Q85 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
If the slope of one of the pair of lines represented by $2 x^2+3 x y+K y^2=0$ is 2 , then the angle between the pair of lines is
- Correct answerA$\frac{\pi}{2}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{4}$
Solution
$\begin{aligned}
& 2 x^2+3 x y+k y^2=0 \\
& \Rightarrow \frac{2}{k} x^2+\frac{3}{k} x y+y^2=0 ....(i)
\end{aligned}$
Let slope of other line be $m$ then two lines are $y=2 x, y$ $\begin{aligned} & =m x \\ & (y-m x)(y-2 x)=0 \\ & \Rightarrow 2 m x^2-(m+2) x y+y^2=0 ...(ii) \end{aligned}$
Comparing (i) and (ii) $\begin{aligned} & 2 m=\frac{2}{k} \text { and } m+2=\frac{-3}{k} \\ & \Rightarrow \frac{2 m}{m+2}=\frac{-2}{3} \Rightarrow m=\frac{-1}{2}, k=-2 \end{aligned}$
So, angle between the lines is $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{2+\frac{1}{2}}{1-1}\right|=\infty \Rightarrow \theta=\frac{\pi}{2}$
Let slope of other line be $m$ then two lines are $y=2 x, y$ $\begin{aligned} & =m x \\ & (y-m x)(y-2 x)=0 \\ & \Rightarrow 2 m x^2-(m+2) x y+y^2=0 ...(ii) \end{aligned}$
Comparing (i) and (ii) $\begin{aligned} & 2 m=\frac{2}{k} \text { and } m+2=\frac{-3}{k} \\ & \Rightarrow \frac{2 m}{m+2}=\frac{-2}{3} \Rightarrow m=\frac{-1}{2}, k=-2 \end{aligned}$
So, angle between the lines is $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{2+\frac{1}{2}}{1-1}\right|=\infty \Rightarrow \theta=\frac{\pi}{2}$
Hence, the correct answer is option: 'A'.