If the slopes of both the lines given by x^2 | vedclass

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(B) The equation of the pair of lines is $x^2 + 2hxy + 6y^2 = 0$. Let the slopes be $m_1$ and $m_2$. Then $m_1 + m_2 = -\frac{2h}{6} = -\frac{h}{3}$ a

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$\frac{1}{5 \sqrt{2}}$

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(B) The equation of the pair of lines is $x^2 + 2hxy + 6y^2 = 0$. Let the slopes be $m_1$ and $m_2$. Then $m_1 + m_2 = -\frac{2h}{6} = -\frac{h}{3}$ and $m_1 m_2 = \frac{1}{6}$.Since $m_1, m_2 > 0$,$h$ must be negative.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \frac{1}{7}$.Substituting $m_1 m_2 = \frac{1}{6}$,we get $\frac{|m_1 - m_2|}{1 + 1/6} = \frac{1}{7} \implies |m_1 - m_2| = \frac{1}{7} 	imes \frac{7}{6} = \frac{1}{6}$.Using $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$,we have $(\frac{1}{6})^2 = (-\frac{h}{3})^2 - 4(\frac{1}{6}) \implies \frac{1}{36} = \frac{h^2}{9} - \frac{2}{3} \implies \frac{h^2}{9} = \frac{1}{36} + \frac{24}{36} = \frac{25}{36} \implies h^2 = \frac{25}{4} \implies h = -\frac{5}{2}$ (since $h The lines are $x^2 - 5xy + 6y^2 = 0$,which factorizes to $(x - 2y)(x - 3y) = 0$. The lines are $L_1: x - 2y = 0$ and $L_2: x - 3y = 0$.The perpendicular distances from $(1, 1)$ to $L_1$ and $L_2$ are $d_1 = \frac{|1 - 2|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{5}}$ and $d_2 = \frac{|1 - 3|}{\sqrt{1^2 + (-3)^2}} = \frac{2}{\sqrt{10}}$.The product of the perpendiculars is $d_1 d_2 = \frac{1}{\sqrt{5}} 	imes \frac{2}{\sqrt{10}} = \frac{2}{\sqrt{50}} = \frac{2}{5 \sqrt{2}} = \frac{\sqrt{2}}{5}$.Note: Given the options,there might be a typo in the question's target value. Based on standard evaluation,the result is $\frac{\sqrt{2}}{5}$.

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then the product of the perpendiculars from the point $(1, 1)$ to these lines is:

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