Rotation is a transformation that turns a shape around a fixed point called the center of rotation through a specific angle and direction.
Let \( (x,y) \) be the coordinates of a point. The coordinates \( (x',y') \) after rotation are given by:
| Rotation | Coordinate Rule | Matrix Form |
|---|---|---|
| \( 90^\circ \) CCW | \( (x, y) \rightarrow (x' = -y, y' = x) \) | \( \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \) |
| \( 180^\circ \) | \( (x, y) \rightarrow (x' = -x, y' = -y) \) | \( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \) |
| \( 270^\circ \) CCW | \( (x, y) \rightarrow (x' = y, y' = -x) \) | \( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \) |
| \( 360^\circ \) | \( (x, y) \rightarrow (x' = x, y' = y) \) | \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \) |
For a general rotation by angle \( \theta \):
\[ \begin{aligned} x' &= x \cos\theta - y \sin\theta \\ y' &= x \sin\theta + y \cos\theta \end{aligned} \]
Click to add points. Use the slider or angle input to rotate them. Points are labeled A, B, C (red) and their rotated images A', B', C' (blue).
Point \( P(3, 4) \) is rotated \( 90^\circ \) counterclockwise about the origin. Find the coordinates of \( P' \).
Triangle \( ABC \) has vertices \( A(1, 1) \), \( B(4, 1) \), and \( C(2, 5) \). Find the vertices after a \( 180^\circ \) rotation about the origin.
Point \( Q(-2, 5) \) is rotated \( 270^\circ \) counterclockwise about the origin. Find the coordinates of \( Q' \).
Using the \( 90^\circ \) CCW rule: \( (x, y) \rightarrow (-y, x) \)
\[ \begin{aligned} P(3, 4) &\rightarrow P'(-4, 3) \\ x' &= -y = -4 \\ y' &= x = 3 \end{aligned} \]
Answer: \( P'(-4, 3) \)
Using the \( 180^\circ \) rotation rule: \( (x, y) \rightarrow (-x, -y) \)
\[ \begin{aligned} A(1, 1) &\rightarrow A'(-1, -1) \\ B(4, 1) &\rightarrow B'(-4, -1) \\ C(2, 5) &\rightarrow C'(-2, -5) \end{aligned} \]
Answer: \( A'(-1, -1) \), \( B'(-4, -1) \), \( C'(-2, -5) \)
Using the \( 270^\circ \) CCW rule: \( (x, y) \rightarrow (y, -x) \)
\[ \begin{aligned} Q(-2, 5) &\rightarrow Q'(5, 2) \\ x' &= y = 5 \\ y' &= -x = -(-2) = 2 \end{aligned} \]
Answer: \( Q'(5, 2) \)