Answer:
Explanation:
when rotate counterclock wise 90 degrees:
(x,y)⇒(-y,x)
points (2,2) (4,2) (2,-1) (4,-1)
CCW90 degrees (-2,2) (-2,4) (1,2) (1,4)
CCW 270 degrees (2,-2) (2,-4) (-1,-2) (-1,-4)
(x,y)⇒(y,-x) the shape will be in quadrant 3 and 4
Quadrants I and II.
O: (-2, 2)
N: (-2, 4)
M: (1, 4)
P: (1, 2)
When rotating about the origin 90 degrees in a counterclockwise direction, focus on the coordinates of one point on the preimage at a time. So, the x-coordinate on the image will be the opposite of the y-coordinate of the preimage, and the y coordinate of the image will be the x-coordinate of the preimage. That sounds complicated so here is an example from the problem.
O is a point on the preimage. Its coordinates are (2, 2). To find the x-coordinate, take the opposite of the image's y-coordinate. The y-coordinate is 2, so it will be a -2 x-coordinate on the image. To find the y-coordinate on the image, take the x-coordinate of the preimage (O). The x-coordinate of O is 2, so the y-coordinate of the image will be 2. Combine those together and after a 90 degree counterclockwise rotation, you get a point of (-2, 2)
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