The joint equation of the pair of lines through the origin and making an equilateral triangle with the line $x=3$ is

• A

  $3 x^2-y^2=0$
• B

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

  $x^2-3 y^2=0$
• D

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

Solution

Slope of line $\mathrm{OA}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ and Slope of line $\mathrm{OB}=\tan \left(-30^{\circ}\right)=\frac{-1}{\sqrt{3}}$ $\therefore$ Equation of $\mathrm{OA}$ is $\mathrm{y}=\frac{1}{\sqrt{3}} \mathrm{x}$ and equation of $\mathrm{OB}$ is $\mathrm{y}=\frac{1}{\sqrt{3}} \mathrm{x}$ Hence required equation is $$ (x-\sqrt{3 y})(x+\sqrt{3 y})=0 \text { i.e. } x^2-3 y^2=0 $$

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