The joint equation of pair of lines through the origin and making an equilateral triangle with the line…
The joint equation of pair of lines through the origin and making an equilateral triangle with the line $y=3$ is
$x^2+3 y^2=0$
$3 x^2-y^2=0$
$x^2+3 y^2=0$
$3 x^2+y^2=0$
Solution
Let $\triangle \mathrm{OAB}$ be the required triangle. Since $\triangle \mathrm{OAB}$ is an equilateral triangle.
Slope of line $\mathrm{OA}=\tan 60^{\circ}=\sqrt{3}$ and slope of line $\mathrm{OB}=\tan$
$120^{\circ}=-\sqrt{3}$
$\therefore$ Equation of $\mathrm{OA}$ is $\mathrm{y}=\sqrt{3} \mathrm{x}$ i.e. $\sqrt{3} \mathrm{x}-\mathrm{y}=0$ and equation of $O B$ is $y=-\sqrt{3} \mathrm{x}$ i.e.
$\sqrt{3} \mathrm{x}+\mathrm{y}=0$
Hence required joint equation is
$(\sqrt{3} x-y)(\sqrt{3} x+y)=0 \text { i.e. } 3 x^2-y^2=0$