The angle between the pair of straight lines | vedclass

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(C) The given equation is $x^2 - 7xy + 4y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = -7$ (so $h = -7/2

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$	an^{-1}\left(\frac{\sqrt{33}}{5}ight)$

Vedclass · Answer

(C) The given equation is $x^2 - 7xy + 4y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1$,$2h = -7$ (so $h = -7/2$),and $b = 4$. The angle $	heta$ between the pair of straight lines is given by the formula $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-7/2)^2 - (1)(4)}}{1 + 4} ight|$. $	an 	heta = \left| \frac{2\sqrt{49/4 - 4}}{5} ight| = \left| \frac{2\sqrt{33/4}}{5} ight| = \left| \frac{2 	imes \frac{\sqrt{33}}{2}}{5} ight| = \frac{\sqrt{33}}{5}$. Therefore,$	heta = 	an^{-1}\left(\frac{\sqrt{33}}{5}ight)$.

The angle between the pair of straight lines $x^2 + 4y^2 - 7xy = 0$ is

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