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Warm Up Determine the coordinates of the image of P (-3, 5) under each transformation. - PowerPoint PPT Presentation


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Warm Up Determine the coordinates of the image of P (-3, 5) under each transformation. . 1. T <2, -4> (P). (-1 , 1 ). 2. r (270 ,O) (P). (5, 3). 3. R y = 2 (P). (–3, –1). 9.6 Composition of Isometries. SWBATU how identify and draw compositions of transformations.

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Presentation Transcript
slide1

Warm Up

Determine the coordinates of the image of P(-3, 5) under each transformation.

1. T<2, -4>(P)

(-1, 1)

2. r(270,O) (P)

(5, 3)

3. Ry = 2 (P)

(–3, –1)

slide2

9.6 Composition of Isometries

SWBATU how identify and draw compositions of transformations.

slide3

9.4 – Key vocabulary

A composition of transformations is one transformation followed by another.

(T<3,-2>  Ry-axis) (ABCD)

Important Note: read right to left

This would take the quadrilateral ABCD and reflect over the

y-axisand 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.

slide4

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

slide5

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

slide6

K

L

M

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:

Ry-axis  r(180,O) (KLM)

slide7

M’

M”

L’

L”

K”

K’

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).

slide8

L

J

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.

Notation:

r(180,O) Rx-axis (JKL)

slide9

J(1, –2) J’(1, 2), K(4, –2) K’(4, 2), and L(3, 0) L’(3, 0).

J’(1, 2) J”(-1, -2),

K’(4, 2) K”(-4, -2), and

L’(3, 0) L”(-3, 0).

L

J

K

Example 2 Continued

Step 1 The reflection image of (x, y) is (x, -y).

J’

K’

Step 2 The rotational image of (x, y) is (–x, –y).

L'’

L’

J’’

K’’

slide11

L”

M”

L

M

M’

L’

P”

N”

P

N

N’

P’

Example 3

Find the image of the shape after the transformation

translation:

LMNP L”M”N”P”

slide12

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)