Use the property of the perpendicular bisector of a segment and the triangle inequality.

## Solution

The perpendicular bisector of segment $A B$ divides the plane into two half-planes. Consider the half-plane containing point $B$. We will prove that for any point $M$ in this half-plane, $A M > B M$. Indeed, since points $A$ and $M$ lie in different half-planes with boundary $l$, segment $A M$ intersects line $l$ at some point $D$. By the property of the perpendicular bisector, $A D = B D$, therefore

$$B M < A M$$

If $B M \geq A M$, this would contradict the condition.

## Answer

The half-plane containing the point, with the boundary being the perpendicular bisector of the segment.