If the line y=mx is one of the bisectors of x | vedclass

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(A) Given the equation of the pair of straight lines: $x^2+4xy-y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$b=-1$,and

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$-1+\sqrt{5}$

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(A) Given the equation of the pair of straight lines: $x^2+4xy-y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$b=-1$,and $2h=4$,which implies $h=2$. The equation of the bisectors of the angle between the lines is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$. Substituting the values: $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{2} \Rightarrow \frac{x^2-y^2}{2} = \frac{xy}{2}$. This simplifies to $x^2-xy-y^2=0$. Since $y=mx$ is one of the bisectors,we substitute $y=mx$ into the equation: $x^2-x(mx)-(mx)^2=0$. Dividing by $x^2$ (assuming $x 
eq 0$),we get $1-m-m^2=0$,or $m^2+m-1=0$. Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$. Therefore,$2m = -1 \pm \sqrt{5}$.

If the line $y=mx$ is one of the bisectors of $x^2+4xy-y^2=0$,then the value of $2m$ is:

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