TS EAMCET 2017 Code A Solved Paper

Given equation of pair of straight lines can be rewritten as $x(y-1)-1(y-1)=0$ $(x-1)(y-1)=0$ $x=1$ $y=1$ The three concurrent lines are, $x=1 \;\; .....(1)$ $y=1 \;\; .....(2)$ $x+a y-3=0$ since equation (1) and (2) intersect at only points namely (1,1) therefore, this point also satisfy the equation (3) $1+a-3=0$ $a=2$ Now the pair of line becomes, $2x^{2}-13xy-7y^{2}+x+23y-6=0$ The acute angle between the lines is calculated as $\tan\theta = \left| \;\; \frac{2 \sqrt{h^{2} - ab}}{a+b} \right|$ $= \left| \frac{\;\; 2 \sqrt{ \left(-\;\; \frac{13}{2}\right)^2 + 14 }}{-5} \right|$ $= \left| -3 \right|$ $=3$ The angle is calculated as,\ $\theta = \tan^{-1}(3)$ $= \cos^{-1}\left( \;\; \frac{1}{\sqrt{10}} \right)$