MHT CET · Maths · Pair of Lines
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
- A -6
- B -10
- C 6
- D 10
Answer & Solution
Correct Answer
(B) -10
Step-by-step Solution
Detailed explanation
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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