If the lines x^2-4xy+y^2=0 and x+y=10 contain | vedclass

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(C) The pair of lines $x^2-4xy+y^2=0$ passes through the origin $(0,0)$. The angle $	heta$ between these lines is given by $	an 	heta = \left|\frac

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$\frac{50}{\sqrt{3}}$ sq. units

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(C) The pair of lines $x^2-4xy+y^2=0$ passes through the origin $(0,0)$. The angle $	heta$ between these lines is given by $	an 	heta = \left|\frac{2\sqrt{h^2-ab}}{a+b}ight|$.Here $a=1, h=-2, b=1$,so $	an 	heta = \left|\frac{2\sqrt{(-2)^2-1	imes 1}}{1+1}ight| = \sqrt{3}$,which means $	heta = 60^{\circ}$.Since the lines form an equilateral triangle with the line $x+y=10$,the origin is one vertex of the triangle.The perpendicular distance from the origin $(0,0)$ to the line $x+y-10=0$ is the altitude $h$ of the equilateral triangle.$h = \frac{|0+0-10|}{\sqrt{1^2+1^2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}$.For an equilateral triangle with side length $s$,the altitude $h = \frac{\sqrt{3}}{2}s$,so $s = \frac{2h}{\sqrt{3}} = \frac{2(5\sqrt{2})}{\sqrt{3}} = \frac{10\sqrt{2}}{\sqrt{3}}$.The area of the equilateral triangle is $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4} 	imes \left(\frac{10\sqrt{2}}{\sqrt{3}}ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{200}{3} = \frac{50\sqrt{3}}{3} = \frac{50}{\sqrt{3}}$ sq. units.

If the lines $x^2-4xy+y^2=0$ and $x+y=10$ contain the sides of an equilateral triangle,then the area of the equilateral triangle is

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