To find the pre-image triangle \( \Delta ABC \) from the image triangle \( \Delta A'B'C' \), we need to reverse the translation given by the rule \( (x, y) \rightarrow (x + 1.5, y - 3.5) \).
To find the original coordinates, we apply the inverse of the translation: \[ (x', y') \rightarrow (x' - 1.5, y' + 3.5) \]
Now, let's find the original coordinates of points \( A', B', \) and \( C' \):
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The coordinates of \( A' \) are \((-1, 2)\):
- \( x_A = -1 - 1.5 = -2.5 \)
- \( y_A = 2 + 3.5 = 5.5 \)
- So \( A(-2.5, 5.5) \).
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The coordinates of \( B' \) are \((1, -1)\):
- \( x_B = 1 - 1.5 = -0.5 \)
- \( y_B = -1 + 3.5 = 2.5 \)
- So \( B(-0.5, 2.5) \).
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The coordinates of \( C' \) are \((-5, -1)\):
- \( x_C = -5 - 1.5 = -6.5 \)
- \( y_C = -1 + 3.5 = 2.5 \)
- So \( C(-6.5, 2.5) \).
These calculations provide us with the pre-image points \( A(-2.5, 5.5) \), \( B(-0.5, 2.5) \), and \( C(-6.5, 2.5) \).
Now we can see which graph shows the pre-image triangle \( \Delta ABC \):
- The closest coordinates from the options given that match what we calculated is:
- \( A(-2.5, 5.5) \)
- \( B(-2.5, 2.5) \)
- \( C(-6.5, 2.5) \)
Based on these calculations, the correct triangle is: On a coordinate plane, a triangle has points A (negative 2.5, 5.5), B (negative 2.5, 2.5), C (negative 6.5, 2.5).
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