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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-over-point
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Learn point reflection (180° rotation). Learn half-turn transformations using coordinate formulas with worked examples and visualizations.

---

## Understanding Reflection over a Point

Reflection over a point, often called a half-turn rotation ($$180^\circ$$), is a geometric transformation where each point on an object is mapped to a new position such that the center of reflection becomes the midpoint between the original point and its image.

Visible text: Reflection over a point, often called a half-turn rotation (), is a geometric transformation where each point on an object is mapped to a new position such that the center of reflection becomes the midpoint between the original point and its image.

Suppose the center of reflection is $$P(a,b)$$. If a point $$Q(x,y)$$ is reflected over point $$P$$, its image <InlineMath math="Q'(x',y')" /> will lie on the line passing through $$Q$$ and $$P$$, with $$P$$ as the midpoint of the segment <InlineMath math="QQ'" />.

Visible text: Suppose the center of reflection is . If a point is reflected over point , its image <InlineMath math="Q'(x',y')" /> will lie on the line passing through and , with as the midpoint of the segment <InlineMath math="QQ'" />.

### Rule for Reflection over a Point

If a point $$Q(x, y)$$ is reflected over the point $$P(a, b)$$, its image's coordinates, <InlineMath math="Q'(x', y')" />, are determined by the formula:

Visible text: If a point is reflected over the point , its image's coordinates, <InlineMath math="Q'(x', y')" />, are determined by the formula:

Component: MathContainer
Children:

```math
x' = 2a - x
```

```math
y' = 2b - y
```

Alternatively, it can be written as:

```math
Q'(-x + 2a, -y + 2b)
```

This means the $$x$$-coordinate of the image is twice the $$x$$-coordinate of the center minus the original $$x$$-coordinate, and the same applies to the $$y$$-coordinate.

Visible text: This means the -coordinate of the image is twice the -coordinate of the center minus the original -coordinate, and the same applies to the -coordinate.

## Reflecting a Point over Another Point

Determine the image of a half-turn $$\sigma_{(2,3)}$$ for the point $$(5, 4)$$.

Visible text: Determine the image of a half-turn for the point .

This means we are reflecting point $$Q(5,4)$$ over the center point $$P(2,3)$$.

Visible text: This means we are reflecting point over the center point .

Here, $$x=5$$, $$y=4$$, $$a=2$$, and $$b=3$$.

Visible text: Here, , , , and .

Using the formula:

Component: MathContainer
Children:

```math
x' = 2a - x = 2(2) - 5 = 4 - 5 = -1
```

```math
y' = 2b - y = 2(3) - 4 = 6 - 4 = 2
```

Thus, the image of point $$(5,4)$$ is $$(-1,2)$$.

Visible text: Thus, the image of point is .

Component: LineEquation
Props:
- title: Image of Point $$Q(5,4)$$ over Point{" "}
$$P(2,3)$$
  Visible text: Image of Point over Point{" "}
- description: Visualization of reflecting point $$Q(5,4)$$ over the
center point $$P(2,3)$$ resulting in{" "}
<InlineMath math="Q'(-1,2)" />.
  Visible text: Visualization of reflecting point over the
center point resulting in{" "}
<InlineMath math="Q'(-1,2)" />.
- data: [
{
points: [{ x: 2, y: 3, z: 0 }],
color: getColor("ROSE"),
showPoints: true,
labels: [{ text: "P(2,3) - Center", at: 0, offset: [1.5, -0.5, 0] }],
},
{
points: [{ x: 5, y: 4, z: 0 }],
color: getColor("SKY"),
showPoints: true,
labels: [{ text: "Q(5,4) - Original", at: 0, offset: [0.3, 0.5, 0] }],
},
{
points: [{ x: -1, y: 2, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "Q'(-1,2) - Image", at: 0, offset: [-0.7, -0.5, 0] }],
},
{
points: [
{ x: 5, y: 4, z: 0 },
{ x: -1, y: 2, z: 0 },
],
color: getColor("PINK"),
}, // Line connecting Q to Q'
]
- showZAxis: false

## Reflecting a Line over a Point

Determine the image of a half-turn $$\sigma_{(1,3)}$$ for the line $$l$$ with the equation $$2x - y + 3 = 0$$.

Visible text: Determine the image of a half-turn for the line with the equation .

Take an arbitrary point $$(x,y)$$ on line $$l$$. Its image, <InlineMath math="(x',y')" />, after reflection over point $$P(1,3)$$ is:

Visible text: Take an arbitrary point on line . Its image, <InlineMath math="(x',y')" />, after reflection over point is:

Component: MathContainer
Children:

```math
x' = 2(1) - x = 2 - x \implies x = 2 - x'
```

```math
y' = 2(3) - y = 6 - y \implies y = 6 - y'
```

Substitute <InlineMath math="x = 2 - x'" /> and <InlineMath math="y = 6 - y'" /> into the equation of line $$l$$:

Visible text: Substitute <InlineMath math="x = 2 - x'" /> and <InlineMath math="y = 6 - y'" /> into the equation of line :

Component: MathContainer
Children:

```math
2(2 - x') - (6 - y') + 3 = 0
```

```math
4 - 2x' - 6 + y' + 3 = 0
```

```math
-2x' + y' + 1 = 0
```

Replacing <InlineMath math="x'" /> and <InlineMath math="y'" /> back to $$x$$ and $$y$$, the equation of the image line <InlineMath math="l'" /> is:

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

```math
-2x + y + 1 = 0
```

Alternatively, it can be written as $$2x - y - 1 = 0$$.

