MHT CET · Maths · Pair of Lines
The joint equation of pair of lines which bisects the angle between the lines \(x^2+3 x y+2 y^2=0\) is
- A \(3 x^2-r 4 e g h 2 x y-3 y^2=0\)
- B \(3 x^2+2 x y-3 y^2=0\)
- C \(2 x^2-3 x y-2 y^2=0\)
- D \(2 x^2+3 x y-2 y^2=0\)
Answer & Solution
Correct Answer
(B) \(3 x^2+2 x y-3 y^2=0\)
Step-by-step Solution
Detailed explanation
Equation of angle bisector of \(\mathrm{ax}^2+2 \mathrm{hxy}+\mathrm{by}^2=0\) is given by
\(\begin{aligned} & \frac{x-y}{a-b}=\frac{x y}{h} \\ & \Rightarrow \frac{x^2-y^2}{1-2}=\frac{x y}{\frac{3}{2}} \\ & \Rightarrow-3 x^2+3 y^2=2 x y \\ & \Rightarrow 3 x^2+2 x y-3 y^2=0\end{aligned}\)
\(\begin{aligned} & \frac{x-y}{a-b}=\frac{x y}{h} \\ & \Rightarrow \frac{x^2-y^2}{1-2}=\frac{x y}{\frac{3}{2}} \\ & \Rightarrow-3 x^2+3 y^2=2 x y \\ & \Rightarrow 3 x^2+2 x y-3 y^2=0\end{aligned}\)
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