Question

the vertices of trapezoid abcd are a(2, -2), b(5, -2), c(8, -7), and d(1, -7). graph abcd and abcd, its image after a reflection across the y - axis. which graph shows abcd and abcd? \\(\circ\\) a. graph a \\(\circ\\) b. graph b \\(\circ\\) c. graph c

Explanation:

Step1: Recall reflection over y - axis rule

The rule for reflecting a point \((x,y)\) across the \(y\) - axis is \((x,y)\to(-x,y)\).

Step2: Find coordinates of \(A',B',C',D'\)

• For point \(A(2,-2)\), after reflection across \(y\) - axis: \(A'(-2,-2)\)
• For point \(B(5,-2)\), after reflection across \(y\) - axis: \(B'(-5,-2)\)
• For point \(C(8,-7)\), after reflection across \(y\) - axis: \(C'(-8,-7)\)
• For point \(D(1,-7)\), after reflection across \(y\) - axis: \(D'(-1,-7)\)

Now, we analyze the original trapezoid \(ABCD\): \(A(2,-2)\), \(B(5,-2)\), \(C(8,-7)\), \(D(1,-7)\). The top base \(AB\) is from \(x = 2\) to \(x=5\) (length \(5 - 2=3\)) and the bottom base \(CD\) is from \(x = 1\) to \(x = 8\) (length \(8 - 1 = 7\)) at \(y=-7\) and \(y = - 2\) respectively.

After reflection, \(A'(-2,-2)\), \(B'(-5,-2)\), \(C'(-8,-7)\), \(D'(-1,-7)\). The top base \(A'B'\) is from \(x=-2\) to \(x=-5\) (length \(\vert-2-(-5)\vert=3\)) and the bottom base \(C'D'\) is from \(x=-1\) to \(x=-8\) (length \(\vert-1-(-8)\vert = 7\)) at \(y=-2\) and \(y=-7\) respectively.

Looking at the options:

• Option A: The two trapezoids are not reflections (different positions and orientations).
• Option B: The red trapezoid (image) has vertices at \((-2,-2),(-5,-2),(-8,-7),(-1,-7)\) (matching our calculated \(A',B',C',D'\)) and the blue trapezoid (original) has vertices at \((2,-2),(5,-2),(8,-7),(1,-7)\) (matching \(A,B,C,D\)).
• Option C: The reflection is across the \(x\) - axis (since \(y\) - coordinates would be positive if it were a reflection across \(x\) - axis, but our reflection keeps \(y\) - coordinates the same), so it is incorrect.

Answer:

B