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

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
\(
(x-\sqrt{3 y})(x+\sqrt{3 y})=0 \text { i.e. } x^2-3 y^2=0
\)
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