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

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Solution

The correct option is C

$∴$ x = a+m passes through $(\frac{- 2}{m}, - 2)$

$\Rightarrow$ $\frac{- 2}{m}$ = a + m

$\Rightarrow$ $m^{2} + a + 2$ = 0

As 'm' has real value,

$\Rightarrow$ $a^{2} \geq 8 \Rightarrow | a | \geq 2 \sqrt{2}$

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