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The joint equation of bisectors of the angle between the lines represented by
$3 x^{2}+2 x y-y^{2}=0$ is
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$3 x^{2}+2 x y-y^{2}=0$ is
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The correct answer is:
$x^{2}-4 x y-y^{2}=0$
(D)
Equation of lines is $3 x^{2}+2 x y-y^{2}=0 .$ Comparing with $a x^{2}+2 h x y+b y^{2}=0$, we write $a=3 . h=1 . b=-1$
Combined equation of pair of angle bisectors of given lines is
$\begin{aligned} & \frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h} \\ \therefore & \frac{x^{2}-y^{2}}{3-(-1)}=\frac{x y}{1} \Rightarrow \frac{x^{2}-y^{2}}{4}=x y \\ & x^{2}-4 x y-y^{2}=0 \end{aligned}$
Equation of lines is $3 x^{2}+2 x y-y^{2}=0 .$ Comparing with $a x^{2}+2 h x y+b y^{2}=0$, we write $a=3 . h=1 . b=-1$
Combined equation of pair of angle bisectors of given lines is
$\begin{aligned} & \frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h} \\ \therefore & \frac{x^{2}-y^{2}}{3-(-1)}=\frac{x y}{1} \Rightarrow \frac{x^{2}-y^{2}}{4}=x y \\ & x^{2}-4 x y-y^{2}=0 \end{aligned}$
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