The angle between the pair of lines x^{2}+2xy | vedclass

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(B) The given equation of the pair of lines is $x^{2}+2xy-y^{2}=0$. Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$h=1$,and

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$\frac{\pi}{2}$

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(B) The given equation of the pair of lines is $x^{2}+2xy-y^{2}=0$. Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$h=1$,and $b=-1$. We know that the lines represented by $ax^{2}+2hxy+by^{2}=0$ are perpendicular if $a+b=0$. Here,$a+b = 1 + (-1) = 0$. Since the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero,the lines are perpendicular to each other. Therefore,the angle between the lines is $\frac{\pi}{2}$.

The angle between the pair of lines $x^{2}+2xy-y^{2}=0$ is

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