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

• A

  $x^2+\sqrt{3}xy-y^2=0$
• B

  $x^2-\sqrt{3}xy-y^2=0$
• Correct answer
  C

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

  $3x^2-y^2=0$

Solution

As given, both line passes through $\left(0, 0\right) \mathrm{a}\mathrm{n}\mathrm{d} {\theta }_1=\frac{\pi }{6}, {\theta }_2=\frac{\pi }{3}$
$\therefore$ Equation of first line is.
$y-0=\tan \left(\frac{\pi }{6}\right)\left(x-0\right)$
$\Rightarrow y=\frac{1}{\sqrt{3}}x$
$\Rightarrow x-\sqrt{3}y=0$ ...... (i)
Equation of second line is
$y-0=\tan \left(\frac{\pi }{3}\right)(x-0)$
$\Rightarrow y=\sqrt{3} x$
$\Rightarrow \sqrt{3} x-y=0$ ...... (ii)
Hence, joint equation of these line is
$\left(x-\sqrt{3}y\right)\left(\sqrt{3}x-y\right)=0$
$\Rightarrow \sqrt{3}{x}^2-xy-3xy+\sqrt{3}{y}^2=0$
$\Rightarrow \sqrt{3}{x}^2+\sqrt{3}{y}^2-4xy=0$

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