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

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

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

⇒ m2 + am + 2 = 0

Since m is real, a2  ≥ 8,  |a| ≥ 2√2

Hence the least value of |a| is 2√2.