Perpendicular bisector

In a plane, the perpendicular bisector of a line segment $AB$ is a line $l$ such that $AB$ and $l$ are perpendicular and $l$ passes through the midpoint of $AB$ .

In 3-D space, for each plane passing through $AB$ that is not normal to $AB$ there is a distinct perpendicular bisector. The set of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting $AB$ .

In a triangle, the perpendicular bisectors of all three sides intersect at the circumcenter.

Locus

The perpendicular bisector of $\displaystyle AB$ is also the locus of points equidistant from $\displaystyle A$ and $\displaystyle B$ .

To prove this, we must prove that every point on the perpendicular bisector is equidistant from $\displaystyle A$ and $\displaystyle B$ , and also that every point equidistant from $\displaystyle A$ and $\displaystyle B$ .

The first part we prove as follows: Let $\displaystyle P$ be a point on the perpendicular bisector of $\displaystyle AB$ , and let $\displaystyle M$ be the midpoint of $\displaystyle AB$ . Then we observe that the (possibly degenerate) triangles $\displaystyle APM$ and $\displaystyle BPM$ are congruent, by side-angle-side congruence. Hence the segments $\displaystyle PA$ and $\displaystyle PB$ are congruent, meaning that $\displaystyle P$ is equidistant from $\displaystyle A$ and $\displaystyle B$ .

To prove the second part, we let $\displaystyle P$ be any point equidistant from $\displaystyle A$ and $\displaystyle B$ , and we let $\displaystyle M$ be the midpoint of the segment $\displaystyle AB$ . If $\displaystyle P$ and $\displaystyle M$ are the same point, then we are done. If $\displaystyle P$ and $\displaystyle M$ are not the same point, then we observe that the triangles $\displaystyle PAM$ and $\displaystyle PBM$ are congruent by side-side-side congruence, so the angles $\displaystyle PAM$ and $\displaystyle PBM$ are congruent. Since these angles are supplementary angles, each of them must be a right angle. Hence $\displaystyle PM$ is the perpendicular bisector of $\displaystyle AB$ , and we are done.

Page

Toolbox

Search

Perpendicular bisector

Locus