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(C) The first quadrant is the region between the positive $x$-axis $(	heta = 0)$ and the positive $y$-axis $(	heta = \frac{\pi}{2})$.Since the lines

Vedclass · Accepted Answer

$\sqrt{3}x^2-4xy+\sqrt{3}y^2=0$

Vedclass · Answer

(C) The first quadrant is the region between the positive $x$-axis $(	heta = 0)$ and the positive $y$-axis $(	heta = \frac{\pi}{2})$.Since the lines trisect the first quadrant,they divide the $90^\circ$ angle into three equal parts of $30^\circ$ each.Thus,the angles made by the lines with the positive $x$-axis are $	heta_1 = 30^\circ = \frac{\pi}{6}$ and $	heta_2 = 60^\circ = \frac{\pi}{3}$.The equation of the first line is $y = 	an(\frac{\pi}{6})x$ $\Rightarrow y = \frac{1}{\sqrt{3}}x$ $\Rightarrow x - \sqrt{3}y = 0$.The equation of the second line is $y = 	an(\frac{\pi}{3})x$ $\Rightarrow y = \sqrt{3}x$ $\Rightarrow \sqrt{3}x - y = 0$.The joint equation is $(x - \sqrt{3}y)(\sqrt{3}x - y) = 0$.Expanding this,we get $\sqrt{3}x^2 - xy - 3xy + \sqrt{3}y^2 = 0$.$\Rightarrow \sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.

The joint equation of the lines passing through the origin and trisecting the first quadrant is . . . . . .

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