# Nakafa Framework: LLM

URL: /en/subject/high-school/11/mathematics/geometric-transformation/rotation
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/11/mathematics/geometric-transformation/rotation/en.mdx

Output docs content for large language models.

---

import { getColor } from "@repo/design-system/lib/color";
import { LineEquation } from "@repo/design-system/components/contents/line-equation";

export const metadata = {
  title: "Rotation",
  description: "Master geometric rotation transformations with step-by-step formulas, examples, and visual guides. Learn to rotate points and lines around the origin.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## 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 (<InlineMath math="\theta" />):** 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 <InlineMath math="C" /> and a directed angle <InlineMath math="\theta" />. Rotation with center <InlineMath math="C" /> by an angle <InlineMath math="\theta" />, denoted by <InlineMath math="\rho_{C,\theta}" /> or <InlineMath math="R(C,\theta)" />, is defined as a transformation that maps:

1. Point <InlineMath math="C" /> to itself (<InlineMath math="C'=C" />).
2. Any point <InlineMath math="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 <InlineMath math="\vec{CP}" /> and ray <InlineMath math="\vec{CP'}" /> is <InlineMath math="\theta" />.

## Rotation about the Origin

A common special case is rotation about the origin <InlineMath math="O(0,0)" />.

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

<div className="flex flex-col gap-4">
  <BlockMath math="x' = x \cos \theta - y \sin \theta" />
  <BlockMath math="y' = x \sin \theta + y \cos \theta" />
</div>

## Rotating a Point by 90°

A point <InlineMath math="B(0,4)" /> is rotated about the origin <InlineMath math="(0,0)" /> by <InlineMath math="90^\circ" />. Determine its image.

Here, <InlineMath math="x=0" />, <InlineMath math="y=4" />, and <InlineMath math="\theta = 90^\circ" />.

We know <InlineMath math="\cos 90^\circ = 0" /> and <InlineMath math="\sin 90^\circ = 1" />.

Using the formulas:

<div className="flex flex-col gap-4">
  <BlockMath math="x' = (0) \cos 90^\circ - (4) \sin 90^\circ = 0 \cdot 0 - 4 \cdot 1 = -4" />
  <BlockMath math="y' = (0) \sin 90^\circ + (4) \cos 90^\circ = 0 \cdot 1 + 4 \cdot 0 = 0" />
</div>

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

<LineEquation
  title={
    <>
      Rotation of Point <InlineMath math="B(0,4)" /> by{" "}
      <InlineMath math="90^\circ" /> about the Origin
    </>
  }
  description={
    <>
      Visualization of rotating point <InlineMath math="B(0,4)" /> to{" "}
      <InlineMath math="B'(-4,0)" /> by <InlineMath math="90^\circ" />{" "}
      counter-clockwise around the origin <InlineMath math="O(0,0)" />.
    </>
  }
  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 90°

Determine the image of the line <InlineMath math="y=2x" /> rotated about the origin <InlineMath math="(0,0)" /> by <InlineMath math="90^\circ" />.

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

<div className="flex flex-col gap-4">
  <BlockMath math="x' = x \cos 90^\circ - y \sin 90^\circ = x(0) - y(1) = -y" />
  <BlockMath math="y' = x \sin 90^\circ + y \cos 90^\circ = x(1) + y(0) = x" />
</div>

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 <InlineMath math="y=2x" />:

<div className="flex flex-col gap-4">
  <BlockMath math="(-x') = 2(y')" />
  <BlockMath math="-x' = 2y'" />
</div>

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

<LineEquation
  title={
    <>
      Rotation of Line <InlineMath math="y=2x" /> by{" "}
      <InlineMath math="90^\circ" /> about the Origin
    </>
  }
  description={
    <>
      Original line <InlineMath math="y=2x" /> rotated{" "}
      <InlineMath math="90^\circ" /> results in image line{" "}
      <InlineMath math="y = -\frac{1}{2}x" />.
    </>
  }
  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 <InlineMath math="A(3,0)" /> is rotated about the origin <InlineMath math="(0,0)" /> by <InlineMath math="90^\circ" />. Determine its image.
2.  Determine the image of the line <InlineMath math="y=4x" /> rotated about the origin <InlineMath math="(0,0)" /> by <InlineMath math="90^\circ" />.
3.  Point <InlineMath math="P(-2, -5)" /> is rotated about the origin <InlineMath math="O(0,0)" /> by <InlineMath math="180^\circ" />. Determine the coordinates of its image!

### Key Answers

1.  Point <InlineMath math="A(3,0)" />, <InlineMath math="\theta = 90^\circ" />. <InlineMath math="x=3, y=0" />.

    <div className="flex flex-col gap-4">
      <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" />
    </div>

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

2.  Line <InlineMath math="y=4x" />, <InlineMath math="\theta = 90^\circ" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="x' = -y" />
      <BlockMath math="y' = x" />
    </div>

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

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

    Image line equation: <InlineMath math="-x = 4y" /> or <InlineMath math="y = -\frac{1}{4}x" />.

3.  Point <InlineMath math="P(-2,-5)" />, <InlineMath math="\theta = 180^\circ" />. <InlineMath math="x=-2, y=-5" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="\cos 180^\circ = -1" />
      <BlockMath math="\sin 180^\circ = 0" />
    </div>

    <div className="flex flex-col gap-4">
      <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" />
    </div>

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