The joint equation of the pair of lines through the origin and making an equilateral triangle with the line…
- $3 x^2-y^2=0$
- $\sqrt{3} x^2-2 x y+y^2=0$
- $x^2-3 y^2=0$
- $x^2+2 x y-\sqrt{3} x^2=0$
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
$(x-\sqrt{3 y})(x+\sqrt{3 y})=0 \text { i.e. } x^2-3 y^2=0$Asked in: MHT CET 2021 (20 Sep Shift 1)