Question

question 2 of 10
which sequence of transformations will result in an image that maps onto itself?
a. rotate 180 degrees counterclockwise about the origin, and then reflect across the y - axis.
b. reflect over the y - axis, and then reflect again over the y - axis.
c. rotate 180 degrees counterclockwise about the origin, and then reflect across the x - axis.
d. reflect over the y - axis, and then reflect over the x - axis.

Explanation:

Step1: Analyze option A

Let \((x,y)\) be a point. Rotating \((x,y)\) \(180^{\circ}\) counter - clockwise about the origin gives \((-x,-y)\). Reflecting \((-x,-y)\) across the \(y\) - axis gives \((x,-y)
eq(x,y)\) (unless \(y = 0\)).

Step2: Analyze option B

Let \((x,y)\) be a point. Reflecting \((x,y)\) over the \(y\) - axis gives \((-x,y)\). Reflecting \((-x,y)\) over the \(y\) - axis gives \((x,y)\). So the transformation maps the point (and thus the image) onto itself.

Step3: Analyze option C

Let \((x,y)\) be a point. Rotating \((x,y)\) \(180^{\circ}\) counter - clockwise about the origin gives \((-x,-y)\). Reflecting \((-x,-y)\) across the \(x\) - axis gives \((-x,y)
eq(x,y)\) (unless \(x = 0\)).

Step4: Analyze option D

Let \((x,y)\) be a point. Reflecting \((x,y)\) over the \(y\) - axis gives \((-x,y)\). Reflecting \((-x,y)\) over the \(x\) - axis gives \((-x,-y)
eq(x,y)\) (unless \(x=y = 0\)).

Answer:

B. Reflect over the \(y\) - axis, and then reflect again over the \(y\) - axis.