If the pair of lines 2x^2 + 3xy + y^2 = 0 mak | vedclass

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(B) The given equation of the pair of lines is $2x^2 + 3xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 2$,$2

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$1/3$

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(B) The given equation of the pair of lines is $2x^2 + 3xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 2$,$2h = 3$ (so $h = 3/2$),and $b = 1$. Let $m_1 = 	an 	heta_1$ and $m_2 = 	an 	heta_2$ be the slopes of the lines. We know that $m_1 + m_2 = -2h/b = -3/1 = -3$ and $m_1 m_2 = a/b = 2/1 = 2$. The angle between the lines is given by $|	an(	heta_1 - 	heta_2)| = |(m_1 - m_2) / (1 + m_1 m_2)|$. Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$,we have $(m_1 - m_2)^2 = (-3)^2 - 4(2) = 9 - 8 = 1$. Thus,$|m_1 - m_2| = \sqrt{1} = 1$. Therefore,$|	an(	heta_1 - 	heta_2)| = 1 / (1 + 2) = 1/3$.

If the pair of lines $2x^2 + 3xy + y^2 = 0$ makes angles $\theta_1$ and $\theta_2$ with the positive direction of the $X$-axis,then $|\tan(\theta_1 - \theta_2)| = $

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