RGpIcE5M 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’ R΄R΄ Q΄Q΄ S΄S΄ Q R S P X˚X˚

Draw a 120˚ rotation of ABC about P. STEP 2 STEP 3Draw A′ so that PA′ = PA. Draw a ray to form a 120˚ angle with PA.

Draw a rotation STEP 4Repeat 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˚Image A’( ),B’( ),C’( ) Rule_______________

 These rules apply for counterclockwise rotations about the origin  a 90 o rotation (a,b) (-b,a)  a 180 o rotation (a,b) (-a,-b)  a 270 o rotation (a,b) (b,-a)

What are the Rotation Matrices? = A’ B’ C’ Image Matrix X PreImage Matrix 90˚Rotation Matrix A B C

180˚Rotation: Rotation Matrices? 180˚rotation matrix X Pre-Image matrix A B C = A΄ B΄ C ΄ Image Matrix

270˚ Rotation: Rotation Matrices? 270˚rotation matrix X Pre-Image matrix A B C = A΄ B΄ C΄ Image Matrix

Graph JKL with vertices J(3, 0), K(4, 3), and L(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 1Write 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 2Multiply by the matrix for a 180 rotation. o

Use matrices to rotate a figure STEP 3Graph the preimage EFGH. Graph the image E ′ F ′ G ′ H ′.

Standardized Test Practice SOLUTION By 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 + 15y5y = 5y5y = 3(3) + 1 = y2 The correct answer is B.

More Standardized Test Practice 6. Find the value of r in the rotation of the triangle. The correct answer is B. ANSWER