If the bisectors of the pair of lines x^2-2 m | vedclass

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(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of

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$mn+1=0$

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(A) The given equation is $x^2-2 m x y-y^2=0$. Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$. The equation of the bisectors of the angles between the pair of lines is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$. Substituting the values,we get $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-m}$. This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-m}$,which implies $-m(x^2-y^2)=2 x y$. Rearranging the terms,we get $m x^2+2 x y-m y^2=0$. Dividing by $m$ (assuming $m 
eq 0$),we get $x^2+\frac{2}{m} x y-y^2=0$. Comparing this with the given equation of the bisectors $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$. This implies $mn = -1$,or $mn+1=0$.

If the bisectors of the pair of lines $x^2-2 m x y-y^2=0$ are represented by $x^2-2 n x y-y^2=0$,then

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