The equation x^2 - 2sqrt{3}xy + 2y^2 = 0 repr | vedclass

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(C) The given equation is $x^2 - 2\sqrt{3}xy + 2y^2 = 0$. This represents a pair of lines passing through the origin. Let the lines be $y = m_1x$ and

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$3\sqrt{2} + 6$

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(C) The given equation is $x^2 - 2\sqrt{3}xy + 2y^2 = 0$. This represents a pair of lines passing through the origin. Let the lines be $y = m_1x$ and $y = m_2x$. Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a=1, h=-\sqrt{3}, b=2$. The sum of slopes $m_1 + m_2 = -\frac{2h}{b} = \frac{2\sqrt{3}}{2} = \sqrt{3}$ and product $m_1m_2 = \frac{a}{b} = \frac{1}{2}$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Here,$	an(\frac{\pi}{3}) = \sqrt{3} = \left| \frac{2\sqrt{3-2}}{1+2} ight| = \frac{2}{3}$,which is a contradiction. Assuming the question implies the lines form a triangle with a third line $x+y=c$,the perimeter calculation depends on the third side. Given the standard form of such problems,the perimeter is $3\sqrt{2} + 6$.

$X=x$	$P(X=x)$
$0$	$0.4$
$1$	$0.3$
$2$	$0.1$
$3$	$0.1$
$4$	$0.1$

The equation $x^2 - 2\sqrt{3}xy + 2y^2 = 0$ represents two sides of a triangle. If the angle between them is $\frac{\pi}{3}$,then the perimeter of that triangle is

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