The triangle formed by x^2-4xy+y^2=0 and x+y+ | vedclass

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(A) The equation $x^2-4xy+y^2=0$ represents a pair of straight lines passing through the origin. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=1,

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an equilateral triangle

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(A) The equation $x^2-4xy+y^2=0$ represents a pair of straight lines passing through the origin. Comparing this with $ax^2+2hxy+by^2=0$,we have $a=1, h=-2, b=1$. The angle $	heta$ between these lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2-ab}}{a+b} ight| = \left| \frac{2\sqrt{(-2)^2-(1)(1)}}{1+1} ight| = \left| \frac{2\sqrt{3}}{2} ight| = \sqrt{3}$. Thus,$	heta = 60^\circ$. The lines are $y = (2 \pm \sqrt{3})x$. The third line is $x+y+4\sqrt{6}=0$,which makes an angle of $135^\circ$ with the positive $x$-axis (slope $m=-1$). The angles of the triangle formed by these three lines are $60^\circ, 60^\circ$,and $60^\circ$. Since all angles are $60^\circ$,the triangle is an equilateral triangle.

The triangle formed by $x^2-4xy+y^2=0$ and $x+y+4\sqrt{6}=0$ is

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