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(B) The given pair of lines is $2x^2 + xy - y^2 - x + 2y - 1 = 0$.
Factorizing the homogeneous part $2x^2 + xy - y^2 = (2x - y)(x + y)$.
Let the lin

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$-2$

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(B) The given pair of lines is $2x^2 + xy - y^2 - x + 2y - 1 = 0$.
Factorizing the homogeneous part $2x^2 + xy - y^2 = (2x - y)(x + y)$.
Let the lines be $(2x - y + c_1)(x + y + c_2) = 0$.
Comparing with the given equation,we find the lines are $(2x - y + 1)(x + y - 1) = 0$.
The slopes of these lines are $m_1 = 2$ and $m_2 = -1$.
The lines perpendicular to these passing through $(1, 1)$ will have slopes $m_1' = -1/2$ and $m_2' = 1$.
The equations of these lines are $(y - 1) = -1/2(x - 1) \Rightarrow x + 2y - 3 = 0$ and $(y - 1) = 1(x - 1) \Rightarrow x - y = 0$.
The combined equation is $(x + 2y - 3)(x - y) = 0$.
Expanding this,we get $x^2 + xy - 2y^2 - 3x + 3y = 0$.
Comparing this with $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,we have $a = 1, 2h = 1, b = -2, 2g = -3$.
Thus,$\frac{b}{a} = \frac{-2}{1} = -2$.

If the equation of the pair of lines passing through $(1, 1)$ and perpendicular to the pair of lines $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,then $\frac{b}{a} =$

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