The least positive value of $t$, so that the lines $x=t+\alpha, y+16=0 \quad$ and $\quad y=\alpha x$ are concurrent, is

Consider the given equation of lines, $$ \begin{array}{r} x-(t+\alpha)=0 \\ y+16=0 \end{array} $$ and $$ -\alpha x+y=0 $$ Since, these lines are concurrent, therefore the system of equations is consistent. Now, $$ \left|\begin{array}{ccc} 1 & 0 & -(t+\alpha) \\ 0 & 1 & 16 \\ -\alpha & 1 & 0 \end{array}\right|=0 $$ $$ \begin{array}{lr} \Rightarrow & 1(0-16)-(t+\alpha)(0+\alpha)=0 \\ \Rightarrow & -16-\alpha(t+\alpha)=0 \\ \Rightarrow & \alpha(t+\alpha)+16=0 \\ \Rightarrow & \alpha^{2}+t \alpha+16=0 \end{array} $$ Clearly, $\alpha$ should be real. $\therefore t^{2}-4 \times 16 \geq 0 \Rightarrow t^{2}-64 \geq 0$ Hence, least positive value of $t$ is 8.