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Question: The angle between the lines \(x^{2} - xy - 6y^{2} - 7x + 31y - 18 = 0\) is...

The angle between the lines x2−xy−6y2−7x+31y−18=0x^{2} - xy - 6y^{2} - 7x + 31y - 18 = 0 is

A

45o45^{o}

B

60o60^{o}

C

90o90^{o}

D

30o30^{o}

Answer

45o45^{o}

Explanation

Solution

Angle between the lines is θ=tan⁡−1∣2h2−aba+b∣\theta = \tan^{- 1}\left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right|

=tan⁡−1∣2(−12)2−1×(−6)1+(−6)∣= \tan^{- 1}\left| \frac{2\sqrt{\left( \frac{- 1}{2} \right)^{2} - 1 \times ( - 6)}}{1 + ( - 6)} \right|=tan⁡−1∣214+61+(−6)∣=tan⁡−1∣−1∣=tan⁡−1(1)=π4\tan^{- 1}\left| \frac{2\sqrt{\frac{1}{4} + 6}}{1 + ( - 6)} \right| = \tan^{- 1}| - 1| = \tan^{- 1}(1) = \frac{\pi}{4}, 45o45^{o}