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EXAMPLE 2

The endpoints of FG are F (–1, 2) and G (1, 2) . Reflect the segment in the line y = x . Graph the segment and its image. EXAMPLE 2. Graph a reflection in y = x.

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EXAMPLE 2

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  1. The endpoints of FG are F(–1, 2) and G(1, 2). Reflect the segment in the line y = x. Graph the segment and its image. EXAMPLE 2 Graph a reflection in y = x

  2. The slope of GG′ will also be –1. From G, move 0.5 units right and 0.5 units down to y = x. Then move 0.5 units right and 0.5 units down to locate G′(2, 1). The slope of y = xis 1. The segment from Fto its image, FF ′ , is perpendicular to the line of reflection y = x, so the slope of FF ′ will be –1 (because 1(–1) = –1). FromF, move 1.5 units right and 1.5 units down to y = x. From that point, move 1.5 units right and 1.5 units down to locateF′(2,–1). EXAMPLE 2 Graph a reflection in y = x SOLUTION

  3. Reflect FGfrom Example 2 in the line y = –x. Graph FGand its image. (a, b) (–b, –a) F(–1, 2) F ′(–2, 1) G(1, 2) G ′(–2, –1) EXAMPLE 3 Graph a reflection in y = –x SOLUTION Use the coordinate rule for reflecting in y = –x.

  4. SOLUTION for Examples 2 and 3 GUIDED PRACTICE Graph ABC with vertices A(1, 3), B(4, 4), and C(3, 1). Reflect ABC in the lines y = –x and y = x. Graph each image.

  5. Slope of y = – xis –1. The slope of FF′ is 1. The product of their slopes is –1 making them perpendicular. for Examples 2 and 3 GUIDED PRACTICE ′ 5. In Example 3, verify that FF is perpendicular to y = –x. SOLUTION

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