If the pair of straight lines xy-x-y+1=0 and | vedclass

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(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$. This represents two lines: $x=1$ and $y=1$. Since these lines are concurre

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$\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$

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(B) The given pair of lines $xy-x-y+1=0$ can be factored as $(x-1)(y-1)=0$. This represents two lines: $x=1$ and $y=1$. Since these lines are concurrent with $x+ay-3=0$,the point of intersection $(1,1)$ must satisfy the equation $x+ay-3=0$. Substituting $(1,1)$ into $x+ay-3=0$,we get $1+a(1)-3=0$,which implies $a=2$. Substituting $a=2$ into the second pair of lines,we get $2x^2-13xy-7y^2+x+23y-6=0$. For a general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,the angle $	heta$ between the lines is given by $	an 	heta = \left|\frac{2\sqrt{H^2-AB}}{A+B}ight|$. Here $A=2, 2H=-13, B=-7$. Thus,$	an 	heta = \left|\frac{2\sqrt{(-13/2)^2 - (2)(-7)}}{2-7}ight| = \left|\frac{2\sqrt{169/4 + 14}}{-5}ight| = \left|\frac{2\sqrt{225/4}}{-5}ight| = \left|\frac{2(15/2)}{-5}ight| = |-3| = 3$. Since $	an 	heta = 3$,we have $\cos 	heta = \frac{1}{\sqrt{1+3^2}} = \frac{1}{\sqrt{10}}$. Therefore,$	heta = \cos^{-1}\left(\frac{1}{\sqrt{10}}ight)$.

If the pair of straight lines $xy-x-y+1=0$ and the line $x+ay-3=0$ are concurrent,then the acute angle between the pair of lines $ax^2-13xy-7y^2+x+23y-6=0$ is

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