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The joint equation of the lines through the origin trisecting angles in first and third equadrant is
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Verified Answer
The correct answer is:
$\sqrt{3}\left(x^{2}+y^{2}\right)-4 x y=0$
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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