# Warm Up Determine the coordinates of the image of P(-3, 5) under each transformation. 1. T (P) 2. r (270,O) (P) (-1, 1) (5, 3) 3. R y = 2 (P)(–3, –1)

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Warm Up Determine the coordinates of the image of P(-3, 5) under each transformation. 1. T (P) 2. r (270,O) (P) (-1, 1) (5, 3) 3. R y = 2 (P)(–3, –1)

SWBATU how identify and draw compositions of transformations. 9.6 Composition of Isometries

A composition of transformations is one transformation followed by another. (T  R y-axis ) (ABCD) Important Note: read right to left This would take the quadrilateral ABCD and reflect over the y-axis and then translate that result right 3 and down 2. a glide reflection is the composition of a translation and a reflection across a line parallel to the translation vector. The combination above is not a glide reflection. 9.4 – Key vocabulary

The glide reflection that maps ∆JKL to ∆J’K’L’ is the composition of a translation along followed by a reflection across line l.

The image after each transformation is congruent to the previous image. By the Transitive Property of Congruence, the final image is congruent to the preimage. This leads to the following theorem. Theorem

Example 1: Drawing Compositions of Isometries Draw the result of the composition of Isometries. ∆KLM has vertices K(4, –1), L(5, –2), and M(1, –4). Rotate ∆KLM 180° about the origin and then reflect it across the y-axis. Notation: R y-axis  r (180,O) (KLM) K L M

Example 1 Continued Step 1 The rotational image of (x, y) is (–x, –y). K(4, –1)  K’(–4, 1), L(5, –2)  L’(–5, 2), and M(1, –4)  M’(–1, 4). Step 2 The reflection image of (x, y) is (–x, y). K’(–4, 1)  K”(4, 1), L’(–5, 2)  L”(5, 2), and M’(–1, 4)  M”(1, 4). K L M M’ K’ L’ L” M” K”

Example 2 ∆JKL has vertices J(1,–2), K(4, –2), and L(3, 0). Reflect ∆JKL across the x-axis and then rotate it 180° about the origin. L KJ Notation: r (180,O)  R x-axis (JKL)

L KJ L'’ J’’ K’’ K’ J’ L’ Example 2 Continued Step 2 The rotational image of (x, y) is (–x, –y). J’(1, 2)J”(-1, -2), K’(4, 2)K”(-4, -2), and L’(3, 0)L”(-3, 0). Step 1 The reflection image of (x, y) is (x, -y). J(1, –2) J’(1, 2), K(4, –2) K’(4, 2), and L(3, 0) L’(3, 0).

Theorem

Example 3 Find the image of the shape after the transformation LMNP  L”M”N”P”translation: LM PN L’ M’ P’ N’ L”M” P”N”

You Try PQR has vertices P(5, –2), Q(1, –4), and R(–3, 3). P”(3, 1), Q”(–1, 3), R”(–5, –4) P”(5, 2), Q”(1, 4), R”(-3, -3)

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