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

• Correct answer
  A

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

  $x^2-\sqrt{3} x y-y^2=0$
• C

  $3 x^2-y^2=0$
• D

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

Solution

$\begin{aligned} & y=\tan 30^{\circ} x \text { and } y=\tan 60^{\circ} x \\ & \Rightarrow y=\frac{1}{\sqrt{3}} x \text { and } y=\sqrt{3} \cdot x \\ & \Rightarrow x-\sqrt{3} y=0 \text { and } \sqrt{3} x-y=0\end{aligned}$
Joint equation $(x-\sqrt{3} y)(\sqrt{3} x-y)=0$ $\Rightarrow \sqrt{3} x^2-4 x y+\sqrt{3} y^2=0$

Hence, the correct answer is option: 'A'.