Problem: If the lines x=a + m, y=-2 and y = mx are concurrent, the least value of |a| is

(A) 0

(B) √2

(C) 2√2

(D) none of these

Ans:(C)

Sol: Since the lines are concurrent

$ \large \left| \begin{array}{ccc} 1 & 0 & -a-m \\ 0 & 1 & 2 \\ m & -1 & 0 \end{array} \right| = 0 $

⇒ m^{2} + am + 2 = 0

Since m is real, a^{2} ≥ 8, |a| ≥ 2√2

Hence the least value of |a| is 2√2.