MHT CET · Maths · Pair of Lines
The joint equation of pair of lines through the origin and making equilateral triangle with the line \(y=4\) is
- A \(3 x^{2}+y^{2}=0\)
- B \(3 x^{2}-y^{2}=0\)
- C \(x^{2}-y^{2}=0\)
- D \(x^{2}-3 y^{2}=0\)
Answer & Solution
Correct Answer
(B) \(3 x^{2}-y^{2}=0\)
Step-by-step Solution
Detailed explanation

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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