MHT CET · Maths · Pair of Lines
The joint equation of pair of lines through the origin and making an equilateral triangle with the line \(y=3\) is
- A \(x^2+3 y^2=0\)
- B \(3 x^2-y^2=0\)
- C \(x^2+3 y^2=0\)
- D \(3 x^2+y^2=0\)
Answer & Solution
Correct Answer
(B) \(3 x^2-y^2=0\)
Step-by-step Solution
Detailed explanation
Let \(\triangle \mathrm{OAB}\) be the required triangle. Since \(\triangle \mathrm{OAB}\) is an equilateral triangle.
Slope of line \(\mathrm{OA}=\tan 60^{\circ}=\sqrt{3}\) and slope of line \(\mathrm{OB}=\tan\)
\(
120^{\circ}=-\sqrt{3}
\)
\(\therefore\) Equation of \(\mathrm{OA}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\) i.e. \(\sqrt{3} \mathrm{x}-\mathrm{y}=0\) and equation of \(O B\) is \(y=-\sqrt{3} \mathrm{x}\) i.e.
\(
\sqrt{3} \mathrm{x}+\mathrm{y}=0
\)
Hence required joint equation is
\(
(\sqrt{3} x-y)(\sqrt{3} x+y)=0 \text { i.e. } 3 x^2-y^2=0
\)
Slope of line \(\mathrm{OA}=\tan 60^{\circ}=\sqrt{3}\) and slope of line \(\mathrm{OB}=\tan\)
\(
120^{\circ}=-\sqrt{3}
\)
\(\therefore\) Equation of \(\mathrm{OA}\) is \(\mathrm{y}=\sqrt{3} \mathrm{x}\) i.e. \(\sqrt{3} \mathrm{x}-\mathrm{y}=0\) and equation of \(O B\) is \(y=-\sqrt{3} \mathrm{x}\) i.e.
\(
\sqrt{3} \mathrm{x}+\mathrm{y}=0
\)
Hence required joint equation is
\(
(\sqrt{3} x-y)(\sqrt{3} x+y)=0 \text { i.e. } 3 x^2-y^2=0
\)
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