Question

If the line's $2x+3y-5=0$ and $(t^2+1)x+(2t-3)y+5=0$ are perpendicular to each other, then absolute value of sum of all possible real values of $t$ is

Answer 1

Given lines
$l_1:2x+3y-5=0l_2:(t^2+1)x+(2t-3)y+5=0$
Slope of both the lines is
$m_1=-\frac{2}{3},m_2=-\frac{t^2+1}{2t-3}$
So,
$m_1\times m_2=-1\Rightarrow \frac{2}{3}\times \left(\frac{t^2+1}{2t-3}\right)=-1\Rightarrow 2t^2+2=9-6t$
$\Rightarrow 2t^2+6t-7=0$
$\Rightarrow t_1+t_2=-3$
$\therefore |t_1+t_2|=3$