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: Understanding the Concept
The equation represents a pair of lines. Factorise to get the two lines and their slopes.

Step 2: Key Formula or Approach
The equation factorises as \((2x-y+1)(x+y-1)=0\). Check: the product gives \(2x^2+xy-y^2-x+2y-1\).

Step 3: Detailed Explanation
Lines: \(2x-y+1=0\) with slope 2, and \(x+y-1=0\) with slope \(-1\).
\[ \tan\theta=\left|\frac{2-(-1)}{1+2(-1)}\right|=\left|\frac{3}{-1}\right|=3 \]
With \(\tan\theta=3\), \(\cos\theta=\dfrac{1}{\sqrt{1+9}}=\dfrac{1}{\sqrt{10}}\).
So \(\theta=\cos^{-1}\dfrac{1}{\sqrt{10}}\).

Final Answer:
The acute angle is \(\cos^{-1}\frac{1}{\sqrt{10}}\), option (C). \[ \boxed{\cos^{-1}\dfrac{1}{\sqrt{10}}\ \text{(C)}} \]