Perpendicular bisector of a given segment

Example

Given the segment $AB$ with $A(-2;-4)$ and $B(4;1)$, write its perpendicular bisector.

Locus condition: $\sqrt{(x+2)^2 + (y+4)^2} = \sqrt{(x-4)^2 + (y-1)^2}.$ Both sides are distances, hence $\ge 0$ “by construction”: no existence conditions are needed and one can square. $(x+2)^2 + (y+4)^2 = (x-4)^2 + (y-1)^2.$ Expanding the squares, the $x^2$ and $y^2$ terms cancel (this is why the locus is a line!) and there remains: $4x + 4 + 8y + 16 = -8x + 16 - 2y + 1 \implies 12x + 10y + 3 = 0.$ Written in explicit form: $\boxed{y = -\tfrac{6}{5}x - \tfrac{3}{10}}$.

The segment $AB$, its midpoint $M$ and the perpendicular bisector (blue line), perpendicular to $AB$ and passing through $M$.

Links

Topics: Loci
Concepts: Perpendicular bisector of a segment · Distance between points · Equation of the line · Cartesian plane · Line
Skills: Analytic geometry · Solving equations