Question

The measure of the acute angle between the pair of lines \(2x^2+xy-y^2-x+2y-1 = 0\) is:

• A. \(tan^{-1}\frac{1}{3}\)
• B. \(tan^{-1}1\)
• C. \(cos^{-1}\frac{1}{\sqrt{10}}\)
• D. \(cos^{-1}\sqrt{10}\)

Hint:

Factor into two lines, find slopes and use the angle formula.

Answer 1

The Correct Option is C

Step 1: Use the pair-of-lines formula
For $ax^2+2hxy+by^2$ here $a=2$, $2h=1$, $b=-1$. $\tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}$.

Step 2: Substitute
$h^2-ab=\tfrac14+2=\tfrac94$, so $2\sqrt{h^2-ab}=3$ and $|a+b|=1$. Thus $\tan\theta=3$ and $\cos\theta=1/\sqrt{10}$, option (C).

Final Answer:
Since $\tan\theta=3$, $\theta=\cos^{-1}(1/\sqrt{10})$, option (C). \[ \boxed{\cos^{-1}\dfrac{1}{\sqrt{10}}} \]