Question
Suppose m is the line with equation y = −4 and ΔA'B'C is mapped to ΔA"B"C" by applying the glide reflection T⟨3, 0⟩ ∘ Rm.
What are the coordinates of ΔA″B″C″?
The graph with X-coordinate marks -4, -2, 0, 2, 4 and Y-coordinate mark -4, -2, -1, 0, 2, 4. -The triangle ABC with coordinates A at (3, 5), B at (1, 2), C at (5, 1). The triangle A'B'C' with coordinates A' at (-4, -5), B' at (-5, -2), C' at (-2, -1).
A. A″(−1, −3), B″(−2, −6), C″(1, −7)
B. A″(−7, −3), B″(−8, −6), C″(−5, −7)
C. A″(−2, −5), B″(0, −2), C″(−3, −1)
D. A″(−8, −3), B″(−6, 4), C″(−9, 0)
What are the coordinates of ΔA″B″C″?
The graph with X-coordinate marks -4, -2, 0, 2, 4 and Y-coordinate mark -4, -2, -1, 0, 2, 4. -The triangle ABC with coordinates A at (3, 5), B at (1, 2), C at (5, 1). The triangle A'B'C' with coordinates A' at (-4, -5), B' at (-5, -2), C' at (-2, -1).
A. A″(−1, −3), B″(−2, −6), C″(1, −7)
B. A″(−7, −3), B″(−8, −6), C″(−5, −7)
C. A″(−2, −5), B″(0, −2), C″(−3, −1)
D. A″(−8, −3), B″(−6, 4), C″(−9, 0)
Answers
GPT-4o mini
To find the coordinates of ΔA″B″C″ after applying the transformations on ΔA'B'C, we need to perform the following two steps:
1. **Reflect ΔA'B'C over the line \( y = -4 \)**.
2. **Translate (glide) the resulting points by the vector \( \langle 3, 0 \rangle \)**.
### Step 1: Reflect over the line \( y = -4 \)
The reflection formula across the line \( y = k \) is given by:
\[
y' = k - (y - k) = 2k - y
\]
For our case where \( k = -4 \), it becomes:
\[
y' = -4 - (y + 4) = -4 - y - 4 = -y - 8
\]
Now, let's apply this reflection to each of the points A', B', and C':
- **For A'(-4, -5)**:
\[
y' = -(-5) - 8 = 5 - 8 = -3 \quad \Rightarrow \quad A'' = (-4, -3)
\]
- **For B'(-5, -2)**:
\[
y' = -(-2) - 8 = 2 - 8 = -6 \quad \Rightarrow \quad B'' = (-5, -6)
\]
- **For C'(-2, -1)**:
\[
y' = -(-1) - 8 = 1 - 8 = -7 \quad \Rightarrow \quad C'' = (-2, -7)
\]
After reflection, we have:
- A''(-4, -3)
- B''(-5, -6)
- C''(-2, -7)
### Step 2: Translate by \( \langle 3, 0 \rangle \)
Now we'll apply the translation to each of the reflected points.
- **For A''(-4, -3)**:
\[
A''' = (-4 + 3, -3 + 0) = (-1, -3)
\]
- **For B''(-5, -6)**:
\[
B''' = (-5 + 3, -6 + 0) = (-2, -6)
\]
- **For C''(-2, -7)**:
\[
C''' = (-2 + 3, -7 + 0) = (1, -7)
\]
Finally, we have the coordinates of ΔA'''B'''C''' as:
- A'''(-1, -3)
- B'''(-2, -6)
- C'''(1, -7)
### Conclusion
The coordinates of ΔA''B''C'' are A''(-1, -3), B''(-2, -6), C''(1, -7), which matches with option **A**.
The answer is:
**A. A″(−1, −3), B″(−2, −6), C″(1, −7)**
1. **Reflect ΔA'B'C over the line \( y = -4 \)**.
2. **Translate (glide) the resulting points by the vector \( \langle 3, 0 \rangle \)**.
### Step 1: Reflect over the line \( y = -4 \)
The reflection formula across the line \( y = k \) is given by:
\[
y' = k - (y - k) = 2k - y
\]
For our case where \( k = -4 \), it becomes:
\[
y' = -4 - (y + 4) = -4 - y - 4 = -y - 8
\]
Now, let's apply this reflection to each of the points A', B', and C':
- **For A'(-4, -5)**:
\[
y' = -(-5) - 8 = 5 - 8 = -3 \quad \Rightarrow \quad A'' = (-4, -3)
\]
- **For B'(-5, -2)**:
\[
y' = -(-2) - 8 = 2 - 8 = -6 \quad \Rightarrow \quad B'' = (-5, -6)
\]
- **For C'(-2, -1)**:
\[
y' = -(-1) - 8 = 1 - 8 = -7 \quad \Rightarrow \quad C'' = (-2, -7)
\]
After reflection, we have:
- A''(-4, -3)
- B''(-5, -6)
- C''(-2, -7)
### Step 2: Translate by \( \langle 3, 0 \rangle \)
Now we'll apply the translation to each of the reflected points.
- **For A''(-4, -3)**:
\[
A''' = (-4 + 3, -3 + 0) = (-1, -3)
\]
- **For B''(-5, -6)**:
\[
B''' = (-5 + 3, -6 + 0) = (-2, -6)
\]
- **For C''(-2, -7)**:
\[
C''' = (-2 + 3, -7 + 0) = (1, -7)
\]
Finally, we have the coordinates of ΔA'''B'''C''' as:
- A'''(-1, -3)
- B'''(-2, -6)
- C'''(1, -7)
### Conclusion
The coordinates of ΔA''B''C'' are A''(-1, -3), B''(-2, -6), C''(1, -7), which matches with option **A**.
The answer is:
**A. A″(−1, −3), B″(−2, −6), C″(1, −7)**