We are asked to consider a situation where a preimage is rotated 360 degrees about the origin. Let's consider an example figure.
Let's rotate this figure 360^(∘) about the origin.
As we can see, the image of the rotation is exactly the same as the preimage. This is because 360^(∘) is a complete angle, so the rotation maps every point onto itself. Alternatively, we could see a rotation by 360^(∘) as a rotation by 270^(∘) followed by a rotation by 90^(∘).
270^(∘) + 90^(∘) = 360^(∘)
A rotation by 270^(∘) maps a point with coordinates (x, y) onto the point with coordinates (y, - x). A rotation by 90^(∘) maps a point with coordinates (x, y) onto the point with coordinates (- y, x).
Rotation by270^(∘)&: ( x, y) → ( y, - x)
Rotation by90^(∘)&: ( x, y) → (- y, x)
Let's see what happens to a point with coordinates (a, b) when we rotate it by 270^(∘) and then by 90^(∘). A rotation by 270^(∘) switches the coordinates around and multiplies the second coordinate of the image by - 1. Then, after the first rotation, the point is mapped onto the point (b, - a).
( a, b) → ( b, - a)
A rotation by 90^(∘) also swaps the coordinates around. Additionally, it multiplies the first coordinate of the image by - 1. Let's see what happens when we rotate a point with the coordinates (b, - a) by 90^(∘).
( b, - a) → (- ( - a), b)
Note that - (- a)) is the same as a.
(- (- a), b) = (a, b)
This confirms that for any point, the image of a rotation by 360^(∘) about the origin is the same point as the original.
