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The joint equation of pair of lines through the origin and making equilateral triangle with the line $y=4$ is
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Verified Answer
The correct answer is:
$3 x^{2}-y^{2}=0$
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
$$
(y-\sqrt{3} x)(y+\sqrt{3} x)=0 \text { i.e. } y^{2}-3 x^{2}=0 \Rightarrow 3 x^{2}-y^{2}=0
$$
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