Rotate the triangle 180 degrees clockwise around center C.

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1 Math 350 Section 5.1 Answers to lasswork lasswork 1: erform the rotations indicated below: Results in bold: Rotate the triangle 90 degrees clockwise around center. Rotate the triangle 180 degrees clockwise around center. Rotate the triangle 70 degrees clockwise around center. lasswork : To test your ability to perform a reflection, create the reflection F(E) of the figure E at above right across the diagonal mirror line shown. Mirror Line, or Line of Reflection lasswork 3: In the figure at right locate the image F(A) of the point A across the mirror line. (1) onstruct the perpendicular to the mirror line that passes through A. () A is the point on this perpendicular that is the same distance from the mirror line as A is. A lasswork : In each of the geoboards below reflect the triangle across M. o this by using the above procedure to reflect each vertex of the triangle, and the join the reflected vertices to create the reflected triangle. (Note that it is possible for the mirror line to pass through the figure being reflected.) M M M

2 lasswork 5: Translate the triangle at right 3 units to the right and units down. lasswork 6: For each pair of figures, find a rigid motion that moves one on top of the other. If the answer is a rotation, locate the center and state the amount of rotation. If the answer is a reflection, sketch the mirror line. If the answer is a translation, give the number of units moved in each direction. Reflection; mirror line is shown Rotation 90 degrees about Translation right up 1 Glide Reflection: Translate right units, then reflect across the mirror line shown lasswork 8: Here are four congruent copies of a triangle, W, X, Y, and Z. Which pairs have the same orientation? W and X have the same, Y and Z have the same. Z Which pairs have opposite orientation? W and Y, W and Z, X and Y, X and Z. W X Y

3 lasswork 7 Shown below are two pairs of congruent figures. For each, decide which type of single rigid motion will map one figure onto the other: If there is a rotation that maps one onto the other, describe a procedure for precisely locating the center of the rotation, and estimate the amount of rotation. If there is a reflection that maps one onto the other, describe a procedure for precisely locating the mirror line of the reflection. (a) Following the procedure on the top of page 100, we draw segments AA', BB', and '. These segments are parallel and not of equal length. If we construct the perpendicular bisector of each (most easily done by folding) we find they are the same, shown as EF in the drawing. So the motion is the reflection that uses EF as the mirror line. E B A A B F (b) Following the procedure on the top of page 100, we draw segments AA', BB', ', and '. These are not all parallel and are of different lengths, indicating a possible rotation. If we construct the perpendicular bisectors of these segments (most easily done by folding) we see that they have a common intersection point, shown as E in the diagram. This is the center of the rotation. The amount of the rotation about this center is the measure of AEA', about 10 degrees. B A E B A

4 lasswork 9: In the coordinate system below, let T be rotation 90 degrees clockwise about the origin, and let S be reflection across the vertical axis. If is the triangle shown, with vertices (1, ), (1, 3), and (, ), determine and then S(): Answer: Let A = (1, ), B = (1, 3), and = (, ), Rotating these 90 degrees clockwise about the origin, we obtain A = T(A) = (, -1), B = T(B) = (3, -1), and = T() = (, -). onnecting these gives. Reflecting across the vertical axis gives the vertices A = S(, -1) = (, -1), B = S(3, -1) = (-3, -1), and = S(, -) = (, -). lotting these images and connecting them gives the triangle S() shown at right. S() - What type of rigid motion is S T? escribe it completely. That is: if it is a rotation give the coordinates of the center and the amount of rotation. If it is a translation give the vector of the translation in (x, y) form. If it is a reflection give the equation of the mirror line. Answer: onnect each vertex of triangle with its image in the triangle S(), as shown in the diagram at right. These segments are parallel, so the composition S T is a translation or reflection. Since the segments are not all the same length S T cannot be a translation, so it is a reflection. Notice that all the segments have the same perpendicular bisector; this must be the mirror line of the reflection. The mirror line is y = -x. From the above we infer that: In this case the composition of a rotation and a reflection was a reflection. (This is true in general if the center of the rotation is on the mirror line of the reflection; if not the composition is a glide reflection.) Mirror line S() -

