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(B) The equation of the pair of lines is $ax^2 + hxy + by^2 = 0$,where $h = 1/2$.The angle $	heta$ between the lines is given by $	an 	heta = \left

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$a = 1, b = -6$

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(B) The equation of the pair of lines is $ax^2 + hxy + by^2 = 0$,where $h = 1/2$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Given $	heta = 45^\circ$,so $	an 45^\circ = 1$.$1 = \left| \frac{2\sqrt{(1/2)^2 - ab}}{a + b} ight| = \left| \frac{2\sqrt{1/4 - ab}}{a + b} ight| = \left| \frac{\sqrt{1 - 4ab}}{a + b} ight|$.Squaring both sides,we get $(a + b)^2 = 1 - 4ab$.$a^2 + 2ab + b^2 = 1 - 4ab \Rightarrow a^2 + 6ab + b^2 = 1$.Checking option $(b)$: $a = 1, b = -6$.$1^2 + 6(1)(-6) + (-6)^2 = 1 - 36 + 36 = 1$. This satisfies the condition.

The angle between the lines represented by the equation $ax^2 + xy + by^2 = 0$ will be $45^\circ$,if

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