If one of the lines given by k x^{2}+x y-y^{2 | vedclass

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(C) The given equation of the pair of lines is $k x^{2}+x y-y^{2}=0$. Dividing by $x^{2}$,we get $k + \frac{y}{x} - (\frac{y}{x})^{2} = 0$. Let $m = \

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$0, 2$

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(C) The given equation of the pair of lines is $k x^{2}+x y-y^{2}=0$. Dividing by $x^{2}$,we get $k + \frac{y}{x} - (\frac{y}{x})^{2} = 0$. Let $m = \frac{y}{x}$ be the slope of the lines. Then the auxiliary equation is $-m^{2} + m + k = 0$. Since one of the lines bisects the angle between the coordinate axes,its slope $m$ must be $\pm 1$. Case $1$: If $m = 1$,then $-(1)^{2} + 1 + k = 0 \implies -1 + 1 + k = 0 \implies k = 0$. Case $2$: If $m = -1$,then $-(-1)^{2} + (-1) + k = 0 \implies -1 - 1 + k = 0 \implies k = 2$. Thus,the values of $k$ are $0$ and $2$.

If one of the lines given by $k x^{2}+x y-y^{2}=0$ bisects the angle between the coordinate axes,then the values of $k$ are

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