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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/rotation
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Learn geometric rotation transformations with formula practice, examples, and visual guides. Learn to rotate points and lines around the origin.

---

## Understanding Rotation

Rotation is a geometric transformation that turns every point of an object around a specific center point by a certain angle. This transformation preserves the congruence (shape and size) of the object, but its orientation can change.

Key aspects of rotation:

- **Center of Rotation (C):** The fixed point around which the rotation occurs.
- **Angle of Rotation ($$\theta$$):** The amount of turn. If the angle is positive, the rotation is counter-clockwise. If the angle is negative, the rotation is clockwise.

Visible text: - **Center of Rotation (C):** The fixed point around which the rotation occurs.
- **Angle of Rotation ():** The amount of turn. If the angle is positive, the rotation is counter-clockwise. If the angle is negative, the rotation is 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:

Visible text: Given a center point and a directed angle . Rotation with center by an angle , denoted by or , is defined as a transformation that maps:

1. Point $$C$$ to itself (<InlineMath math="C'=C" />).
2. Any point $$P$$ to a point <InlineMath math="P'" /> such that <InlineMath math="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 <InlineMath math="\vec{CP'}" /> is $$\theta$$.

Visible text: 1. Point to itself (<InlineMath math="C'=C" />).
2. Any point to a point <InlineMath math="P'" /> such that <InlineMath math="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 and ray <InlineMath math="\vec{CP'}" /> is .

## Rotation about the Origin

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

Visible text: A common special case is rotation about the origin .

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

Visible text: If a point is rotated about the origin by an angle , its image coordinates <InlineMath math="P'(x',y')" /> can be calculated using the following formulas:

Component: MathContainer
Children:

```math
x' = x \cos \theta - y \sin \theta
```

```math
y' = x \sin \theta + y \cos \theta
```

## 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.

Visible text: A point is rotated about the origin by . Determine its image.

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

Visible text: Here, , , and .

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

Visible text: We know and .

Using the formulas:

Component: MathContainer
Children:

```math
x' = (0) \cos 90^\circ - (4) \sin 90^\circ = 0 \cdot 0 - 4 \cdot 1 = -4
```

```math
y' = (0) \sin 90^\circ + (4) \cos 90^\circ = 0 \cdot 1 + 4 \cdot 0 = 0
```

Thus, the image of point $$B(0,4)$$ is <InlineMath math="B'(-4,0)" />.

Visible text: Thus, the image of point is <InlineMath math="B'(-4,0)" />.

Component: LineEquation
Props:
- title: Rotation of Point $$B(0,4)$$ by{" "}
$$90^\circ$$ about the Origin
  Visible text: Rotation of Point by{" "}
 about the Origin
- description: Visualization of rotating point $$B(0,4)$$ to{" "}
<InlineMath math="B'(-4,0)" /> by $$90^\circ$${" "}
counter-clockwise around the origin $$O(0,0)$$.
  Visible text: Visualization of rotating point to{" "}
<InlineMath math="B'(-4,0)" /> by {" "}
counter-clockwise around the origin .
- data: [
{
points: [{ x: 0, y: 0, z: 0 }],
color: getColor("ROSE"),
showPoints: true,
labels: [{ text: "O(0,0)", at: 0, offset: [0.3, -0.3, 0] }],
}, // Center of Rotation
{
points: [{ x: 0, y: 4, z: 0 }],
color: getColor("SKY"),
showPoints: true,
labels: [{ text: "B(0,4) - Original", at: 0, offset: [0.3, 0.3, 0] }],
}, // Original Point
{
points: [{ x: -4, y: 0, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "B'(-4,0) - Image", at: 0, offset: [-0.7, 0.3, 0] }],
}, // Image Point
{
points: [
{ x: 0, y: 0, z: 0 },
{ x: 0, y: 4, z: 0 },
],
color: getColor("PURPLE"),
}, // Line OB
{
points: [
{ x: 0, y: 0, z: 0 },
{ x: -4, y: 0, z: 0 },
],
color: getColor("PURPLE"),
}, // Line OB'
]
- showZAxis: false
- cameraPosition: [2, 2, 15]

## Rotating a Line by a Quarter Turn

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

Visible text: Determine the image of the line rotated about the origin by .

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

Visible text: Let be any point on the line . Its image, <InlineMath math="P'(x',y')" />, after a rotation about the origin is:

Component: MathContainer
Children:

```math
x' = x \cos 90^\circ - y \sin 90^\circ = x(0) - y(1) = -y
```

```math
y' = x \sin 90^\circ + y \cos 90^\circ = x(1) + y(0) = x
```

From this, we get <InlineMath math="y = -x'" /> and <InlineMath math="x = y'" />.

