Question

point p was rotated by 135° (the center of rotation is indicated). d c b a center p which point is the image of p? choose 1 answer: a a b b c c d d

Explanation:

Step1: Recall Rotation Properties

Rotation preserves distance from the center, so \( OP = OP' \) (where \( O \) is the center, \( P' \) is the image). Also, the angle between \( \overrightarrow{OP} \) and \( \overrightarrow{OP'} \) is \( 135^\circ \).

Step2: Analyze Angle and Position

Visually, the center is fixed. The angle of \( 135^\circ \) (a counterclockwise or clockwise rotation? Assuming standard position, counterclockwise from \( P \)'s initial direction). The point \( C \) would correspond to a \( 135^\circ \) rotation (checking the arc between \( P \) and the candidates). Wait, no—wait, the initial position of \( P \) is to the left of the center. Rotating \( 135^\circ \) (let's assume counterclockwise, as \( 135^\circ \) is in the second quadrant relative to the center). Wait, actually, looking at the diagram: the center, \( P \) is left, \( B, A \) are right, \( C, D \) are top. The angle from \( P \) (left) to \( C \): let's imagine the center. The vector from center to \( P \) is left; rotating \( 135^\circ \) counterclockwise would point towards the top-left? No, wait, \( 135^\circ \) from the left direction (vector \( \langle -1, 0 angle \)) rotated \( 135^\circ \) counterclockwise: the new direction is \( \langle -1, 0 angle \) rotated \( 135^\circ \), which is \( \cos(135^\circ) = -\frac{\sqrt{2}}{2} \), \( \sin(135^\circ) = \frac{\sqrt{2}}{2} \), so direction is \( \langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} angle \)? No, wait, initial vector is \( \vec{OP} = (P_x - O_x, P_y - O_y) \), which is left (negative x). Rotating that vector by \( 135^\circ \) counterclockwise: the rotation matrix is \( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \). So \( \theta = 135^\circ \), \( \cos\theta = -\frac{\sqrt{2}}{2} \), \( \sin\theta = \frac{\sqrt{2}}{2} \). The initial vector is \( (-k, 0) \) (k > 0). Applying rotation: \( x' = -\frac{\sqrt{2}}{2}(-k) - \frac{\sqrt{2}}{2}(0) = \frac{\sqrt{2}k}{2} \), \( y' = \frac{\sqrt{2}}{2}(-k) + \frac{\sqrt{2}}{2}(0) = -\frac{\sqrt{2}k}{2} \)? No, that can't be. Wait, maybe clockwise rotation? \( \theta = -135^\circ \), rotation matrix \( \begin{pmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \). Then \( x' = -\frac{\sqrt{2}}{2}(-k) + \frac{\sqrt{2}}{2}(0) = \frac{\sqrt{2}k}{2} \), \( y' = -\frac{\sqrt{2}}{2}(-k) + -\frac{\sqrt{2}}{2}(0) = \frac{\sqrt{2}k}{2} \). Wait, no, maybe I'm overcomplicating. Visually, the point \( C \) is above the center, and the angle from \( P \) (left) to \( C \) is \( 135^\circ \) (since from left, turning \( 135^\circ \) towards top). Alternatively, the key is that the rotation of \( 135^\circ \) maps \( P \) to \( C \)? Wait, no—wait, the options: \( A, B \) are right, \( C, D \) are top. Wait, maybe the rotation is clockwise? \( 135^\circ \) clockwise from \( P \)'s position. The initial direction of \( P \) is left (vector \( \langle -1, 0 angle \)). Rotating \( 135^\circ \) clockwise: the angle becomes \( -135^\circ \), so the direction is \( \cos(-135^\circ) = -\frac{\sqrt{2}}{2} \), \( \sin(-135^\circ) = -\frac{\sqrt{2}}{2} \)? No, that's down-left. Wait, maybe the diagram is simpler: the center, \( P \) is left, \( B \) and \( A \) are right (smaller angle, like \( 45^\circ \) or \( 90^\circ \)), \( C \) and \( D \) are top ( \( 135^\circ \) or \( 180^\circ \)). So a \( 135^\circ \) rotation (either direction) would land on \( C \)? Wait, no, the correct answer is \( C \)? Wait, no—wait, let's think again. The problem says "the center of rotation is indicated". So the rotation is around the center. So the distance from center to \( P \) must equal distance from center to the image. So \( OP = OC \), \( OP = OB \), etc. Now, the angle between \( OP \) and \( OC \): if \( P \) is left, \( C \) is top, the angle between left (180°) and top (90°) is 90°, no—wait, maybe I got the direction wrong. Wait, \( P \) is to the left of the center, so vector \( \vec{OP} = (P - O) = (-a, 0) \) (a > 0). Rotating this vector by \( 135^\circ \) counterclockwise: the new vector is \( (-a\cos135^\circ - 0\sin135^\circ, -a\sin135^\circ + 0\cos135^\circ) \)? Wait, no, rotation matrix is \( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \), so applying to \( (-a, 0) \): \( x' = -a\cos\theta - 0\sin\theta = -a\cos135^\circ = -a(-\frac{\sqrt{2}}{2}) = \frac{a\sqrt{2}}{2} \), \( y' = -a\sin\theta + 0\cos\theta = -a\sin135^\circ = -a(\frac{\sqrt{2}}{2}) = -\frac{a\sqrt{2}}{2} \). That's down-right, which is not \( C \). Wait, maybe clockwise rotation: \( \theta = -135^\circ \). Then \( x' = -a\cos(-135^\circ) - 0\sin(-135^\circ) = -a(-\frac{\sqrt{2}}{2}) = \frac{a\sqrt{2}}{2} \), \( y' = -a\sin(-135^\circ) + 0\cos(-135^\circ) = -a(-\frac{\sqrt{2}}{2}) = \frac{a\sqrt{2}}{2} \). So that's up-right, but \( B \) and \( A \) are right. Wait, maybe the diagram is such that \( P \) is left, \( B \) is at \( 45^\circ \) right, \( A \) at \( 90^\circ \) right, \( C \) at \( 135^\circ \) top, \( D \) at \( 180^\circ \) top. Wait, no, the key is that \( 135^\circ \) rotation (from \( P \)'s position) would align with \( C \). Wait, maybe the correct answer is \( C \), so option C: \( C \).

Answer:

C. \( C \)