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(C) The general equation of a pair of straight lines passing through the origin is $ax^2+2hxy+by^2=0$.If one of these lines bisects the angle between

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$\{-10,2\}$

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(C) The general equation of a pair of straight lines passing through the origin is $ax^2+2hxy+by^2=0$.If one of these lines bisects the angle between the coordinate axes,its equation is $y=x$ or $y=-x$.Substituting $y=x$ into the equation $ax^2+2hxy+by^2=0$,we get $ax^2+2hx^2+bx^2=0$,which implies $a+2h+b=0$,or $a+b=-2h$. Squaring both sides gives $(a+b)^2=4h^2$.Similarly,if $y=-x$,we get $ax^2-2hx^2+bx^2=0$,which implies $a-2h+b=0$,or $a+b=2h$. Squaring both sides gives $(a+b)^2=4h^2$.Thus,the condition for one of the lines to bisect the angle between the axes is $(a+b)^2=4h^2$.For the given equation $4x^2+6xy+ky^2=0$,we have $a=4$,$2h=6$ (so $h=3$),and $b=k$.Substituting these values into the condition $(a+b)^2=4h^2$:$(4+k)^2=4(3)^2$$(4+k)^2=36$$4+k = \pm 6$If $4+k=6$,then $k=2$.If $4+k=-6$,then $k=-10$.Therefore,$k \in \{-10, 2\}$.

If one of the lines in the pair of straight lines given by $4x^2+6xy+ky^2=0$ bisects the angle between the coordinate axes,then $k \in$

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