Rotation

Center and Angle of a Rotation

Rotation turns every point of a figure through the same angle around a fixed center. The figure changes position and turns, while distances, angles, shape, size, and the order of its vertices are preserved.

A rotation needs two pieces of information:

• Center of Rotation (C): The fixed point around which the rotation occurs.
• Angle of Rotation ($\theta$): A directed angle that gives both the amount and direction of the turn. A positive angle rotates counter-clockwise, while a negative angle rotates clockwise.

Definition of Rotation

Given a center point $C$ and a directed angle $\theta$. Rotation with center $C$ by an angle $\theta$, denoted by $\rho_{C,\theta}$ or $R(C,\theta)$, is defined as a transformation that maps:

1. Point $C$ to itself ($C'=C$).
2. Any point $P$ to a point $P'$ such that $CP = CP'$ (the distance from the center to the point is equal to the distance from the center to the image) and the angle formed by ray $\vec{CP}$ and ray $\vec{CP'}$ is $\theta$.

Rotation about the Origin

A common special case is rotation about the origin $O(0,0)$.

If a point $P(x,y)$ is rotated about the origin $O(0,0)$ by an angle $\theta$, its image coordinates $P'(x',y')$ can be calculated using the following formulas:

Rotating a Point by a Quarter Turn

A point $B(0,4)$ is rotated about the origin $(0,0)$ by $90^\circ$. Determine its image.

Here, $x=0$, $y=4$, and $\theta = 90^\circ$.

We know $\cos 90^\circ = 0$ and $\sin 90^\circ = 1$.

Using the formulas:

The image of point $B(0,4)$ is $B'(-4,0)$.

Rotating a Line by a Quarter Turn

Determine the image of the line $y=2x$ rotated about the origin $(0,0)$ by $90^\circ$.

Let $P(x,y)$ be any point on the line $y=2x$. Its image, $P'(x',y')$, after a $90^\circ$ rotation about the origin is:

From this, we get $y = -x'$ and $x = y'$.

Substitute $x = y'$ and $y = -x'$ into the original line equation $y=2x$:

Replacing $x'$ and $y'$ back to $x$ and $y$, the equation of the image line is $-x = 2y$ or $y = -\frac{1}{2}x$.

Exercises

The problems rotate a point, a line, and a second point about the origin. The line is rotated through its equation, which moves every point of the line together.

1. A point $A(3,0)$ is rotated about the origin $(0,0)$ by $90^\circ$. Determine its image.
2. Determine the image of the line $y=4x$ rotated about the origin $(0,0)$ by $90^\circ$.
3. Point $P(-2, -5)$ is rotated about the origin $O(0,0)$ by $180^\circ$. Determine the coordinates of its image!

Solutions

1. Point $A(3,0)$, $\theta = 90^\circ$. $x=3, y=0$.

   $$\begin{aligned} x' &= 3 \cos 90^\circ - 0 \sin 90^\circ \\ &= 3(0) - 0(1) \\ &= 0 \end{aligned}$$

   $$\begin{aligned} y' &= 3 \sin 90^\circ + 0 \cos 90^\circ \\ &= 3(1) + 0(0) \\ &= 3 \end{aligned}$$

   Thus, its image is $A'(0,3)$.

2. Line $y=4x$, $\theta = 90^\circ$.

   $$x' = -y$$

   $$y' = x$$

   So $y = -x'$ and $x = y'$.

   Substitute into the line equation: $(-x') = 4(y')$.

   Image line equation: $-x = 4y$ or $y = -\frac{1}{4}x$.

3. Point $P(-2,-5)$, $\theta = 180^\circ$. $x=-2, y=-5$.

   $$\cos 180^\circ = -1$$

   $$\sin 180^\circ = 0$$
   $$\begin{aligned} x' &= (-2) \cos 180^\circ - (-5) \sin 180^\circ \\ &= (-2)(-1) - (-5)(0) \\ &= 2 - 0 \\ &= 2 \end{aligned}$$

   $$\begin{aligned} y' &= (-2) \sin 180^\circ + (-5) \cos 180^\circ \\ &= (-2)(0) + (-5)(-1) \\ &= 0 + 5 \\ &= 5 \end{aligned}$$

   Thus, the image of point $P$ is $P'(2,5)$.