If theta is the acute angle between the lines | vedclass

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(C) The given equation is $2x^2 + 7xy + 3y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$, we have $a = 2$, $2h = 7 \implies h = \frac{7}{2}$, a

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$-\frac{1}{9}$

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(C) The given equation is $2x^2 + 7xy + 3y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$, we have $a = 2$, $2h = 7 \implies h = \frac{7}{2}$, and $b = 3$.The formula for the acute angle $	heta$ between the lines is $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.$	an 	heta = \left| \frac{2\sqrt{(\frac{7}{2})^2 - (2)(3)}}{2 + 3} ight| = \left| \frac{2\sqrt{\frac{49}{4} - 6}}{5} ight| = \left| \frac{2\sqrt{\frac{25}{4}}}{5} ight| = \left| \frac{2 \cdot \frac{5}{2}}{5} ight| = 1$.Since $	an 	heta = 1$, $	heta = 45^\circ$. Thus, $\sin 	heta = \frac{1}{\sqrt{2}}$ and $\cos 	heta = \frac{1}{\sqrt{2}}$.Substituting these values into the expression: $\frac{2(\frac{1}{\sqrt{2}}) - 3(\frac{1}{\sqrt{2}})}{4(\frac{1}{\sqrt{2}}) + 5(\frac{1}{\sqrt{2}})} = \frac{\frac{2-3}{\sqrt{2}}}{\frac{4+5}{\sqrt{2}}} = \frac{-1}{9}$.

If $\theta$ is the acute angle between the lines represented by the equation $2x^2 + 7xy + 3y^2 = 0$, then the value of $\frac{2 \cos \theta - 3 \sin \theta}{4 \sin \theta + 5 \cos \theta}$ is:

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