MHT CET · Maths · Pair of Lines
The angle between the lines \(x^{2}-x y-6 y^{2}-7 x+31 y-18=0\) is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Given equation is
\(
x^{2}-x y-6 y^{2}-7 x+31 y-18=0
\)
Here, \(a=1, b=-6, h=\frac{-1}{2}\)
\(\theta=\tan ^{-1}\left|\frac{2 \sqrt{\left(-\frac{1}{2}\right)^{2}-1 \times(-6)}}{1+(-6)}\right|\)
\(=\tan ^{-1}\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
\(=\tan ^{-1}|-1|\)
\(=\frac{\pi}{4}\)
\(
x^{2}-x y-6 y^{2}-7 x+31 y-18=0
\)
Here, \(a=1, b=-6, h=\frac{-1}{2}\)
\(\theta=\tan ^{-1}\left|\frac{2 \sqrt{\left(-\frac{1}{2}\right)^{2}-1 \times(-6)}}{1+(-6)}\right|\)
\(=\tan ^{-1}\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
\(=\tan ^{-1}|-1|\)
\(=\frac{\pi}{4}\)
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