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If the equation $3 x^2-k x y-3 y^2=0$ represents the bisectors of angles between the lines $x^2-3 x y-4 y^2=0$, then value of $k$ is
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Verified Answer
The correct answer is:
-10
We have $x^2-3 x y-4 y^2=0$ and comparing it with standard equation, we write
$$
\mathrm{A}=1, \mathrm{H}=\frac{-3}{2}, \mathrm{~B}=-4
$$
Equation of bisector of angle of this line is
$$
\begin{aligned}
& \frac{x^2-y^2}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^2-y^2}{1+4}=\frac{x y}{\left(\frac{-3}{2}\right)} \\
& \therefore-3 x^2+3 y^2=10 x y \Rightarrow 3 x^2+10 x y-3 y^2=0
\end{aligned}
$$
Comparing with given equation, we get $\mathrm{k}=-10$
$$
\mathrm{A}=1, \mathrm{H}=\frac{-3}{2}, \mathrm{~B}=-4
$$
Equation of bisector of angle of this line is
$$
\begin{aligned}
& \frac{x^2-y^2}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^2-y^2}{1+4}=\frac{x y}{\left(\frac{-3}{2}\right)} \\
& \therefore-3 x^2+3 y^2=10 x y \Rightarrow 3 x^2+10 x y-3 y^2=0
\end{aligned}
$$
Comparing with given equation, we get $\mathrm{k}=-10$
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