MHT CET · Maths · Pair of Lines
The acute angle between the lines \(\left(x^2+y^2\right) \sin \theta+2 x y=0\) is
- A \(\theta\)
- B \(\frac{\pi}{2}+\theta\)
- C \(\frac{\pi}{2}-\theta\)
- D \(\frac{\theta}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}-\theta\)
Step-by-step Solution
Detailed explanation
We have \(\left(x^2+y^2\right) \sin \theta+2 x y=0\)
\(\therefore \mathrm{a}=\sin \theta, \mathrm{b}=\sin \theta\) and \(\mathrm{h}=1\) and let \(\alpha\) be the acute angle between the given lines.
Then \(\tan \alpha=\frac{\left|2 \sqrt{(1)^2-(\sin \theta)(\sin \theta)}\right|}{\sin \theta+\sin \theta}\)
\(\begin{aligned}
& =\frac{\left|2 \sqrt{1-\sin ^2 \theta}\right|}{2 \sin \theta}=\frac{2 \cos \theta}{2 \sin \theta}=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right) \\
& \therefore \alpha=\frac{\pi}{2}-\theta
\end{aligned}\)
\(\therefore \mathrm{a}=\sin \theta, \mathrm{b}=\sin \theta\) and \(\mathrm{h}=1\) and let \(\alpha\) be the acute angle between the given lines.
Then \(\tan \alpha=\frac{\left|2 \sqrt{(1)^2-(\sin \theta)(\sin \theta)}\right|}{\sin \theta+\sin \theta}\)
\(\begin{aligned}
& =\frac{\left|2 \sqrt{1-\sin ^2 \theta}\right|}{2 \sin \theta}=\frac{2 \cos \theta}{2 \sin \theta}=\cot \theta=\tan \left(\frac{\pi}{2}-\theta\right) \\
& \therefore \alpha=\frac{\pi}{2}-\theta
\end{aligned}\)
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