Pearson Unit 2 Topic 8: Transformational Geometry 8-2: Reflections Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (3)(A) Describe and perform transformations of figures in a plane using coordinate notation. (1)(E) Create and use representations to organize, record, and communicate mathematical ideas. (1)(D) Communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication. (3)(C) Identify the sequence of transformations that will carry a given pre-image onto an image on and off the coordinate plane.

Discovery #1 Reflect A(4, 2) across the x-axis. The x-axis has the equation y = 0. What are the coordinates of A’? Write a reflection rule if the line of reflection is the x-axis. If d is the distance from the point to the line of reflection, how far is it from the pre-image to the image? Answers: A’(4, -2). Rule: (x, y) (x, -y) or 𝑅 𝑥−𝑎𝑥𝑖𝑠 𝑥, 𝑦 = 𝑥, −𝑦 The distance from the pre-image to the image is 2d.

Discovery #2 Reflect B(-3, -4) across the y-axis. The y-axis has the equation x = 0. What are the coordinates of B’? Write a reflection rule if the line of reflection is the y-axis. If d is the distance from the point to the line of reflection, how far is it from the pre-image to the image? Answers: B’(3, -4). Rule: (x, y) (-x, y) or 𝑅 𝑦−𝑎𝑥𝑖𝑠 𝑥, 𝑦 = −𝑥, 𝑦 The distance from the pre-image to the image is 2d.

Discovery #3 Graph the equation y = x. Reflect C(4, 1) across y = x. What are the coordinates of C’? Write a reflection rule if the line of reflection is y = x. Answers: C’(1, 4). Rule: (x, y) (y, x) or 𝑅 𝑦=𝑥 𝑥, 𝑦 = 𝑦, 𝑥

Discovery #4 Graph the equation y = -x. Reflect D(-4, -1) across y = -x. What are the coordinates of D’? Write a reflection rule if the line of reflection is y = -x. Answers: D’(1, 4). Rule: (x, y) (-y, -x) or 𝑅 𝑦=−𝑥 𝑥, 𝑦 = −𝑦, −𝑥

Reflection Rules ACROSS THE LINE y = -x (x, y) (-y, -x)

More Reflection Rules The line of reflection is the perpendicular bisector of the segment from P to P’. Reflections are a rigid transformation. Reflections preserve distance. Reflections preserve angle measure. Reflections map each point of the preimage to one and only one corresponding point of its image. Reflections can be written as a motion rule or as a coordinate (or function) rule.

Example: 1 Identify the line of reflection and the equation of the line of reflection. Write a coordinate form that describes this reflection. The line of reflection is the x-axis. The line of reflection is y = 0. 𝑅 𝑥−𝑎𝑥𝑖𝑠 𝐷𝐸𝐹𝐺 = 𝐷 ′ 𝐸 ′ 𝐹 ′ 𝐺′

Example: 2 X(2, –1), Y(–4, –3), Z(3, 2); y-axis Reflect the figure with the given vertices across the given line. X(2, –1), Y(–4, –3), Z(3, 2); y-axis Graph the pre-image & image. The reflection of (x, y) is (-x, y). Y’ X’ Z’ Z X(2,–1) X’ (-2, -1) Y(–4,–3) Y’ (4, -3) X Y Z(3, 2) Z ’(-3, 2)

Example: 3 R(–2, 2), S(5, 0), T(3, –1); y = x Reflect the figure with the given vertices across the given line. R(–2, 2), S(5, 0), T(3, –1); y = x Graph the pre-image & image. S’ R’ T’ The reflection of (x, y) is (y, x). S R T R(–2, 2) R’ (2, –2) S (5, 0) S’ (0, 5) T(3, –1) T ’(–1, 3)

Example: 4 R(-3, 2), S(0, -1), T(-3, –3); y = -x Reflect the figure with the given vertices across the given line. R(-3, 2), S(0, -1), T(-3, –3); y = -x Graph the pre-image & image. The reflection of (x, y) is (-y, -x). S’ R’ T’ R(–3, 2) R’ (-2, 3) S R T S (0, -1) S’ (1, 0) T(-3, –3) T ’(3, 3)

Example: 5 S(3, 3) S’ (3, –5) T(3, 0) T ’(3, –2) U(–2, 0) U ’(–2, –2) Reflect the rectangle with vertices S(3, 3), T(3, 0), U(–2, 0) and V(–2, 3) across the y = -1. Graph the pre-image and the image. S(3, 3) S’ (3, –5) V S U T T(3, 0) T ’(3, –2) U(–2, 0) U ’(–2, –2) V’ S’ U’ T’ V(–2, 3) V ’(–2, –5)

Example: 6 S(5, 2) S’ (-3, 2) T(5, -1) T ’(-3, –1) U(2, -1) U ’(0, –1) Reflect the rectangle with vertices S(5, 2), T(5, -1), U(2, -1) and V(2, 2) across the x = 1. Graph the pre-image and the image. S(5, 2) S’ (-3, 2) V’ S’ U’ T’ V S U T T(5, -1) T ’(-3, –1) U(2, -1) U ’(0, –1) V(2, 2) V ’(0, 2)

Example: 7 Find of equation of the line of reflection. Step #1: draw a segment from A to A’ (or B to B’ or C to C’, etc.). Step #2: find the slope and midpoint of that segment. Step #3: find the perpendicular slope. Step #4: draw the perpendicular line. Step #5: pick a point on that line and use point- slope form to write the equation of the line of reflection. Slope from A to A’ = -1. Midpoint of AA’ = (3, 5). Perpendicular slope = 1. Since the perpendicular bisector passes through the y-intercept, the equation is y = x + 2.