Question

What are the vertices of ΔA'B'C produced by T−3, 6 (ΔABC) = 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, 0), B at (3, -2), C at (0, -3).
A. A′(0, 6), B′(0, 4), C′(−3, 3)
B. A′(6, 6), B′(6, 4), C′(3, 3)
C. A′(0, −6), B′(0, −8), C′(−3, 9)
D. A′(6, −6), B′(6, −8), C′(3, 9)

Answer 1

To find the vertices of triangle ΔA'B'C produced by the translation T(-3, 6) applied to triangle ΔABC with vertices A(3, 0), B(3, -2), and C(0, -3), we will apply the translation to each vertex of the triangle.

The translation T(-3, 6) means that we will subtract 3 from the x-coordinates and add 6 to the y-coordinates of each vertex of the original triangle.

1. **For vertex A(3, 0)**:
- New x-coordinate: \(3 - 3 = 0\)
- New y-coordinate: \(0 + 6 = 6\)
- So, A' = (0, 6).

2. **For vertex B(3, -2)**:
- New x-coordinate: \(3 - 3 = 0\)
- New y-coordinate: \(-2 + 6 = 4\)
- So, B' = (0, 4).

3. **For vertex C(0, -3)**:
- New x-coordinate: \(0 - 3 = -3\)
- New y-coordinate: \(-3 + 6 = 3\)
- So, C' = (-3, 3).

Putting it all together, we find:
- A' = (0, 6)
- B' = (0, 4)
- C' = (-3, 3)

Thus, the correct answer is:
**A. A′(0, 6), B′(0, 4), C′(−3, 3)**