Straight-Line Graphs · 0606 Topic 7
Perpendicular Bisector
Written by Teacher Rig
8 years teaching IGCSE Add Math · Updated 12 June 2026
The perpendicular bisector of a segment AB passes through its midpoint at right angles. It’s the most-assembled composite in coordinate geometry: three ingredients, three marking points, one fixed order.
The routine
Find the perpendicular bisector of , .
- Midpoint: , formula (B/M)
- Gradient of : perpendicular gradient , the negative reciprocal, stated (M)
- Line through the midpoint: , point-gradient build (A)
Three visible ingredients; cramming them into one unexplained line risks all three marks on a single slip. The classic self-sabotage is using point or in step 3 instead of the midpoint, the bisector passes through neither endpoint.
Why the bisector matters: it’s an equidistance machine
Every point on the perpendicular bisector is equidistant from and . That property is what the harder questions actually use:
- “Find the point on the -axis equidistant from and ” → find the bisector, set
- ” is equidistant from and ; lies on ; find ” → bisector line, a simultaneous solve
- Circle geometry: the centre of a circle is equidistant from every chord’s endpoints, so the centre lies on the perpendicular bisector of any chord, the standard route to finding centres in the new circle topic. Two chords → two bisectors → intersect → centre.
If a question mentions equidistance from two fixed points, the perpendicular bisector is the method, whether or not the words appear.
Common mistakes
- Built through an endpoint instead of the midpoint
- Gradient of used directly (the bisector is perpendicular, flip and negate)
- Midpoint computed with differences instead of sums
- Final answer not in the demanded form (integer coefficients etc.)
- The equidistance property unrecognised, so the harder variants never start
Full topic context: Straight-Line Graphs notes.