Rotation Transformation

A rotation moves every point in the plane the same angular distance about a fixed point (the center of rotation). The shape and size are preserved — only the orientation changes.

The Rotation Formulas

For a counterclockwise rotation by angle $\theta$ about the origin:

$x' = x\cos\theta - y\sin\theta$

$y' = x\sin\theta + y\cos\theta$

In matrix form:

$\begin{pmatrix}x'\\y'\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$

For the special case $\theta = 90°$ ($\cos = 0$, $\sin = 1$):

$x' = -y, \qquad y' = x$

So the rotation $(x, y) \to (-y, x)$ rotates 90° counterclockwise. Conveniently, you can do this by hand: just swap and negate.

Step-by-Step Solution

Given: A triangle with vertices $A = (1, 1)$, $B = (4, 1)$, $C = (1, 3)$. Rotate it 90° counterclockwise about the origin.

Find: The image vertices.

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Step 1 — Apply the formula $(x, y) \to (-y, x)$ to each vertex.

$A = (1, 1) \to (-1, 1) = A'$

$B = (4, 1) \to (-1, 4) = B'$

$C = (1, 3) \to (-3, 1) = C'$

Step 2 — Verify with the matrix multiplication for $A$.

$\begin{pmatrix}\cos 90° & -\sin 90°\\\sin 90° & \cos 90°\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}0(1) + (-1)(1)\\1(1) + 0(1)\end{pmatrix} = \begin{pmatrix}-1\\1\end{pmatrix} \;\;\checkmark$

Same answer.

Step 3 — Verify the side lengths are preserved.

$|AB| = 4 - 1 = 3$ (horizontal segment).

$|A'B'| = \sqrt{(-1-(-1))^{2} + (4-1)^{2}} = \sqrt{0 + 9} = 3$ ✓

The lengths match. Rotations preserve all distances.

Step 4 — Verify the angles are preserved.

The right angle at $A$ (since $AB$ is horizontal and $AC$ is vertical) becomes the right angle at $A'$ (now $A'B'$ is vertical and $A'C'$ is horizontal).

Step 5 — Other special rotations.

• 180° about origin: $(x, y) \to (-x, -y)$
• 270° (or −90°): $(x, y) \to (y, -x)$
• 360°: identity, no change

Step 6 — Composition of rotations.

Two consecutive rotations of $\theta_1$ and $\theta_2$ about the same point are equivalent to a single rotation of $\theta_1 + \theta_2$. Two 90° rotations make a 180° rotation, etc. (This isn't true if the centers differ — that's a more complex operation.)

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Answer: Rotating $(1,1)$, $(4,1)$, $(1,3)$ by 90° counterclockwise about the origin yields:

$\boxed{\;A' = (-1, 1),\quad B' = (-1, 4),\quad C' = (-3, 1)\;}$

The triangle is preserved in size and shape but rotated to a new position. The map $(x, y) \to (-y, x)$ does the work — and it generalizes to $(x, y) \to (x\cos\theta - y\sin\theta,\, x\sin\theta + y\cos\theta)$ for any angle $\theta$.

Try It

• Adjust the rotation angle $\theta$.
• Watch the triangle rotate smoothly through 0° to 360°.
• The original triangle is faint, the rotated is bright.
• Try $\theta = 180°$ — the triangle ends up at $(-1, -1)$, $(-4, -1)$, $(-1, -3)$.