AP EAMCET · Maths · Pair of Lines
If the pair of lines \(a x^2-7 x y-3 y^2=0\) and \(2 x^2+x y-6 y^2=0\) have exactly one line in common and ' \(a\) ' is an integer, then the equation of the pair of bisectors of the angles between the lines \(a x^2-7 x y-3 y^2=0\) is
- A \(7 x^2+18 x y-7 y^2=0\)
- B \(x^2-16 x y-y^2=0\)
- C \(7 x^2-9 x y-7 y^2=0\)
- D \(x^2-8 x y-y^2=0\)
Answer & Solution
Correct Answer
(A) \(7 x^2+18 x y-7 y^2=0\)
Step-by-step Solution
Detailed explanation
\(2 x^2+x y-6 y^2 = (2x-3y)(x+2y) = 0\) If \(2x-3y=0\) is the common line, substitute \(x=\frac{3y}{2}\) into \(a x^2-7 x y-3 y^2=0\): \(a \left(\frac{3y}{2}\right)^2 - 7 \left(\frac{3y}{2}\right) y - 3 y^2 = 0\)…
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