If the equation 3 x^2-k x y-3 y^2=0 represent | vedclass

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(B) The equation of the pair of lines is $x^2-3 x y-4 y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=1, H=-\frac{3}{2}, B=-4$. The equation

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-$10$

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(B) The equation of the pair of lines is $x^2-3 x y-4 y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=1, H=-\frac{3}{2}, B=-4$. The equation of the angle bisectors is given by $\frac{x^2-y^2}{A-B} = \frac{xy}{H}$. Substituting the values,we get $\frac{x^2-y^2}{1-(-4)} = \frac{xy}{-3/2}$. $\frac{x^2-y^2}{5} = -\frac{2xy}{3}$. $3(x^2-y^2) = -10xy$. $3x^2+10xy-3y^2=0$. Comparing this with the given equation $3x^2-kxy-3y^2=0$,we have $-k=10$,which implies $k=-10$.

If the equation $3 x^2-k x y-3 y^2=0$ represents the bisectors of the angles between the lines $x^2-3 x y-4 y^2=0$,then the value of $k$ is

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