WGU C972 SDM1 Task 4 Passed (2026) – Counterexamples | College Geometry

C972
College Geometry
SDM1 TASK 4 – Passed
Counterexamples
Western Governors University

, SDM1 Task 4 – Counterexamples

A. A composi)on of transforma)ons is used to prove that the result of performing two
isometries is not always commuta)ve.

The original composi)on of transforma)ons for (𝑷 ∘ 𝑸)&𝑹𝑶,𝟗𝟎° ∘ 𝑻𝟒,'𝟏 ).
Preimage ∆𝐴𝐵𝐶 was graphed at:
A=(3,4) B=(3,8) C=(7,4)

Let Isometry 𝑃 = 𝑅),*+° , which is a counterclockwise rota)on by 90° about point 𝐹(0,0).
Let Isometry 𝑄 = 𝑇,,'- , is a transla)on of 4 units to the right and 1 unit down.

When we have two isometries, the transforma)on is done from right to leD. Therefore, for
the original composi)on of transforma)ons (𝑷 ∘ 𝑸)&𝑹𝑶,𝟗𝟎° ∘ 𝑻𝟒,'𝟏 ), the isometry, Q,
transla)on is evaluated first, and the image is then rotated, isometry P.




For 𝑃 ∘ 𝑄, the transla)on of ∆𝐴𝐵𝐶: 𝐴(3,4) 𝐵(3,8) 𝐶(7,4) is 4 units to the right and 1 unit
down. The result is ∆𝐴. 𝐵. 𝐶 . : 𝐴. (7,3) 𝐵. (7,7) 𝐶′(11,3). ∆𝐴. 𝐵. 𝐶 . : 𝐴. (7,3) 𝐵. (7,7) 𝐶′(11,3)




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