Question
What set of reflections would carry hexagon ABCDEF onto itself?
Hexagon ABCDEF on the coordinate plane with point A at 0, 1, point B at negative 1, 0, point C at negative 2, 1, point D at negative 2, 3, point E at negative 1, 4, and point F at 0, 3.
x‒axis, y = x, x‒axis, y = x
y-axis, x‒axis, y-axis
x‒axis, y-axis, y-axis
y = x, x‒axis, y = x, y-axis
Answers
Answered by
GPT-4o mini
To determine the set of reflections that would carry hexagon ABCDEF onto itself, we start by analyzing the symmetry of the hexagon based on the given coordinates of its vertices:
- A (0, 1)
- B (-1, 0)
- C (-2, 1)
- D (-2, 3)
- E (-1, 4)
- F (0, 3)
### Step 1: Identify Symmetry
Next, we will check for lines of symmetry that could reflect the shape of the hexagon back onto itself.
1. **Vertical Symmetry (y-axis)**:
- By reflecting across the y-axis, we can see that points on the left mirror points on the right:
- A (0, 1) remains A (0, 1),
- B (-1, 0) reflects to C (1, 0) which is not a vertex. So, no symmetry here.
2. **Horizontal Symmetry (x-axis)**:
- Reflecting across the x-axis, the points transform as follows:
- A (0, 1) becomes A' (0, -1), not a vertex.
- So, no symmetry here as well.
3. **Diagonal Symmetry (y = x)**:
- Points undergo the transformation:
- A (0, 1) -> A' (1, 0)
- B (-1, 0) -> B' (0, -1), not a vertex.
- This reflection doesn't hold.
4. **Other Diagonal (y = -x)**:
- Not relevant given our defined corners; this doesn't preserve vertices either.
### Step 2: Check Composite Reflections and Other Lines
- Given that there are no immediate vertical, horizontal, or main diagonal points of reflection, consider combinations or less obvious axes.
1. **Combination of reflections**:
- One valid combination is:
- Reflect over the **y-axis**, then the **x-axis**, and finally back across the **y-axis** can yield the necessary adjustments to achieve reflection across vertices but care would need to be retained on the original placement.
### Conclusion
Given that the analyses did not yield a straight-forward single axis that suffices, the set that makes feasible reflection alignments involves:
- Reflecting across **y = x**, then **x-axis**, then another **y = x**, and next through **y-axis** serves as potent sequences.
Thus, the answer is: **y = x, x-axis, y = x, y-axis**. Each of which holds transformational integrity to return ABCDEF back to its coordinates in respective reflections and rotations.
- A (0, 1)
- B (-1, 0)
- C (-2, 1)
- D (-2, 3)
- E (-1, 4)
- F (0, 3)
### Step 1: Identify Symmetry
Next, we will check for lines of symmetry that could reflect the shape of the hexagon back onto itself.
1. **Vertical Symmetry (y-axis)**:
- By reflecting across the y-axis, we can see that points on the left mirror points on the right:
- A (0, 1) remains A (0, 1),
- B (-1, 0) reflects to C (1, 0) which is not a vertex. So, no symmetry here.
2. **Horizontal Symmetry (x-axis)**:
- Reflecting across the x-axis, the points transform as follows:
- A (0, 1) becomes A' (0, -1), not a vertex.
- So, no symmetry here as well.
3. **Diagonal Symmetry (y = x)**:
- Points undergo the transformation:
- A (0, 1) -> A' (1, 0)
- B (-1, 0) -> B' (0, -1), not a vertex.
- This reflection doesn't hold.
4. **Other Diagonal (y = -x)**:
- Not relevant given our defined corners; this doesn't preserve vertices either.
### Step 2: Check Composite Reflections and Other Lines
- Given that there are no immediate vertical, horizontal, or main diagonal points of reflection, consider combinations or less obvious axes.
1. **Combination of reflections**:
- One valid combination is:
- Reflect over the **y-axis**, then the **x-axis**, and finally back across the **y-axis** can yield the necessary adjustments to achieve reflection across vertices but care would need to be retained on the original placement.
### Conclusion
Given that the analyses did not yield a straight-forward single axis that suffices, the set that makes feasible reflection alignments involves:
- Reflecting across **y = x**, then **x-axis**, then another **y = x**, and next through **y-axis** serves as potent sequences.
Thus, the answer is: **y = x, x-axis, y = x, y-axis**. Each of which holds transformational integrity to return ABCDEF back to its coordinates in respective reflections and rotations.