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(C) The pair of straight lines is given by $ax^2+2hxy+by^2=0$.If one of the lines bisects the angle between the coordinate axes,its equation is $y=x$

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$\{-10,2\}$

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(C) The pair of straight lines is given by $ax^2+2hxy+by^2=0$.If one of the lines bisects the angle between the coordinate axes,its equation is $y=x$ or $y=-x$.This implies that the line satisfies the equation $y^2=x^2$,or $x^2-y^2=0$.The condition for one of the lines of $ax^2+2hxy+by^2=0$ to be $y=mx$ is $am^2+2hm+b=0$.For $y=x$,$m=1$,so $a+2h+b=0$. For $y=-x$,$m=-1$,so $a-2h+b=0$.Combining these,we get $(a+b)^2 = (2h)^2 = 4h^2$.Given $4x^2+6xy+ky^2=0$,we have $a=4, 2h=6 \Rightarrow h=3, b=k$.Substituting into $(a+b)^2=4h^2$:$(4+k)^2 = 4(3)^2 = 36$.$4+k = \pm 6$.If $4+k=6$,then $k=2$.If $4+k=-6$,then $k=-10$.Thus,$k \in \{-10, 2\}$.

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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