Triangle A has vertices (2;2) (6;2) (6;4)
Reflect triangle A in the line y=4

To reflect triangle \( A \) with vertices \( (2, 2) \), \( (6, 2) \), and \( (6, 4) \) over the line \( y = 4 \), we proceed as follows:

1. **Identify the line of reflection**: The line \( y = 4 \) is horizontal. Reflection across this line means that any point \( (x, y) \) will reflect to a point \( (x, 2 \times 4 - y) = (x, 8 - y) \).

2. **Reflect each vertex**:

   - For the point \( (2, 2) \):
     \[
     y' = 8 - 2 = 6
     \]
     The reflected point is \( (2, 6) \).

   - For the point \( (6, 2) \):
     \[
     y' = 8 - 2 = 6
     \]
     The reflected point is \( (6, 6) \).

   - For the point \( (6, 4) \):
     \[
     y' = 8 - 4 = 4
     \]
     The reflected point remains \( (6, 4) \) as it lies on the line of reflection.

3. **Summary of the reflected vertices**:
   - The reflected triangle \( A' \) has vertices \( (2, 6) \), \( (6, 6) \), and \( (6, 4) \).

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**Related questions to explore:**
1. How can you determine the reflection of a point across other lines, like \( y = x \)?
2. What happens if the triangle is reflected across a vertical line, like \( x = 3 \)?
3. How can you use the distance formula to verify the reflection of each point?
4. How would the reflection change if we used a different line, such as \( y = 0 \)?
5. Can you reflect any shape using similar principles?

**Tip:** When reflecting across a horizontal line \( y = c \), only the \( y \)-coordinates change while \( x \)-coordinates remain constant.

Triangle A has vertices (2,2), (6,2), and (6,4). Reflect triangle A in the line y=4.

Understand how to reflect a triangle with vertices (2,2), (6,2), and (6,4) across the line y=4. This geometry problem includes key concepts such as reflections, transformations, and applying the reflection formula for a horizontal line. Suitable for students in Grades 8-10.

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