If the lines x^2+kxy+y^2=0 and x+y=1 form the | vedclass

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(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The

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$16$

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(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The equation of the third line is $x+y=1$,which can be written as $y=-x+1$,so its slope is $m_3=-1$.Since the lines form an equilateral triangle,the angle between any two lines is $60^{\circ}$.The angle between the line $y=mx$ and the line $x+y=1$ (slope $-1$) is given by:$	an 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} ight|$$\sqrt{3} = \left| \frac{m+1}{1-m} ight|$Squaring both sides:$3 = \frac{(m+1)^2}{(1-m)^2}$$3(1-2m+m^2) = m^2+2m+1$$3-6m+3m^2 = m^2+2m+1$$2m^2-8m+2 = 0$$m^2-4m+1 = 0$Since $m = y/x$,we substitute this into the quadratic equation:$(y/x)^2 - 4(y/x) + 1 = 0$$y^2 - 4xy + x^2 = 0$Comparing this with $x^2+kxy+y^2=0$,we get $k=-4$.Therefore,$k^2 = (-4)^2 = 16$.

If the lines $x^2+kxy+y^2=0$ and $x+y=1$ form the sides of an equilateral triangle,then the value of $k^2$ is

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