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(C) The equation $x^2+4xy+y^2=0$ represents a pair of lines passing through the origin. Let these lines be $y = m_1x$ and $y = m_2x$. Comparing with $

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Equilateral Triangle

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(C) The equation $x^2+4xy+y^2=0$ represents a pair of lines passing through the origin. Let these lines be $y = m_1x$ and $y = m_2x$. Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=2, b=1$. The slopes $m_1, m_2$ satisfy $m_1+m_2 = -2h/b = -4$ and $m_1m_2 = a/b = 1$. The angle $	heta$ between these lines is given by $	an 	heta = |\frac{2\sqrt{h^2-ab}}{a+b}| = |\frac{2\sqrt{4-1}}{1+1}| = \sqrt{3}$. Thus,$	heta = 60^\circ$. Since the lines pass through the origin and the angle between them is $60^\circ$,the triangle formed by these two lines and the line $x-y=4$ is an equilateral triangle because the line $x-y=4$ makes an angle of $45^\circ$ with the axes,but checking the intersection points shows the triangle is equilateral.

The equations $x-y=4$ and $x^2+4xy+y^2=0$ represent the sides of a/an

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