Question: Choose the correct answer.
How was Figure transformed to become Figure A'
A
A'
1.
[ ? ]
2.
[ ]
Translate
Dilate
Graph shape: two congruent triangles on the coordinate plane; the right triangle has vertices $(1,1)$, $(1,4)$, $(3,1)$ and the left triangle has vertices $(-6,2)$, $(-2,2)$, $(-2,8)$, with position_hint = center.
1. **State the problem:** Determine how the original figure (triangle with vertices $(-6,2)$, $(-2,2)$, $(-2,8)$) was transformed to become Figure A' (triangle with vertices $(1,1)$, $(1,4)$, $(3,1)$).
2. **Identify the transformation types:** The options are translation or dilation.
3. **Check for translation:** Translation moves every point by the same vector. Calculate the vector from one vertex of the original to the corresponding vertex of the image.
For example, from $(-6,2)$ to $(1,1)$:
$$\Delta x = 1 - (-6) = 7$$
$$\Delta y = 1 - 2 = -1$$
Check if the same vector applies to other vertices:
From $(-2,2)$ to $(1,4)$:
$$\Delta x = 1 - (-2) = 3$$
$$\Delta y = 4 - 2 = 2$$
Since the vectors differ, it is not a translation.
4. **Check for dilation:** Dilation scales distances from a center point by a scale factor.
Calculate side lengths of original triangle:
- Between $(-6,2)$ and $(-2,2)$:
$$\sqrt{(-2 + 6)^2 + (2 - 2)^2} = \sqrt{4^2 + 0} = 4$$
- Between $(-2,2)$ and $(-2,8)$:
$$\sqrt{(-2 + 2)^2 + (8 - 2)^2} = \sqrt{0 + 6^2} = 6$$
Calculate side lengths of image triangle:
- Between $(1,1)$ and $(1,4)$:
$$\sqrt{(1 - 1)^2 + (4 - 1)^2} = \sqrt{0 + 3^2} = 3$$
- Between $(1,4)$ and $(3,1)$:
$$\sqrt{(3 - 1)^2 + (1 - 4)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6$$
The side lengths are not proportional (4 to 3 and 6 to approx 3.6), so it is not a dilation.
5. **Conclusion:** Since the figure is congruent but shifted, and the vectors differ, the transformation is a translation but not by a single vector, so likely a combination or incorrect options.
**Final answer:** The figure was translated.