Alyssia Agnitti
November 22, 2010
MAE 501
Notes for 11/17/10
In class on Monday, we started by going over number five from the homework which
was:
Students explained that a translation may be decomposed into a product of reflections
through two parallel lines. Explain how to decompose the translation 𝑇(3,4) (𝑥, 𝑦) into a product
of two reflections.
There were two different ways to decompose this translation which were shown in class.
The first way was to draw a line from (𝑥, 𝑦) to (𝑥 ′ , 𝑦 ′ ) and construct a perpendicular bisector in
order to find the midpoint, 𝑀, of the line. Then construct perpendicular bisectors for each
bisected line on either side of point 𝑀 in order to find the midpoints, 𝑃 and 𝐷, of these lines. The
perpendicular lines at points 𝑃 and 𝐷 form the two lines of reflection which is a product of the
translation 𝑇(3,4) (𝑥, 𝑦).
(𝑥 ′ , 𝑦 ′ )
𝐷
𝑀
𝑃
(𝑥, 𝑦)
The second way uses the slope of the line from (𝑥, 𝑦) to (𝑥 ′ , 𝑦 ′ ) in order to decompose
the translation into a product of two reflections. Since the slope of this line is 𝑚 = 4/3 then the
lines perpendicular to this one have the slope 𝑚 = −3/4. After using the distance formula, we
find that the length of the line from (𝑥, 𝑦) to (𝑥 ′ , 𝑦 ′ ) is 5 which makes the length between (𝑥, 𝑦)
and the midpoint, 𝑀, 2.5. Since we decided as a class that you can slide parallel lines as long as
the distance between the lines is preserved, then you can move lines, 𝑛 and 𝑙, in order to produce
the two line reflections which is a product of the translation 𝑇(3,4) (𝑥, 𝑦).
𝑛
(𝑥 ′ , 𝑦 ′ )
𝑙
2.5
𝑀
2.5
(𝑥, 𝑦)
Alyssia Agnitti
November 22, 2010
MAE 501
Notes for 11/17/10
Can you slide parallel lines the same way that you can rotate the angles of intersecting
lines? The answer is yes because while rotating intersecting lines, the angle matters and while
sliding parallel lines, the distance matters between the lines. Therefore, they are analogous
statements. This is why we did not have to give an equation to the example above because we
only needed to know the slope of line (𝑥, 𝑦) (𝑥 ′ , 𝑦 ′ ) and the distance between the parallel lines
that are perpendicular to the line (𝑥, 𝑦) (𝑥 ′ , 𝑦 ′ ).
In the previous class, we wanted to figure out how to find the line of a single reflection
without performing the three reflections through three parallel lines. Someone stated that in order
to find the line of the single reflection, you need to take the distance, 𝑥, from the line of the
second reflection, 𝑙2 , and the line of the third reflection, 𝑙3 , and measure that distance from the
line of the first reflection, 𝑙1. The order of the lines does not matter, but the order of reflections
matter. Therefore, 𝑙1, 𝑙2 , and 𝑙3 signifies the order of reflections in which the object crosses the
lines. This picture below will better illustrate where the line of the single reflection is located.
Line of single reflection
𝑥
𝑙3
𝑥
𝑙1
𝑙2
Here is a proof of this assumption which was shown in class.
𝑧
𝐴′′′
𝑥
(𝑥 − 𝑧)
Line of single reflection
𝑥
𝑥
(𝑥 − 𝑧)
𝑧
𝑙3
𝐴
𝐴′
𝐴′′
𝑙1
z
𝛼
𝛼
𝑙2
Alyssia Agnitti
November 22, 2010
MAE 501
Notes for 11/17/10
Proof:
Based on the picture, let 𝑑( 𝐴 𝐴′′′ ) = (𝑥 − 𝑧) + (𝑥 − 𝑧), 𝑑(𝐴𝐴′ ) = 2𝑧, and
′ ′′ )
𝑑(𝐴 𝐴 = 2𝛼 where d denotes the distance between the given two points. In order to prove our
assumption of the distances between the lines of reflection, then we have to show the order of
reflections of point A and their distances from the lines of reflections. If we prove that the
equation [𝑑( 𝐴 𝐴′ ) + 𝑑(𝐴′ 𝐴′′ ) + 𝑑(𝐴′′ 𝐴′′′ )] − [𝑑(𝐴𝐴′ ) + 𝑑(𝐴′ 𝐴′′ )] = 𝑑( 𝐴 𝐴′′′ ) = 2𝑥 − 2𝑧 is
true, which is shown in the picture, after the three reflections, then we have proved where the
line of single reflection is located. Since the 𝑑(𝐴′′ 𝐴′′′ ) counts certain distances twice, then we
have to subtract off the extra distances to produce the equation of the 𝑑( 𝐴 𝐴′′′ ). So, we have
[𝑑( 𝐴 𝐴′ ) + 𝑑(𝐴′ 𝐴′′ ) + 𝑑(𝐴′′ 𝐴′′′ )] − [𝑑(𝐴𝐴′ ) + 𝑑(𝐴′ 𝐴′′ )] = 2𝑧 + 2𝛼 + (𝑥 − 𝑧) + (𝑥 − 𝑧) −
2𝑧 − 2𝛼 = 2𝑥 − 2𝑧. Therefore, 𝑑( 𝐴 𝐴′′′ ) = 2𝑥 − 2𝑧. This means that we found the correct line
of the single reflection.
In class, someone conjectured that if two isometries are equivalent on three non-collinear
points then they are equivalent. Here is a proof from class of this statement.
Proof of Conjecture: First of all, since any isometry can be decomposed into a product of
reflections as we proved in previous classes, then the composition of reflections is equivalent to
an isometry.
Additionally, take points 𝐴, 𝐵, 𝑎𝑛𝑑 𝐶 and the image of these points
′
′
′
𝐴 , 𝐵 , 𝑎𝑛𝑑 𝐶 . Let 𝑋 be another point and the distances between 𝐴, 𝐵, 𝑜𝑟 𝐶 and point 𝑋 be the
radii of each constructed circle in the image. Since two of the circles intersect at two different
points, we need a third circle in order to find point 𝑋 ′ . Therefore, the conjecture is true.
(Not to scale)
C
A
Distance of the
radius
𝐴′
𝐶′
𝑋′
𝐵′
B
X
A definition for congruence is: two objects are congruent if we can take one onto another
via an isometry. Therefore, prove how the congruence triangle theorems, SSS, SAS, AAS, and
ASA, hold. For example, if we assume that there exists an isometry that takes AB to 𝐴′ 𝐵′, BC to
𝐵′𝐶′, and CA to 𝐶′𝐴′, then we want to show that there exists an isometry that takes
∆𝐴𝐵𝐶 𝑡𝑜 ∆𝐴′𝐵′𝐶′.