Perpendicular Bisector

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 \((90^\circ)\).

Why every point is equidistant

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 \(B\).

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.

Example

Find the perpendicular bisector of \(A(1,2)\) and \(B(5,10)\).

\(M=(3,6),\quad m_{AB}=\dfrac{10-2}{5-1}=2,\quad m_{\perp}=-\dfrac{1}{2}\)

\(\begin{aligned} y-6&=-\dfrac{1}{2}(x-3) \\ x+2y&=15 \end{aligned}\)