The measure of the acute angle theta between | vedclass

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(C) The general equation of a pair of lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing $2x^2 + xy - y^2 - x + 2y - 1 = 0$ with the general

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$\cos^{-1} \frac{1}{\sqrt{10}}$

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(C) The general equation of a pair of lines is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing $2x^2 + xy - y^2 - x + 2y - 1 = 0$ with the general form, we get $a = 2$, $2h = 1 \implies h = \frac{1}{2}$, and $b = -1$.The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(\frac{1}{2})^2 - (2)(-1)}}{2 + (-1)} ight| = \left| \frac{2\sqrt{\frac{1}{4} + 2}}{1} ight| = 2\sqrt{\frac{9}{4}} = 2 	imes \frac{3}{2} = 3$.Thus, $	heta = 	an^{-1}(3)$. Since the question asks for the acute angle, we check the options. Note that $	an 	heta = 3 \implies \sin 	heta = \frac{3}{\sqrt{10}}$ and $\cos 	heta = \frac{1}{\sqrt{10}}$.Therefore, $	heta = \cos^{-1} \frac{1}{\sqrt{10}}$. The correct option is $C$.

The measure of the acute angle $\theta$ between the pair of lines represented by the equation $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is:

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