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


Slope of the line x = [latex] \sqrt{3}y = \frac{1}{\sqrt{3}} \Rightarrow 30° [/latex]with the positive x-axis.

As it is an equilateral triangle the other vertex must be on x =[latex] – \sqrt{3}y[/latex]

Let the side be (x,y).

[latex] \Rightarrow x^{2} + y^{2} = a^{2}[/latex] [latex] \Rightarrow 4y^{2} = a^{2}[/latex] [latex] \Rightarrow y = \pm \frac{a}{2}[/latex]

Similarly, x = [latex] \pm \frac{\sqrt{3}a}{2}[/latex]

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

[latex] \left ( \frac{\sqrt{3}a}{2},-\frac{a}{2} \right)[/latex]

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

[latex] \left ( -\frac{\sqrt{3}a}{2},-\frac{a}{2} \right)[/latex]

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