Editing Bisection (section)

== Perpendicular line segment bisector ==
{{anchor|Line segment bisector}}

=== Definition ===
[[File:Mittelsenkr-ab-e.svg|thumb|upright=0.8|Perpendicular bisector of a line segment]]
*The [[perpendicular]] bisector of a line segment is a line which meets the segment at its [[midpoint]] perpendicularly.
*The perpendicular bisector of a line segment <math>AB</math> also has the property that each of its points <math>X</math> is [[equidistant]] from segment AB's endpoints:
'''(D)'''<math>\quad |XA| = |XB|</math>.
 
The proof follows from <math>|MA|=|MB|</math> and [[Pythagoras' theorem]]:
:<math>|XA|^2=|XM|^2+|MA|^2=|XM|^2+|MB|^2=|XB|^2 \; .</math>

Property '''(D)''' is usually used for the construction of a perpendicular bisector:

=== Construction by straight edge and compass ===
[[File:Mittelsenkr-ab-konstr-e.svg|thumb|upright=0.8|Construction by straight edge and compass]]
In classical geometry, the bisection is a simple [[compass and straightedge construction]], whose possibility depends on the ability to draw [[arc (geometry)|arc]]s of equal radii and different centers:

The segment <math>AB</math> is bisected by drawing intersecting circles of equal radius <math>r>\tfrac 1 2 |AB|</math>, whose centers are the endpoints of the segment. The line determined by the points of intersection of the two circles is the perpendicular bisector of the segment.<br> 
Because the construction of the bisector is done without the knowledge of the segment's midpoint <math>M</math>, the construction is used for determining <math>M</math> as the intersection of the bisector and the line segment.

This construction is in fact used when constructing a ''line perpendicular to a given line'' <math>g</math> at a ''given point'' <math>P</math>: drawing a circle whose center is <math>P</math> such that it intersects the line <math>g</math> in two points <math>A,B</math>, and the perpendicular to be constructed is the one bisecting segment <math>AB</math>.

=== Equations ===
If <math>\vec a,\vec b</math> are the position vectors of two points <math>A,B</math>, then its midpoint is <math>M: \vec m=\tfrac{\vec a+\vec b}{2}</math> and vector <math>\vec a -\vec b</math> is a [[normal vector]] of the perpendicular line segment bisector. Hence its vector equation is <math>(\vec x-\vec m)\cdot(\vec a-\vec b)=0</math>. Inserting  <math>\vec m =\cdots</math> and expanding the equation leads to the vector equation

'''(V)''' <math>\quad \vec x\cdot(\vec a-\vec b)=\tfrac 1 2 (\vec a^2-\vec b^2) .</math>

With <math>A=(a_1,a_2),B=(b_1,b_2)</math> one gets the equation in coordinate form:

'''(C)''' <math>\quad (a_1-b_1)x+(a_2-b_2)y=\tfrac 1 2 (a_1^2-b_1^2+a_2^2-b_2^2) \; .</math>

Or explicitly:<br>
'''(E)'''<math>\quad y = m(x - x_0)  +y_0</math>, <br>
where <math>\; m = - \tfrac{b_1 - a_1}{b_2 - a_2}</math>, <math>\;x_0 = \tfrac{1}{2}(a_1 + b_1)\;</math>, and <math>\;y_0 = \tfrac{1}{2}(a_2 + b_2)\;</math>.

=== Applications ===
Perpendicular line segment bisectors were used solving various geometric problems:
#Construction of the center of a [[Thales's theorem|Thales' circle]],
#Construction of the center of the [[Excircle]] of a triangle,
#[[Voronoi diagram]] boundaries consist of segments of such lines or planes.

[[File:Mittelloteb-ab-3d-e.svg|thumb|upright=0.8|Bisector plane]]

=== Perpendicular line segment bisectors in space ===
*The [[perpendicular]] bisector of a line segment is a ''plane'', which meets the segment at its [[midpoint]] perpendicularly. 
Its vector equation is literally the same as in the plane case:

'''(V)''' <math>\quad \vec x\cdot(\vec a-\vec b)=\tfrac 1 2 (\vec a^2-\vec b^2) .</math>

With <math>A=(a_1,a_2,a_3),B=(b_1,b_2,b_3)</math> one gets the equation in coordinate form:

'''(C3)''' <math>\quad (a_1-b_1)x+(a_2-b_2)y+(a_3-b_3)z=\tfrac 1 2 (a_1^2-b_1^2+a_2^2-b_2^2+a_3^2-b_3^2) \; .</math>

Property '''(D)''' (see above) is literally true in space, too:<br>
'''(D)''' The perpendicular bisector plane of a segment <math>AB</math> has for any point <math>X</math> the property: <math>\;|XA| = |XB|</math>.