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Under a rotation about the origin, A(5, −1) is mapped to the point A'(1, 5) . What is the image of B(−4, 6) under this rotation? Use words and coordinates to explain.

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The image of B(-4, 6) under the rotation is B'(6, 4).

Let O be the origin and let r be the angle of rotation.

Since A(-5, 1) is mapped to A'(1, 5), we can see that r must be 90 degrees counterclockwise.

To find the image of B(-4, 6), we can use the following rule:

For a rotation of angle r about the origin, the image of point P(x, y) is P'(x', y') where:

x' = rcos(θ)x + rsin(θ)y

y' = -rsin(θ)x + rcos(θ)y

In this case, r = 90 degrees, so cos(r) = 0 and sin(r) = 1. Plugging these values into the above equations, we get:

x' = 0(-4) + 1(6) = 6

y' = -1(-4) + 0(6) = 4

Therefore, the image of B(-4, 6) under the rotation is B'(6, 4).

User Rahulg
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