AP EAMCET · Maths · Pair of Lines
The angle between the lines joining the origin to the points of intersection of \(x+2 y+1=0\) and \(2 x^2-2 x y+3 y^2+2 x-y-1=0\) is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(1 = -(x+2y)\) \(2x^2-2xy+3y^2+(2x-y)(-(x+2y))-(-1)(-(x+2y))^2=0\) \(2x^2-2xy+3y^2-(2x^2+3xy-2y^2)-(x^2+4xy+4y^2)=0\) \(-x^2-9xy+y^2=0\) \(A=-1, B=1\) \(A+B = -1+1=0\) \(\theta = \frac{\pi}{2}\)
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