Visible text: Alternatively, it can be written as .

Component: LineEquation
Props:
- title: Image of Line $$2x-y+3=0$$ over Point{" "}
$$P(1,3)$$
  Visible text: Image of Line over Point{" "}
- description: Original line $$2x-y+3=0$$ reflected over point{" "}
$$P(1,3)$$ results in image line{" "}
$$2x-y-1=0$$.
  Visible text: Original line reflected over point{" "}
 results in image line{" "}
.
- data: [
{
points: [{ x: 1, y: 3, z: 0 }],
color: getColor("ROSE"),
showPoints: true,
labels: [{ text: "P(1,3) - Center", at: 0, offset: [0.5, -0.5, 0] }],
},
{
// Original Line: 2x - y + 3 = 0 => y = 2x + 3
points: Array.from({ length: 11 }, (_, i) => {
const xVal = i - 5;
return { x: xVal, y: 2 * xVal + 3, z: 0 };
}),
color: getColor("SKY"),
labels: [{ text: "2x-y+3=0", at: 5, offset: [-2, 0.5, 0] }],
},
{
// Image Line: 2x - y - 1 = 0 => y = 2x - 1
points: Array.from({ length: 11 }, (_, i) => {
const xVal = i - 5;
return { x: xVal, y: 2 * xVal - 1, z: 0 };
}),
color: getColor("EMERALD"),
labels: [{ text: "2x-y-1=0", at: 5, offset: [2, -0.5, 0] }],
},
]
- showZAxis: false
- cameraPosition: [1, 2, 15]

## Exercises

1.  Determine the image of a half-turn $$\sigma_{(-2,3)}$$ for the point $$(3, -4)$$.
2.  Point $$K(-5, 1)$$ is reflected over the origin $$O(0,0)$$. Determine the coordinates of its image!
3.  Determine the image of a half-turn $$\sigma_{(1,-3)}$$ for the line $$l$$ with the equation $$2x + 5y + 6 = 0$$.

Visible text: 1. Determine the image of a half-turn for the point .
2. Point is reflected over the origin . Determine the coordinates of its image!
3. Determine the image of a half-turn for the line with the equation .

### Key Answers

1.  Center $$P(-2,3)$$, point $$Q(3,-4)$$. So $$a=-2, b=3, x=3, y=-4$$.

    <MathContainer>
      <BlockMath math="x' = 2a - x = 2(-2) - 3 = -4 - 3 = -7" />
      <BlockMath math="y' = 2b - y = 2(3) - (-4) = 6 + 4 = 10" />
    </MathContainer>

    Thus, the image of point $$Q(3,-4)$$ is $$(-7,10)$$.

2.  Center $$O(0,0)$$, point $$K(-5,1)$$. So $$a=0, b=0, x=-5, y=1$$.

    <MathContainer>
      <BlockMath math="x' = 2(0) - (-5) = 0 + 5 = 5" />
      <BlockMath math="y' = 2(0) - 1 = 0 - 1 = -1" />
    </MathContainer>

    Thus, the image of point $$K(-5,1)$$ is <InlineMath math="K'(5,-1)" />.

3.  Center $$P(1,-3)$$. Line $$2x + 5y + 6 = 0$$.

    <MathContainer>
      <BlockMath math="x' = 2(1) - x = 2 - x \implies x = 2 - x'" />
      <BlockMath math="y' = 2(-3) - y = -6 - y \implies y = -6 - y'" />
    </MathContainer>

    Substitute into the line equation:

    <MathContainer>
      <BlockMath math="2(2 - x') + 5(-6 - y') + 6 = 0" />
      <BlockMath math="4 - 2x' - 30 - 5y' + 6 = 0" />
      <BlockMath math="-2x' - 5y' - 20 = 0" />
    </MathContainer>

    Image line equation: $$-2x - 5y - 20 = 0$$ or $$2x + 5y + 20 = 0$$.

Visible text: 1. Center , point . So .

 <MathContainer>
 <BlockMath math="x' = 2a - x = 2(-2) - 3 = -4 - 3 = -7" />
 <BlockMath math="y' = 2b - y = 2(3) - (-4) = 6 + 4 = 10" />
 </MathContainer>

 Thus, the image of point is .

2. Center , point . So .

 <MathContainer>
 <BlockMath math="x' = 2(0) - (-5) = 0 + 5 = 5" />
 <BlockMath math="y' = 2(0) - 1 = 0 - 1 = -1" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="K'(5,-1)" />.

3. Center . Line .

 <MathContainer>
 <BlockMath math="x' = 2(1) - x = 2 - x \implies x = 2 - x'" />
 <BlockMath math="y' = 2(-3) - y = -6 - y \implies y = -6 - y'" />
 </MathContainer>

 Substitute into the line equation:

 <MathContainer>
 <BlockMath math="2(2 - x') + 5(-6 - y') + 6 = 0" />
 <BlockMath math="4 - 2x' - 30 - 5y' + 6 = 0" />
 <BlockMath math="-2x' - 5y' - 20 = 0" />
 </MathContainer>

 Image line equation: or .