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(A) The given pair of straight lines passes through $(1,1)$.Let the slopes of the lines be $m_1 = 	an 	heta$ and $m_2 = 	an(90^\circ - 	heta) = \c

Vedclass · Accepted Answer

$\frac{1}{2} \sin^{-1}\left(\frac{2}{a+2}\right)$

Vedclass · Answer

(A) The given pair of straight lines passes through $(1,1)$.Let the slopes of the lines be $m_1 = 	an 	heta$ and $m_2 = 	an(90^\circ - 	heta) = \cot 	heta$.The equations of the lines are $(y-1) = 	an 	heta(x-1)$ and $(y-1) = \cot 	heta(x-1)$.The combined equation is $[(y-1) - 	an 	heta(x-1)][(y-1) - \cot 	heta(x-1)] = 0$.Expanding this,we get $(y-1)^2 - (x-1)(y-1)(	an 	heta + \cot 	heta) + (x-1)^2 = 0$.Simplifying,$x^2 + y^2 - (	an 	heta + \cot 	heta)xy + (	an 	heta + \cot 	heta - 2)x + (	an 	heta + \cot 	heta - 2)y + (2 - (	an 	heta + \cot 	heta)) = 0$.Comparing this with the given equation $x^2 - (a+2)xy + y^2 + a(x+y-1) = 0$,we identify the coefficient of $xy$:$	an 	heta + \cot 	heta = a+2$.Using $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin 2	heta}$.Thus,$\frac{2}{\sin 2	heta} = a+2$,which implies $\sin 2	heta = \frac{2}{a+2}$.Therefore,$	heta = \frac{1}{2} \sin^{-1}\left(\frac{2}{a+2}ight)$.

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