MHT CET · Maths · Pair of Lines
The measure of the angle between the lines \(x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0\) is
- A \(\frac{\pi}{2}-\alpha\)
- B \(\frac{\pi}{2}+\alpha\)
- C \(\alpha\)
- D \(\pi-\alpha\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}-\alpha\)
Step-by-step Solution
Detailed explanation
We have \(x^{2}+2 x y \operatorname{cosec} \alpha+y^{2}=0\)
Comparing it with standard form, we get \(\mathrm{a}=1, \mathrm{~h}=\operatorname{cosec} \alpha, \mathrm{b}=1\)
Let \(\theta\) be the angle between the lines.
\(\tan \theta =\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \)
\( =\left|\frac{2 \sqrt{\operatorname{cosec}^{2} \alpha-1}}{1+1}\right|=\left|\frac{2 \sqrt{\cot ^{2} \alpha}}{2}\right| \)
\( \tan \theta =\cot \alpha \Rightarrow \tan \theta=\tan \left(\frac{\pi}{2}-\alpha\right) \)
\( \therefore \theta=\frac{\pi}{2}-\alpha\)
Comparing it with standard form, we get \(\mathrm{a}=1, \mathrm{~h}=\operatorname{cosec} \alpha, \mathrm{b}=1\)
Let \(\theta\) be the angle between the lines.
\(\tan \theta =\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \)
\( =\left|\frac{2 \sqrt{\operatorname{cosec}^{2} \alpha-1}}{1+1}\right|=\left|\frac{2 \sqrt{\cot ^{2} \alpha}}{2}\right| \)
\( \tan \theta =\cot \alpha \Rightarrow \tan \theta=\tan \left(\frac{\pi}{2}-\alpha\right) \)
\( \therefore \theta=\frac{\pi}{2}-\alpha\)
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