt{3}} \Rightarrow 30°$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).

$\Rightarrow x^{2} + y^{2} = a^{2}$ $\Rightarrow 4y^{2} = a^{2}$ $\Rightarrow y = \pm \frac{a}{2}$

Similarly, x = $\pm \frac{\sqrt{3}a}{2}$

Hence, if the triangle lies in the 1st quadrant and 4th quadrant vertex is

$\left ( \frac{\sqrt{3}a}{2},-\frac{a}{2} \right)$

If the triangle lies in the 2nd and 3rd quadrant, the vertex is

$\left ( -\frac{\sqrt{3}a}{2},-\frac{a}{2} \right)$

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