Perpendicular bisector

Find the perpendicular bisector of the segment $AB$ with $A(-1, 0)$ and $B(3, 2)$¹.

From the midpoint and the slope

Passing through $M(1, 1)$, the line is $y - 1 = -2(x - 1)$, that is $y = -2x + 3$.

From the equidistance condition

What the line really is, though, is the set of points equidistant from $A$ and $B$, and that description can produce the equation on its own. If $(x, y)$ is equidistant from $A$ and $B$, then the following holds.

Squaring both sides and expanding, the $x^2$ and $y^2$ terms cancel and only a linear equation survives.

This is exactly the line found from the midpoint and the perpendicular slope. It is because the squared terms cancel that a condition about distances turns out to describe a straight line.

Checking at the midpoint

From $M(1, 1)$ the distance to $A(-1, 0)$ is $\sqrt{2^2 + 1^2} = \sqrt{5}$, and to $B(3, 2)$ it is $\sqrt{2^2 + 1^2} = \sqrt{5}$ as well. The midpoint is the one point of the bisector that lies on the segment itself.

Applications

Reading the line as a set of equidistant points opens the door to its applications.

• The three perpendicular bisectors of a triangle meet at the circumcenter
• The boundary between two nearest-shop territories on a map, that is a Voronoi diagram²
• Constructing the center of a circle through three given points

The circumcenter works precisely because that point is equidistant from all three vertices.

Changing the ratio

Ask instead for the points whose distances to $A$ and $B$ stand in the ratio $2 : 1$ rather than $1 : 1$, and the locus is not a line at all but a circle, the circle of Apollonius³.

The squared terms cancel, leaving a straight line, only in the equidistant case.

The large dots mark $A$, $B$ and the midpoint $M$.

1. Bisection, Wikipedia
2. Voronoi diagram, Wikipedia
3. Circles of Apollonius, Wikipedia