Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Perpendicular bisector"

m
(Added stuff about locus)
Line 1: Line 1:
 
In a [[plane]], the '''perpendicular bisector''' of a [[line segment]] <math>AB</math> is a [[line]] <math>l</math> such that <math>AB</math> and <math>l</math> are [[perpendicular]] and <math>l</math> passes through the [[midpoint]] of <math>AB</math>.
 
In a [[plane]], the '''perpendicular bisector''' of a [[line segment]] <math>AB</math> is a [[line]] <math>l</math> such that <math>AB</math> and <math>l</math> are [[perpendicular]] and <math>l</math> passes through the [[midpoint]] of <math>AB</math>.
  
−
In 3-D space, for each plane passing through <math>AB</math> there is a distinct perpendicular bisector.  The [[set]] of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting <math>AB</math>.
+
In 3-D space, for each plane passing through <math>AB</math> that is not [[normal]] to <math>AB</math> there is a distinct perpendicular bisector.  The [[set]] of lines which are perpendicular bisectors of form a plane which is the plane perpendicularly bisecting <math>AB</math>.
  
 
In a [[triangle]], the perpendicular bisectors of all three sides intersect at the [[circumcenter]].
 
In a [[triangle]], the perpendicular bisectors of all three sides intersect at the [[circumcenter]].
 +
 +
== Locus ==
 +
 +
The perpendicular bisector of <math>\displaystyle AB</math> is also the locus of points [[equidistant]] from <math>\displaystyle A</math> and <math>\displaystyle B</math>.
 +
 +
To prove this, we must prove that every point on the perpendicular bisector is equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>, and also that every point equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>.
 +
 +
The first part we prove as follows: Let <math>\displaystyle P</math> be a point on the perpendicular bisector of <math>\displaystyle AB</math>, and let <math>\displaystyle M</math> be the midpoint of <math>\displaystyle AB</math>.  Then we observe that the (possibly [[degenerate]]) [[triangles]] <math>\displaystyle APM</math> and <math>\displaystyle BPM</math> are [[congruent (geometry) | congruent]], by [[side-angle-side congruence]].  Hence the segments <math>\displaystyle PA</math> and <math>\displaystyle PB</math> are congruent, meaning that <math>\displaystyle P</math> is equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>.
 +
 +
To prove the second part, we let <math>\displaystyle P</math> be any point equidistant from <math>\displaystyle A</math> and <math>\displaystyle B</math>, and we let <math>\displaystyle M</math> be the midpoint of the segment <math>\displaystyle AB</math>.  If <math>\displaystyle P</math> and <math>\displaystyle M</math> are the same point, then we are done.  If <math>\displaystyle P</math> and <math>\displaystyle M</math> are not the same point, then we observe that the triangles <math>\displaystyle PAM</math> and <math>\displaystyle PBM</math> are congruent by [[side-side-side congruence]], so the angles <math>\displaystyle PAM</math> and <math>\displaystyle PBM</math> are congruent.  Since these angles are [[supplementary angles]], each of them must be a [[right angle]].  Hence <math>\displaystyle PM</math> is the perpendicular bisector of <math>\displaystyle AB</math>, and we are done.
  
 
[[Category:Definition]]
 
[[Category:Definition]]

Revision as of 18:13, 25 August 2006

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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Perpendicular_bisector&oldid=9890"