The point P (−1,2)

To determine the rotation that maps the point \( P(-1, 2) \) to \( P'(2, 1) \), we can analyze their coordinates.

The transformation of the point can be visualized by considering the movement through the origin. The point \( P(-1, 2) \) is located in Quadrant II of the Cartesian plane, while the point \( P'(2, 1) \) is located in Quadrant I.

To find the angle of rotation, let's first consider the vector from the origin to each point:
- For \( P(-1, 2) \), the vector is \( \langle -1, 2 \rangle \).
- For \( P'(2, 1) \), the vector is \( \langle 2, 1 \rangle \).

Now, we will compute the angle of rotation needed to transition from \( P \) to \( P' \).

1. The angle corresponding to \( P(-1, 2) \): To compute it, we can use the inverse tangent:
\[
\theta_1 = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{2}{-1}\right) = \tan^{-1}(-2)
\]
This angle falls in Quadrant II.

2. The angle corresponding to \( P'(2, 1) \):
\[
\theta_2 = \tan^{-1}\left(\frac{1}{2}\right)
\]
This angle falls in Quadrant I.

3. Now, to find the rotation from \( P \) to \( P' \), we need to determine the difference between the angles. The angle of \( P \) (in degrees) is approximately \( 180° - \tan^{-1}(-2) \), and the angle of \( P' \) is just \( \tan^{-1}\left(\frac{1}{2}\right) \).

4. The angle of rotation \( \Delta \theta = \theta_2 - \theta_1 \).
If we consider the standard position (counter-clockwise is positive):
- From \( P(-1, 2) \) to \( P'(2, 1) \) goes through Quadrants II to I.

Calculating the exact degrees may require a calculator for arctangents. However, from the visual/graphical perspective:
- The angle from \( P \) to \( P' \) is indeed a rotation of \( 90° \) clockwise (or \( -90° \)).

Thus, the answer is:

**−90° rotation**
or
**negative 90 degrees rotation**.