5 lasswork 10: In the coordinate system below, let T be rotation 90 degrees clockwise about the origin, and let S be rotation 180 degrees about point with coordinates (0, ). If is the triangle shown, with vertices (1, ), (1, 3), and (, ), determine and then S(). Answer: The vertices of triangle are (1, ), (1, 3), and (, ). T(1, ) = (, -1), T(1, 3) = (3, -1), and T(, ) = (, -). lotting these images and connecting them gives the triangle shown. The center (0, ) of the rotation S is shown as a dot in the diagram. Applying a 180 degree rotation about his center gives S(, -1) = (, -3), S(3, -1) = (-3, -3), and S(, ) = (, 0). lotting these images and connecting them gives the triangle S() shown. What type of rigid motion is S T? Give as many details as you can. What type of rigid motion is S T? escribe it completely. That is: if it is a rotation give the coordinates of the center and the amount of rotation. If it is a translation give the vector of the translation in (x, y) form. If it is a reflection give the equation of the mirror line. Answer: onnect each vertex of triangle with its image in the triangle S(), as shown in the diagram at right. These segments are obviously not parallel, so the composition S T cannot be a translation or reflection. To see that S T is a rotation, construct the perpendicular bisector of each segment, as shown at right. (See onstruction 1 in section 1.3.) S() These perpendicular bisectors all pass through the point labeled with coordinates (, ). This point is the center of a rotation that takes - each vertex of to the corresponding vertex of S(). You can measure that the angle of rotation is 90 degrees clockwise about (or 70 degrees counterclockwise). From the above we infer that: In this case the composition of a two rotations was a rotation. Another method to locate the center of rotation: One of the segments is from (-3, -3) to (1, 3); it has slope 6/ = 3/ and midpoint (-1, 0). So the perpendicular bisector has slope /3 and goes through (-1, 0). This means its equation is y (0) = /3(x + 1), or y = /3x /3. Another segment is from (, 0) to (, ); it has slope /6 = 1/3 and has midpoint (1, 1). So the perpendicular bisector has slope 3 and goes through (1,1). Its equation is y 1 = -3(x 1), or y = -3x +. Solving the two equations y = /3x /3 and y = -3x + simultaneously gives (, ). S() -

6 lasswork 11 Suppose that in the diagram below F 1 is reflection across mirror line L1 and F is reflection across mirror line L. an we describe the composition of F 1 and F completely? In the diagram F 1 (ΔAB) = ΔA' B' ' and F (ΔA' B' ') = ΔA"B"". Also is the intersection of the two mirror lines. 1. Explain why segments A and A' are congruent, and why the angles they make with L1 are congruent. By the definition of reflection, L1 L1 is the perpendicular bisector of AA' ( AE AE ' ). Since is on L1, then by Theorem 1.9 A A'. Also E is a shared side, so ΔAE ΔAE ' by SSS. By corresponding parts, AE A' E A E B B' '. onstruct segment A". For the same reasons as in 1, segments A' and A" are congruent, and the angles they make with L. are congruent. Thus segments A and A" are congruent. Why does this mean that is on the perpendicular bisector of AA", and the angles between L and A' and A" are congruent? is on the perpendicular bisector of AA" since is equidistant from A and A ( A and A" are congruent). As in 1, ΔA' F ΔA"F by SSS, and by corresponding parts, A' F A" F A' A'' F B'' '' L 3. How does the measure of AA" compare to the measure of the angle between the two mirror lines L1 and L? Why does this mean that F 1 F is a rotation? Justify your conclusion. The measure of the angle between L1 and L = m( EA') + m( A' F). Also m( AA") = m( EA') + m( A' F) = m(angle between L1 and L). Since the point A was arbitrary, any point is rotated by F 1 F through an angle about whose measure is twice the measure of the angle between L1 and L. This means that F 1 F is a rotation of that angle about. You have proved: The composition of two reflections whose mirror lines intersect is a rotation with center at the point of intersection of the two mirror lines, and the angle measure of this rotation is times the measure of the angle between the mirror lines.