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

Compare with $ax^2+2hxy+by^2=0$ and apply $\tan\theta=\dfrac{2\sqrt{h^2-ab}}{|a+b|}$ to get $\theta=\tan^{-1}\!\left(\dfrac{\sqrt{33}}{5}\right)$.

The given equation $x^2-7xy+4y^2=0$ is a homogeneous second-degree equation representing a pair of straight lines through the origin. Writing it as $ax^2+2hxy+by^2=0$:

$a=1,\qquad 2h=-7\Rightarrow h=-\tfrac{7}{2},\qquad b=4.$

The angle $\theta$ between the pair of lines is

$\tan\theta=\frac{2\sqrt{h^2-ab}}{|a+b|}.$

Compute $h^2-ab=\dfrac{49}{4}-4=\dfrac{49-16}{4}=\dfrac{33}{4}$, so $\sqrt{h^2-ab}=\dfrac{\sqrt{33}}{2}$.

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