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(D) The given pair of lines is $x^2 - 4xy + y^2 = 0$. The slopes $m_1, m_2$ of these lines satisfy $m^2 - 4m + 1 = 0$,giving $m = 2 \pm \sqrt{3}$.Thes

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equilateral

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(D) The given pair of lines is $x^2 - 4xy + y^2 = 0$. The slopes $m_1, m_2$ of these lines satisfy $m^2 - 4m + 1 = 0$,giving $m = 2 \pm \sqrt{3}$.These slopes correspond to angles $	heta_1 = 75^\circ$ and $	heta_2 = 15^\circ$ with the $x$-axis.The angle between these two lines is $|75^\circ - 15^\circ| = 60^\circ$.The third line is $x + y + 4 = 0$,which has a slope of $-1$,corresponding to an angle of $135^\circ$ with the $x$-axis.The angles of the triangle formed are $60^\circ$,$180^\circ - 135^\circ + 15^\circ = 60^\circ$,and $180^\circ - 60^\circ - 60^\circ = 60^\circ$.Since all angles are $60^\circ$,the triangle is equilateral.

The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is:

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