What set of reflections would carry trapezoid ABCD onto itself?Trapezoid ABCD is shown. A is at 1, negative 1. B is at 2, negative 2. C is at 3, negative 2. D is at 4, negative 1. y=x, x-axis, y=x, y-axis x-axis, y=x, x-axis, y=x y-axis, x-axis, y-axis x-axis, y-axis, y-axis

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The correct answer is:y=x, x-axis, y=x, y-axis Explanation:A reflection across the line y=x maps every point (x, y) to (y, x); it switches the coordinates but does not negate them.This means for A(1, -1), we would have A'(-1, 1); B(2, -2)β†’B'(-2, 2); C(3, -2)β†’C'(-2, 3); and D(4, -1)β†’D'(-1, 4).A reflection across the x-axis negates the y-coordinate; algebraically, (x, y)β†’(-x, y).This takes our new points A'(-1, 1)β†’A''(-1, -1); B'(-2, 2)β†’B''(-2, -2); C'(-2, 3)β†’C''(-2, -3); and D'(-1, 4)β†’D''(-1, -4).Reflecting again across the line y=x will again switch the x- and y-coordinates:A''(-1, -1)β†’A'''(-1, -1); B''(-2, -2)β†’B'''(-2, -2); C''(-2, -3)β†’C'''(-3, -2); and D''(-1, -4)β†’D'''(-4, -1).Reflecting across the y-axis will negate the x-coordinate:A'''(-1, -1)β†’(1, -1); B'''(-2, -2)β†’(2, -2); C'''(-3, -2)β†’(3, -2); and D'''(-4, -1)β†’(4, -1).These are the same as our original points.
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