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Point [tex]\( P \)[/tex] was rotated about the origin [tex]\((0,0)\)[/tex] by [tex]\(-15^{\circ}\)[/tex].

Which point is the image of [tex]\( P \)[/tex]?

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Sagot :

To determine the image of point [tex]\( P \)[/tex] after it has been rotated about the origin [tex]\((0,0)\)[/tex] by [tex]\(-15^\circ\)[/tex], we follow these steps:

1. Identify the coordinates of point [tex]\( P \)[/tex] before rotation: Given point [tex]\( P \)[/tex] is at the origin, its coordinates are [tex]\( (0, 0) \)[/tex].

2. Define the rotation angle: The rotation angle is [tex]\(-15^\circ\)[/tex].

3. Construct the rotation matrix: Rotation matrices are used to handle transformations involving rotations about the origin. The general form of the rotation matrix for an angle [tex]\(\theta\)[/tex] is:
[tex]\[ R(\theta) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \][/tex]
Here, [tex]\(\theta\)[/tex] is [tex]\(-15^\circ\)[/tex].

4. Apply the rotation matrix to point [tex]\( P \)[/tex]: To find the new position of point [tex]\( P \)[/tex], we multiply the rotation matrix by the coordinate vector of [tex]\( P \)[/tex].

5. Compute the new coordinates: By multiplying the rotation matrix by the coordinates [tex]\((0, 0)\)[/tex], we get:
[tex]\[ \begin{pmatrix} \cos(-15^\circ) & -\sin(-15^\circ) \\ \sin(-15^\circ) & \cos(-15^\circ) \end{pmatrix} \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \cdot \cos(-15^\circ) + 0 \cdot -\sin(-15^\circ) \\ 0 \cdot \sin(-15^\circ) + 0 \cdot \cos(-15^\circ) \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]

Thus, rotating the point [tex]\((0, 0)\)[/tex] by [tex]\(-15^\circ\)[/tex] does not change its position. The coordinates of the image of point [tex]\( P \)[/tex] remain [tex]\((0, 0)\)[/tex].

So, the image of [tex]\( P \)[/tex] after rotation by [tex]\(-15^\circ\)[/tex] is the point:
[tex]\[ \boxed{(0,0)} \][/tex]