MHT CET · Maths · Pair of Lines
The joint equation of the lines through the origin trisecting angles in first and third equadrant is
- A \(\sqrt{3}\left(x^{2}-y^{2}\right)+4 x y=0\)
- B \(\sqrt{3}\left(x^{2}+y^{2}\right)-4 x y=0\)
- C \(\sqrt{3}\left(x^{2}+y^{2}\right)+4 x y=0\)
- D \(\sqrt{3}\left(x^{2}-y^{2}\right)-4 x y=0\)
Answer & Solution
Correct Answer
(B) \(\sqrt{3}\left(x^{2}+y^{2}\right)-4 x y=0\)
Step-by-step Solution
Detailed explanation
Equation line \(\mathrm{L}_{1}\) is \(\mathrm{y}=\tan 30^{\circ} \mathrm{x}\)
\(y=\frac{1}{\sqrt{3}} x \Rightarrow x-\sqrt{3} y=0 ...(1)\)
Equation of line \(\mathrm{L}_{2}\) is \(\mathrm{y}=\tan 60^{\circ} \mathrm{x}\)
\(
y=\sqrt{3} x \Rightarrow \sqrt{3} x-y=0 ...(2)
\)
\(\therefore\) Joint equation is
\(
(x-\sqrt{3} y)(\sqrt{3} x-y)=0 \Rightarrow \sqrt{3}\left(x^{2}+y^{2}\right)-4 x y=0
\)
\(y=\frac{1}{\sqrt{3}} x \Rightarrow x-\sqrt{3} y=0 ...(1)\)
Equation of line \(\mathrm{L}_{2}\) is \(\mathrm{y}=\tan 60^{\circ} \mathrm{x}\)
\(
y=\sqrt{3} x \Rightarrow \sqrt{3} x-y=0 ...(2)
\)
\(\therefore\) Joint equation is
\(
(x-\sqrt{3} y)(\sqrt{3} x-y)=0 \Rightarrow \sqrt{3}\left(x^{2}+y^{2}\right)-4 x y=0
\)
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