The absolute value of the tangent of the diff | vedclass

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(C) Given the equation of the pair of lines: $4x^2 - 24xy + 11y^2 = 0$. Factorizing the quadratic equation: $(2x - y)(2x - 11y) = 0$. Thus,the equatio

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$\frac{4}{3}$

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(C) Given the equation of the pair of lines: $4x^2 - 24xy + 11y^2 = 0$. Factorizing the quadratic equation: $(2x - y)(2x - 11y) = 0$. Thus,the equations of the two lines are $y = 2x$ and $y = \frac{2}{11}x$. The angles $	heta_1$ and $	heta_2$ made by these lines with the $X$-axis are $	an 	heta_1 = 2$ and $	an 	heta_2 = \frac{2}{11}$. We need to find $	an |	heta_1 - 	heta_2|$. Using the formula $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$: $	an(	heta_1 - 	heta_2) = \frac{2 - \frac{2}{11}}{1 + 2 	imes \frac{2}{11}} = \frac{\frac{22 - 2}{11}}{\frac{11 + 4}{11}} = \frac{20}{15} = \frac{4}{3}$. Therefore,the absolute value is $\frac{4}{3}$.

The absolute value of the tangent of the difference of the angles made by the lines $4x^2 - 24xy + 11y^2 = 0$ with the $X$-axis is

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