TS EAMCET 5 May 2018 Shift 1 Solved Paper

Assume the slope of required line be m. The angle between bisector $x+y+1=0$ and the new line is given as, $\tan \theta = \left| \frac{m+1}{1-m} \right|$ …… (1) The angle between bisector $x+y+1=0$ and $2x+3y-4=0$ is given as, $\tan \theta = \left| \frac{-\frac{2}{3} + \frac{1}{3}}{1 + \frac{2}{3} \times 1} \right|$ $=\left|\frac{1}{5}\right|$ …… (2) From equations (1) and equations (2), $\frac{m+1}{1-m} = \left( \frac{1}{5} \right)$ or $\frac{m+1}{1-m} = \left( \frac{-1}{5} \right)$ $5m+5=1-m$ or $5m+5=-1+m$ $6m=-4$ or $4m=-6$ $m=-\frac{2}{3}$ or $m=-\frac{3}{2}$ The equation is expressed as, $x+y=-1$ $2x+3y=4$ Solve the above equation for the value of x and y . $y=6$ $x=-7$ The required equation is expressed as, $y-y_1 = m(x-x_1)$ $(y-6) = \frac{-3}{2}(x+7)$ $3x+2y+9=0$