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Midpoint proof

by Imparcticle
Tags: midpoint, proof
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Imparcticle
#1
Jul18-04, 01:03 AM
P: 566
How would you right a proof for this theorem: "If a segment is given, then it has exactly one midpoint"?
Please note that the numbering of the postulates (P) is based on my geometry book. Also, I'm just in 9th grade geometry, so please don't use differential equations or some other math than basic Euclidean geomtery. This will help me better understand the concept.

This is what I did so far:
Let points A and B be the end points of a line segment, AB.

By P2-3 (For any two points on a line, and a given unit of measure, there is a unique positive number called the measure of the distance between the two points) there is a defined distance between the points A and B on line segment AB

By the "definition of between", a point Q is between points A and B if and only if each of the following conditions hold.
1.) A, Q, B are collinear.
2.) AQ + BQ = AB

Condition 1: A,Q,B are collinear
By P1-1 ( Through any two points there is exactly one line) and P1-3 (There are at least two points on a line), points A, Q and B are on the same line. Therefore, by the definition of collinear (which states that points are collinear if and only if they are on the same line), A,Q and B are collinear.

Condition 2: AQ + BQ = AB, Let Q be the midpoint
Since Q is the only common point of segments AQ and BQ, then AQ and BQ both intersect at Q. Therefore, the end points are A and B. Since A, Q and B are all collinear, they form a line segment AB. Thus Q is in between points A and B in AB.
This is where I kind of get lost....


AQ + BQ = AB
Let A=2, B=8, Q=5
l 2-5 l + l 8-5 l = l 2-8 l
6=6


By the "definition of midpoint", a point Q is the midpoint of a segment AB if and only if Q is between A and B and AQ=BQ.
Does that prove Q is the midpoint? I don't think so because:
1.) I haven't shown that the arithmetic I did in an attempt to show the distances between AQ and BQ are equal to that of AB works for all cases.

2.)I'm not sure I have adequately proven that A, Q and B are on the same line.
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Imparcticle
#2
Jul18-04, 01:08 AM
P: 566
Here is the proof my math tutor wrote:

Let AB be a segment.
(show that AB has exactly one midpoint)

Let M and N be midpoints of AB.

Since M, N are midpoints of AB, M and N are between A and B and AM=MB. Furthermore, M, A and B are collinear and AM + MB = AB
AN + NB = AB
AM + MB = AN + NB
AM + AM = AN + AN (i.e. 2 * AN -> AM = AN

Therefore M and N are the same point.
zeronem
#3
Jul18-04, 01:20 AM
P: 100
Midpoint - Point B is the midpoint of the segment AC if it is between A and C and if AB=BC(that is, the distance from B to A is the same distance from B to C).

Simply this

A---B---C.

B is located at the midpoint. There is equal distance between A and B, as well there is equal distance from B to C. This shows that B is the midpoint of AC. You also notice that B is located halfway in between AC. AC/2 = B.

Find the midpoint of the points in cartesian coordinates: (2,4) and (-7, 8).

The Midpoint of points (2,4) and (-7, 8) would be the point in cartesian coordinates- ((-7 + 2)/2 , (8 + 4)/2) = (-5/2 , 6)

loseyourname
#4
Jul18-04, 01:25 AM
Emeritus
PF Gold
loseyourname's Avatar
P: 3,634
Midpoint proof

For the sake of simplicity, let [itex]AB[/itex] be a line segment on a two dimensional coordinate grid. [itex]A = (x_{1}, y_{1})[/itex] and [itex]B = (x_{2}, y_{2})[/itex]. The length [itex]l[/itex] of line segment [itex]AB[/itex] = [itex]\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}[/itex] by the Pythagorean theorem. A midpoint, by definition, is of half the length of the line segment, so we will define [itex]AQ[/itex] by point [itex]A[/itex] given previously and point [itex]Q = (x_{3}, y_{3})[/itex] such that the length

[tex]\sqrt{(x_{3} - x_{1})^2 + (y_{3} - y_{1})^2} = \sqrt{(x_{2} - x_{3})^2 + (y_{2} - y_{3})^2} = l/2[/tex]

Pick any numbers to fill in these values, then use induction to finish this part of the proof.

I think you adequately proved that the points are on the same line.
Imparcticle
#5
Jul18-04, 01:34 AM
P: 566
What is the difference between the application of the Pythagorean theorem to this proof and using the midpoint formula? I thought about using the midpoint formula but then thought against it.

You guys are just brilliant. Thanks.
loseyourname
#6
Jul18-04, 01:37 AM
Emeritus
PF Gold
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P: 3,634
Midpoint formula follows from the Pythagoream theorem. Either term would work.
Didd
#7
Jul18-04, 04:17 AM
P: 28
"By the "definition of between", a point Q is between points A and B if and only if each of the following conditions hold.
1.) A, Q, B are collinear.
2.) AQ + BQ = AB"

Incorrect! You must also say that |AQ|=|BQ|. The statment you wrote have no gurantee to be Q the mid-point. Find out the reason.
loseyourname
#8
Jul18-04, 03:54 PM
Emeritus
PF Gold
loseyourname's Avatar
P: 3,634
She knows that. That part of the proof was only meant to show that Q is colinear with A and B as well as somewhere between A and B, not necessarily that it is the midpoint.


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