Draw a line segment of length 7 cm. Draw its perpendicular bisector, using ruler and compasses.
Construction: (1) Draw segment AB = 7 cm. (2) With centres A and B and radius > 3.5 cm, draw arcs above and below AB to intersect at P and Q. (3) Join PQ to get the perpendicular bisector of AB.
Goal: We want a line that cuts AB (length 7 cm) into two equal parts and is also at a right angle (90°) to AB. This special line is called the perpendicular bisector.
Points that are the same distance from both ends A and B always lie on the perpendicular bisector of AB. So, we create two points P and Q that are equally far from A and B by drawing arcs with the same radius from A and from B.
( AP = BP )
( AQ = BQ )
This happens because we used equal radii for the arcs from A and from B.
Since P and Q are both the same distance from A and B, they must lie on the locus of points equidistant from A and B — that locus is exactly the perpendicular bisector of AB. So joining P and Q gives that line.
The perpendicular bisector is at a right angle to the segment it bisects. Therefore,
( PQ perp AB )
Let M be the point where PQ meets AB. Because M lies on the perpendicular bisector,
( AM = MB )
Since AB is 7 cm, each half is
( AM = MB = 3.5 ext{cm} )
( AM = MB )
.( PQ perp AB )
.“Equal arcs from the two ends → intersection points are equally distant from both ends → join those points → you get the perpendicular bisector.”