The perpendicular bisector of a line segment AB is the straight line that passes through its midpoint M and meets AB at a right angle (90). Why every point is equidistant Let P be any point on the perpendicular bisector, with P!= M. Join PA and PB. AM=BM because M is the midpoint, PM is a common side, and AMP= BMP=90. Hence AMP BMP by SAS, so PA=PB (corresponding sides). If P=M, the distances are also equal because MA=MB. Thus every point on the perpendicular bisector is equidistant from A and B. Finding its equation Find the midpoint and the perpendicular gradient. Use the midpoint in the point-gradient equation; handle a horizontal or vertical segment directly.

The perpendicular bisector of a line segment \(AB\) is the straight line that passes through its midpoint \(M\) and meets \(AB\) at a right angle \((90^\circ)\).

Let \(P\) be any point on the perpendicular bisector, with \(P\ne M\). Join \(PA\) and \(PB\).

\(AM=BM\) because \(M\) is the midpoint, \(PM\) is a common side, and \(\angle AMP=\angle BMP=90^\circ\).

Hence \(\triangle AMP\equiv\triangle BMP\) by SAS, so \(PA=PB\) (corresponding sides). If \(P=M\), the distances are also equal because \(MA=MB\). Thus every point on the perpendicular bisector is equidistant from \(A\) and \(B\).

Find the midpoint and the perpendicular gradient. Use the midpoint in the point-gradient equation; handle a horizontal or vertical segment directly.

Perpendicular Bisector

Perpendicular Bisector