Trapezoid ABCD is shown. A is at negative 5, 1. B is at negative 4, 3. C is at negative 2, 3. D is at negative 1, 1.
A) x-axis, y=x, y-axis, x-axis
b) x-axis, y-axis, x-axis
c) y=x, x-axis, x-axis
d) y-axis, x-axis, y-axis, x-axis
The best answer is C180 rotation wud take that point to 4th quadrant
reflection in x-axis takes that to 1st quadrant
reflection in y-axis brings it back to 2nd quadrant again. So, the sequence of transformations will bring A back to where it started