Steps of construction:
(1) Draw a line segment AB of the length $9 cm$.
(2) Then draw perpendicular bisector PQ of line segment AB. So PQ is the required locus.
Proof:
(a)Let us take any point on PQ i.e. C.
(b) Now, let us join CA and CB.
Since, C lies on the perpendicular bisector of line segment AB.
Hence, C is equidistant from A and B.
Therefore, we can say that $CA = CB$ Hence, it is proved that perpendicular bisector of line segment AB is the locus of all points which are equidistant from A and B.