MHT CET · Maths · Pair of Lines
The joint equation of the lines passing through the origin and trisecting the first quadrant is
- A \(\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\)
- B \(x^2-\sqrt{3} x y-y^2=0\)
- C \(3 x^2-y^2=0\)
- D \(x^2+\sqrt{3} x y-y^2=0\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\tan 30^{\circ} x \text { and } y=\tan 60^{\circ} x \\ & \Rightarrow y=\frac{1}{\sqrt{3}} x \text { and } y=\sqrt{3} \cdot x \\ & \Rightarrow x-\sqrt{3} y=0 \text { and } \sqrt{3} x-y=0\end{aligned}\)
Joint equation \((x-\sqrt{3} y)(\sqrt{3} x-y)=0\) \(\Rightarrow \sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\)
Joint equation \((x-\sqrt{3} y)(\sqrt{3} x-y)=0\) \(\Rightarrow \sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\)
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