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

The correct option is D $8$

Consider the given equation of lines,
$x-(t+\alpha )=0$
$y+16=0$
and $-\alpha x+y=0$
Since, these lines are concurrent, therefore the system of equations is consistent.
Now,

$\left|\begin{array}{ccc}1& 0& -(t+\alpha )\\ 0& 1& 16\\ -\alpha & 1& 0\end{array}\right|=0$

$\Rightarrow 1(0-16)-(t+\alpha )(0+\alpha )=0$
$\Rightarrow -16-\alpha (t-\alpha )=0$
$\Rightarrow \alpha (t+\alpha )+16=0$
$\Rightarrow {\alpha }^2+t\alpha +16=0$

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 $8$.