AP EAMCET · Maths · Pair of Lines
If the equation of the pair of straight lines passing through the point \((1,1)\) and perpendicular to the pair of lines \(3 \mathrm{x}^2\) \(+11 x y-4 y^2=0\) is \(\mathrm{a}^2+2 \mathrm{~h} x y+\mathrm{b} y^2+2 \mathrm{~g} x+2 \mathrm{f} y+12=0\), then \(2(a-h+b-g+f-12)=\)
- A \(0\)
- B \(-7\)
- C \(-19\)
- D \(13\)
Answer & Solution
Correct Answer
(C) \(-19\)
Step-by-step Solution
Detailed explanation
\(3 x^2+11 x y-4 y^2=0 \Rightarrow(3 x-y)(x+4 y)=0\) Equation of lines are \(3 x-y=0 \ldots\) (i); \(x+4 y=0 \ldots\) (ii) \(\therefore \quad m_1=3, m_2=\frac{-1}{4}\) So lines perpendicular to (i) and (ii) will have slope \(\frac{-1}{2}, 4\) Also the lines passes through…
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