The absolute value of the tangent of the diff | vedclass

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(C) Given the equation of the pair of lines is $4x^2 - 24xy + 11y^2 = 0$. Factorizing the equation: $4x^2 - 22xy - 2xy + 11y^2 = 0$ $\Rightarrow 2x(2x

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$\frac{4}{3}$

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(C) Given the equation of the pair of lines is $4x^2 - 24xy + 11y^2 = 0$. Factorizing the equation: $4x^2 - 22xy - 2xy + 11y^2 = 0$ $\Rightarrow 2x(2x - 11y) - y(2x - 11y) = 0$ $\Rightarrow (2x - y)(2x - 11y) = 0$. Thus,the lines are $2x - y = 0$ and $2x - 11y = 0$. The slopes of these lines are $m_1 = 2$ and $m_2 = \frac{2}{11}$. Let $	heta_1$ and $	heta_2$ be the angles made by these lines with the $X$-axis,so $	an 	heta_1 = 2$ and $	an 	heta_2 = \frac{2}{11}$. We need to find $	an |	heta_1 - 	heta_2| = \left| \frac{	an 	heta_1 - 	an 	heta_2}{1 + 	an 	heta_1 	an 	heta_2} ight|$. Substituting the values: $\left| \frac{2 - \frac{2}{11}}{1 + 2 	imes \frac{2}{11}} ight| = \left| \frac{\frac{22 - 2}{11}}{\frac{11 + 4}{11}} ight| = \left| \frac{20}{15} ight| = \frac{4}{3}$.

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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