MHT CET · Maths · Pair of Lines
The joint equation of pair of lines through the origin, each of which makes an angle of \(30^{\circ}\) with Y -axis, is
- A \(3 x^2-y^2=0\)
- B \(x^2-3 y^2=0\)
- C \(3 x^2+y^2=0\)
- D \(x^2+3 y^2=0\)
Answer & Solution
Correct Answer
(A) \(3 x^2-y^2=0\)
Step-by-step Solution
Detailed explanation
Let OA and OB be the required lines.
\(\therefore\) angles made by OA and OB with X -axis are \(60^{\circ}\) and \(120^{\circ}\) respectively.
\(\therefore\) Their equations are \(y=\sqrt{3} x\) and \(y=-\sqrt{3} x\) i.e., \(\sqrt{3} x-y=0\) and \(\sqrt{3} x+y=0\)
\(\therefore\) The joint equation of the lines is
\((\sqrt{3} x-y)(\sqrt{3} x+y)=0 \Rightarrow 3 x^2-y^2=0\)
\(\therefore\) angles made by OA and OB with X -axis are \(60^{\circ}\) and \(120^{\circ}\) respectively.
\(\therefore\) Their equations are \(y=\sqrt{3} x\) and \(y=-\sqrt{3} x\) i.e., \(\sqrt{3} x-y=0\) and \(\sqrt{3} x+y=0\)
\(\therefore\) The joint equation of the lines is
\((\sqrt{3} x-y)(\sqrt{3} x+y)=0 \Rightarrow 3 x^2-y^2=0\)
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