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(C) The equation $x^2+4xy+y^2=0$ represents a pair of straight lines passing through the origin. Let these lines be $L_1$ and $L_2$. The equation $x-y

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Equilateral Triangle

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(C) The equation $x^2+4xy+y^2=0$ represents a pair of straight lines passing through the origin. Let these lines be $L_1$ and $L_2$. The equation $x-y=4$ represents a third line $L_3$. The lines $L_1$ and $L_2$ are given by $y = m_1x$ and $y = m_2x$,where $m_1, m_2$ are roots of $m^2+4m+1=0$. Thus,$m_1+m_2 = -4$ and $m_1m_2 = 1$. The slopes are $m = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$. The angle between $L_1$ and $L_2$ is $	an 	heta = |\frac{2\sqrt{h^2-ab}}{a+b}| = |\frac{2\sqrt{4-1}}{1+1}| = \sqrt{3}$,so $	heta = 60^\circ$. Since the lines $L_1$ and $L_2$ pass through the origin and the angle between them is $60^\circ$,and the third line $L_3$ intersects them,we check the slopes of the lines formed by the intersection. By calculating the slopes of the sides of the triangle formed by these three lines,it is found that all sides are equal in length or the angles are $60^\circ$,confirming it is an Equilateral Triangle.

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

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