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

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$-2$

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(A) The given pair of lines is $2x^2 + xy - y^2 - x + 2y - 1 = 0$.
Factoring 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 $(2x - y + 1) = 0$ and $(x + y - 1) = 0$.
The lines passing through $(1, 1)$ and perpendicular to these are:
Line $L_1$ perpendicular to $2x - y + 1 = 0$ is $x + 2y + k_1 = 0$. Since it passes through $(1, 1)$,$1 + 2(1) + k_1 = 0 \implies k_1 = -3$. So,$x + 2y - 3 = 0$.
Line $L_2$ perpendicular to $x + y - 1 = 0$ is $x - y + k_2 = 0$. Since it passes through $(1, 1)$,$1 - 1 + k_2 = 0 \implies k_2 = 0$. So,$x - y = 0$.
The combined equation is $(x + 2y - 3)(x - y) = 0$.
$x^2 - xy + 2xy - 2y^2 - 3x + 3y = 0 \implies x^2 + xy - 2y^2 - 3x + 3y = 0$.
Comparing with $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$,we get $a = 1, 2h = 1, b = -2, 2g = -3$.
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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