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
\(x+2y+1=0 \implies 1=-(x+2y)\) \(2x^2-2xy+3y^2+2x(1)-y(1)-1(1)^2=0\) \(2x^2-2xy+3y^2+2x(-(x+2y))-y(-(x+2y))-(-(x+2y))^2=0\) \(2x^2-2xy+3y^2-2x^2-4xy+xy+2y^2-(x^2+4xy+4y^2)=0\) \((2-2-1)x^2+(-2-4+1-4)xy+(3+2-4)y^2=0\) \(-x^2-9xy+y^2=0 \implies x^2+9xy-y^2=0\) \(A=1, B=-1\)…
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