If the acute angles between the pairs of line | vedclass

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(A) The acute angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} igh

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$	heta_1 = 	heta_2$

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(A) The acute angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.For the first pair $3x^2 - 7xy + 4y^2 = 0$,we have $a=3, 2h=-7, b=4$. Thus $h = -3.5$.$	an 	heta_1 = \left| \frac{2\sqrt{(-3.5)^2 - (3)(4)}}{3 + 4} ight| = \left| \frac{2\sqrt{12.25 - 12}}{7} ight| = \frac{2\sqrt{0.25}}{7} = \frac{2(0.5)}{7} = \frac{1}{7}$.So,$	heta_1 = 	an^{-1}(\frac{1}{7})$.For the second pair $6x^2 - 5xy + y^2 = 0$,we have $a=6, 2h=-5, b=1$. Thus $h = -2.5$.$	an 	heta_2 = \left| \frac{2\sqrt{(-2.5)^2 - (6)(1)}}{6 + 1} ight| = \left| \frac{2\sqrt{6.25 - 6}}{7} ight| = \frac{2\sqrt{0.25}}{7} = \frac{2(0.5)}{7} = \frac{1}{7}$.So,$	heta_2 = 	an^{-1}(\frac{1}{7})$.Therefore,$	heta_1 = 	heta_2$.

If the acute angles between the pairs of lines $3x^2 - 7xy + 4y^2 = 0$ and $6x^2 - 5xy + y^2 = 0$ are $\theta_1$ and $\theta_2$ respectively,then:

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