The joint equation of pair of lines through the origin and making equilateral triangle with the line $y=4$ is
- $3 x^{2}+y^{2}=0$
- $3 x^{2}-y^{2}=0$
- $x^{2}-y^{2}=0$
- $x^{2}-3 y^{2}=0$
Solution
Let $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be the required lines.
Since $\Delta \mathrm{OAB}$ is equilateral,
$\mathrm{m} \angle \mathrm{ABO}=60^{\circ}=\mathrm{m} \angle \mathrm{BAO}$
$\therefore$ Slope of line $\mathrm{L}_{2}=\tan 60^{\circ}=\sqrt{3}$ and
Slope of line $L_{1}=\tan \left(\pi-60^{\circ}\right)=-\sqrt{3}$
Hence required equation is
$(y-\sqrt{3} x)(y+\sqrt{3} x)=0 \text { i.e. } y^{2}-3 x^{2}=0 \Rightarrow 3 x^{2}-y^{2}=0$Asked in: MHT CET 2020 (20 Oct Shift 2)