Q114 / 180 · COMEDK 2023 (Memory Based) · Mathematics
Let the equation of pair of lines $y=m_1 x$ and $y=m_2 x$ can be written as
$\left(y-m_1 x\right)\left(y-m_2 x\right)=0$. Then, the equation of the pair of the angle bisector of the line $3 y^2-5 x y-2 x^2=0$ is
- A$x^2+5 x y-y^2=0$
- B$x^2-5 x y+y^2=0$
- C$x^2-x y+y^2=0$
- Correct answerD$x^2+x y-y^2=0$
Solution
$\because$ Equation of angles of bisector of pair of straight line, $a x^2+2 b x y+b y^2$ is
$\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
$\therefore$ For, $3 y^2-5 x y-2 x^2=0$
$a=3, b=-2, h=-5$
So, equation of angle bisector is
$\begin{aligned}
& \frac{x^2-y^2}{3-(-2)}=\frac{x y}{-5} \\
\Rightarrow \quad \frac{x^2-y^2}{5} & =\frac{x y}{-5} \Rightarrow x^2-y^2+x y=0
\end{aligned}$
Hence, the correct answer is option: 'D'.