Substitute <InlineMath math="x = y'" /> and <InlineMath math="y = -x'" /> into the original line equation $$y=2x$$:

Visible text: Substitute <InlineMath math="x = y'" /> and <InlineMath math="y = -x'" /> into the original line equation :

Component: MathContainer
Children:

```math
(-x') = 2(y')
```

```math
-x' = 2y'
```

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

Visible text: Replacing <InlineMath math="x'" /> and <InlineMath math="y'" /> back to and , the equation of the image line is or .

Component: LineEquation
Props:
- title: Rotation of Line $$y=2x$$ by{" "}
$$90^\circ$$ about the Origin
  Visible text: Rotation of Line by{" "}
 about the Origin
- description: Original line $$y=2x$$ rotated{" "}
$$90^\circ$$ results in image line{" "}
$$y = -\frac{1}{2}x$$.
  Visible text: Original line rotated{" "}
 results in image line{" "}
.
- data: [
{
points: [{ x: 0, y: 0, z: 0 }],
color: getColor("ROSE"),
showPoints: false,
labels: [{ text: "O(0,0)", at: 0, offset: [0.5, -0.5, 0] }],
}, // Center of Rotation
{
// Original Line: y = 2x
points: Array.from({ length: 11 }, (_, i) => {
const xVal = (i - 5) * 0.5; // range from -2.5 to 2.5
return { x: xVal, y: 2 * xVal, z: 0 };
}),
color: getColor("PURPLE"),
showPoints: false,
labels: [{ text: "y=2x", at: 6, offset: [1, 0.5, 0] }],
},
{
// Image Line: y = -1/2 x
points: Array.from({ length: 11 }, (_, i) => {
const xVal = (i - 5) * 0.5;
return { x: xVal, y: (-1 / 2) * xVal, z: 0 };
}),
color: getColor("PINK"),
showPoints: false,
labels: [{ text: "y=(-1/2)x", at: 1, offset: [0.3, 0.5, 0] }],
},
]
- showZAxis: false
- cameraPosition: [0, 0, 10]

## Exercises

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!

Visible text: 1. A point is rotated about the origin by . Determine its image.
2. Determine the image of the line rotated about the origin by .
3. Point is rotated about the origin by . Determine the coordinates of its image!

### Key Answers

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

    <MathContainer>
      <BlockMath math="x' = 3 \cos 90^\circ - 0 \sin 90^\circ = 3(0) - 0(1) = 0" />
      <BlockMath math="y' = 3 \sin 90^\circ + 0 \cos 90^\circ = 3(1) + 0(0) = 3" />
    </MathContainer>

    Thus, its image is <InlineMath math="A'(0,3)" />.

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

    <MathContainer>
      <BlockMath math="x' = -y" />
      <BlockMath math="y' = x" />
    </MathContainer>

    So <InlineMath math="y = -x'" /> and <InlineMath math="x = y'" />.

    Substitute into the line equation: <InlineMath math="(-x') = 4(y')" />.

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

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

    <MathContainer>
      
    
    ```math
    \cos 180^\circ = -1
    ```

      
    
    ```math
    \sin 180^\circ = 0
    ```

    </MathContainer>

    <MathContainer>
      <BlockMath math="x' = (-2) \cos 180^\circ - (-5) \sin 180^\circ = (-2)(-1) - (-5)(0) = 2 - 0 = 2" />
      <BlockMath math="y' = (-2) \sin 180^\circ + (-5) \cos 180^\circ = (-2)(0) + (-5)(-1) = 0 + 5 = 5" />
    </MathContainer>

    Thus, the image of point $$P$$ is <InlineMath math="P'(2,5)" />.

Visible text: 1. Point , . .

 <MathContainer>
 <BlockMath math="x' = 3 \cos 90^\circ - 0 \sin 90^\circ = 3(0) - 0(1) = 0" />
 <BlockMath math="y' = 3 \sin 90^\circ + 0 \cos 90^\circ = 3(1) + 0(0) = 3" />
 </MathContainer>

 Thus, its image is <InlineMath math="A'(0,3)" />.

2. Line , .

 <MathContainer>
 <BlockMath math="x' = -y" />
 <BlockMath math="y' = x" />
 </MathContainer>

 So <InlineMath math="y = -x'" /> and <InlineMath math="x = y'" />.

 Substitute into the line equation: <InlineMath math="(-x') = 4(y')" />.

 Image line equation: or .

3. Point , . .

 <MathContainer>
 
 

 
 

 </MathContainer>

 <MathContainer>
 <BlockMath math="x' = (-2) \cos 180^\circ - (-5) \sin 180^\circ = (-2)(-1) - (-5)(0) = 2 - 0 = 2" />
 <BlockMath math="y' = (-2) \sin 180^\circ + (-5) \cos 180^\circ = (-2)(0) + (-5)(-1) = 0 + 5 = 5" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="P'(2,5)" />.