What set of reflections would carry trapezoid abcd onto itself? 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

y=x, x-axis, y=x, y-axis

Step-by-step explanation:

The correct answer is:

y=x, x-axis, y=x, y-axis

When a reflection takes place across the line y=x, then it maps every point (x, y) to (y, x).

This means that the mapping switches the coordinates, but does not negate them.

This means for A(1, -1), we will have A'(-1, 1); B(2, -2)→B'(-2, 2); C(3, -2)→C'(-2, 3); and D(4, -1)→D'(-1, 4).

And we get new points as :

A'(-1, 1)→A''(-1, -1); B'(-2, 2)→B''(-2, -2); C'(-2, 3)→C''(-2, -3); and D'(-1, 4)→D''(-1, -4).

Now again reflecting across the line y=x will again switch the x- and y-coordinates to :

A''(-1, -1)→A'''(-1, -1); B''(-2, -2)→B'''(-2, -2); C''(-2, -3)→C'''(-3, -2); and D''(-1, -4)→D'''(-4, -1).

And we know that while reflecting across the y-axis, the transformation will negate the x-coordinate:

So, new coordinates are :

A'''(-1, -1)→(1, -1); B'''(-2, -2)→(2, -2); C'''(-3, -2)→(3, -2); and D'''(-4, -1)→(4, -1).

Finally, we can see that these points are the same as our original points.

So, option A is the answer.

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