The figure formed by the lines x^2 + 4xy + y^ | vedclass

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(C) The pair of lines represented by $x^2 + 4xy + y^2 = 0$ are given by $y = m_1x$ and $y = m_2x$. Dividing by $x^2$,we get $m^2 + 4m + 1 = 0$. The sl

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An equilateral triangle

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(C) The pair of lines represented by $x^2 + 4xy + y^2 = 0$ are given by $y = m_1x$ and $y = m_2x$. Dividing by $x^2$,we get $m^2 + 4m + 1 = 0$. The slopes are $m_1, m_2 = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}$. The third line is $x - y = 4$,which can be written as $y = x - 4$,so its slope $m_3 = 1$. The angle $	heta$ between two lines with slopes $m_a$ and $m_b$ is given by $	an 	heta = |\frac{m_a - m_b}{1 + m_a m_b}|$. For $m_1 = -2 + \sqrt{3}$ and $m_3 = 1$: $	an 	heta_{13} = |\frac{-2 + \sqrt{3} - 1}{1 + (-2 + \sqrt{3})(1)}| = |\frac{\sqrt{3} - 3}{\sqrt{3} - 1}| = |\frac{\sqrt{3}(1 - \sqrt{3})}{\sqrt{3} - 1}| = \sqrt{3}$. Thus,$	heta_{13} = 60^\circ$. For $m_2 = -2 - \sqrt{3}$ and $m_3 = 1$: $	an 	heta_{23} = |\frac{-2 - \sqrt{3} - 1}{1 + (-2 - \sqrt{3})(1)}| = |\frac{-3 - \sqrt{3}}{-1 - \sqrt{3}}| = |\frac{\sqrt{3}(\sqrt{3} + 1)}{\sqrt{3} + 1}| = \sqrt{3}$. Thus,$	heta_{23} = 60^\circ$. Since two angles are $60^\circ$,the third angle must also be $60^\circ$. Therefore,the triangle is an equilateral triangle.

The figure formed by the lines $x^2 + 4xy + y^2 = 0$ and $x - y = 4$ is

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