To determine if the figures \( \triangle ABC \) and \( \triangle EFG \) are similar after the transformation, first, we need to find the coordinates of the rotated triangle \( \triangle ABC \) after the clockwise rotation about the origin.
Step 1: Rotation
Let’s consider a clockwise rotation of \( \triangle ABC \) by \( \theta \) degrees. A common choice for simplicity is to rotate by \( 90^\circ \). The rotation of a point \((x, y)\) about the origin by \( 90^\circ \) clockwise is given by the formula:
\[ (x, y) \rightarrow (y, -x) \]
Applying this to each vertex of \( \triangle ABC \):
- A(-4, 3) rotates to (3, 4)
- B(-3, 1) rotates to (1, 3)
- C(-1, 2) rotates to (2, 1)
So after a \( 90^\circ \) clockwise rotation, the vertices of the triangle are now:
- A' (3, 4)
- B' (1, 3)
- C' (2, 1)
Step 2: Check Similarity
The triangles \( \triangle ABC \) (original) and \( \triangle A'B'C' \) (rotated) are similar due to the properties of rotations (which preserve shapes and angles). However, to completely confirm similarity, we should check the lengths of the sides.
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Calculate the lengths of sides of \( \triangle ABC \):
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Length of \( AB \): \[ AB = \sqrt{((-4) - (-3))^2 + (3 - 1)^2} = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \]
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Length of \( BC \): \[ BC = \sqrt{((-3) - (-1))^2 + (1 - 2)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
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Length of \( CA \): \[ CA = \sqrt{((-1) - (-4))^2 + (2 - 3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]
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Calculate the lengths of sides of \( \triangle A'B'C' \):
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Length of \( A'B' \): \[ A'B' = \sqrt{(3 - 1)^2 + (4 - 3)^2} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
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Length of \( B'C' \): \[ B'C' = \sqrt{(1 - 2)^2 + (3 - 1)^2} = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \]
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Length of \( C'A' \): \[ C'A' = \sqrt{(2 - 3)^2 + (1 - 4)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Conclusion
The lengths of sides for triangles \( \triangle ABC \) and \( \triangle A'B'C' \) are as follows:
- For \( \triangle ABC \): \( AB = \sqrt{5}, , BC = \sqrt{5}, , CA = \sqrt{10} \)
- For \( \triangle A'B'C' \): \( A'B' = \sqrt{5}, , B'C' = \sqrt{5}, , C'A' = \sqrt{10} \)
Since all corresponding sides are equal, \( \triangle ABC \) and \( \triangle A'B'C' \) are indeed similar. Thus, if you translate \( \triangle A'B'C' \) (the rotated image) to form \( \triangle EFG \), the figures \( \triangle ABC \) and \( \triangle EFG \) also remain similar regardless of the translation.
Therefore, yes, \( \triangle ABC \) and the translated image \( \triangle EFG \) are similar figures.