http://www.youtube.com/watch?v=r5hRGpIcE5M

Today’s Transformation Lesson???

A rotation is an isometry

Center of rotation- a fixed point in which a figure is turned about

Angle of Rotation- the angle formed from rays drawn from the center of rotation to a point and its image. Assume counterclockwise.

If Q is not the center of rotation , then QP=Q΄P and m∠QPQ΄= x˚

If Q is the center of rotation, then Q= Q’

9.4 Performing Rotations

R΄

Q΄

S΄Q

R

SPX˚

Draw a 120˚ rotation of ABC about P.

STEP 2

STEP 3

Draw A′ so that PA′ = PA.

Draw a ray to form a 120˚ angle with PA .

Draw a rotationSTEP 4

Repeat Steps 1– 3 for each vertex. Draw A′B′C′.

Preimage A(1,2), B(3,5), C (5,1)Discovery: Determine coordinate rules for the following counterclockwise rotations:

90˚Image A’( ),B’( ),C’( )Rule______________180˚Image A’( ),B’( ),C’( )Rule______________270˚ImageA’( ),B’( ),C’( )Rule_______________

These rules apply for counterclockwise rotations about the origin

a 90o rotation (a,b) (-b,a) a 180o rotation (a,b) (-a,-b) a 270o rotation (a,b) (b,-a)

Coordinate Rules for Rotations

Use to find the image of a line or polygon rotation about the origin.The rotation matrix must be first when multiplying 90˚Rotation Matrix

What are the Rotation Matrices?

= A’ B’ C’

Image Matrix

X

PreImage Matrix

90˚RotationMatrix

A B C

-2 -5 -1

1 3 5

1 3 5 2 5 1

180˚Rotation:

Rotation Matrices?

180˚rotation matrix

X1 3 5

2 5 1

Pre-Image matrix

A B C

=

A΄ B΄ C ΄

Image Matrix

-1 -3 -5 -2 -5 -1

270˚ Rotation:

Rotation Matrices?

270˚rotation matrix

X1 3 5

2 5 1

Pre-Image matrix

A B C

=

A΄ B΄ C΄

Image Matrix

2 -5 -1-1 3 5

Graph JKL with vertices J(3, 0), K(4, 3), andL(6, 0). Rotate the triangle 90° about the origin.

ANSWER

Rotate a figure 90º about the origin

Rotate a figure using the coordinate rules

Graph the image R′S′T′U′.

SOLUTION

(a, b) (b, –a) R(3, 1) R′(1, –3)

T(5, –3) T′(–3, –5) U(2, –1) U′(–1, –2)

S(5, 1) S′(1, –5)

Graph quadrilateral RSTU with vertices R(3, 1), S(5, 1), T(5, –3), and U(2, –1). Then rotate the quadrilateral 270 about the origin.

o

Graph RSTU. Use the coordinate rule for a 270 rotation to find the images of the vertices.

o

Use matrices to rotate a figure

SOLUTION

STEP 1

Write the polygon matrix:

Trapezoid EFGH has vertices E(–3, 2), F(–3, 4), G(1, 4), and H(2, 2). Find the image matrix for a 180 rotation of EFGH about the origin. Graph EFGH and its image.

o

STEP 2

Multiply by the matrix for a 180 rotation.o

Use matrices to rotate a figureSTEP 3

Graph the preimage EFGH.Graph the image E′F′G′H′.

Standardized Test Practice

SOLUTIONBy Theorem 9.3, the rotation is an isometry, so corresponding side lengths are equal. Then 2x = 6, so x = 3. Now set up an equation to solve for y.

Corresponding lengths in an isometry are equal.Substitute 3 for x.

Solve for y.

3x + 15y =5y = 3(3) + 1

=y 2The correct answer is B.

More Standardized Test Practice6. Find the value of r in the rotation of the triangle.

The correct answer is B.

ANSWER