Question

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

• A. 2
• B. 4
• C. 16
• D. 8

Answer 1

Answer: (D)

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