To perform a dilation of a triangle given its vertices, a center of dilation, and a scale factor, follow these steps:
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Identify the original points of the triangle: You mentioned "triangle 4.2. 4.2" but did not specify the vertices. For demonstration purposes, let's define the triangle's vertices as follows (you can replace these with the actual vertices if needed):
- A(1, 1)
- B(2, 3)
- C(3, 1)
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Identify the center of dilation: You indicated the center is at (4.2, 4.2).
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Identify the scale factor: The scale factor is given as 6.
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Apply the dilation formula: The formula for the dilation of a point \( P(x, y) \) from a center of dilation \( O(x_0, y_0) \) by a scale factor \( k \) is: \[ P'(x', y') = \left( x_0 + k(x - x_0), y_0 + k(y - y_0) \right) \]
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Applying the formula to each vertex:
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For point A(1, 1): \[ A'(x', y') = \left( 4.2 + 6(1 - 4.2), 4.2 + 6(1 - 4.2) \right) \] \[ A' = \left( 4.2 + 6(-3.2), 4.2 + 6(-3.2) \right) = \left( 4.2 - 19.2, 4.2 - 19.2 \right) = \left( -15, -15 \right) \]
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For point B(2, 3): \[ B'(x', y') = \left( 4.2 + 6(2 - 4.2), 4.2 + 6(3 - 4.2) \right) \] \[ B' = \left( 4.2 + 6(-2.2), 4.2 + 6(-1.2) \right) = \left( 4.2 - 13.2, 4.2 - 7.2 \right) = \left( -9, -3 \right) \]
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For point C(3, 1): \[ C'(x', y') = \left( 4.2 + 6(3 - 4.2), 4.2 + 6(1 - 4.2) \right) \] \[ C' = \left( 4.2 + 6(-1.2), 4.2 + 6(-3.2) \right) = \left( 4.2 - 7.2, 4.2 - 19.2 \right) = \left( -3, -15 \right) \]
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Final Dilated Points:
- A'(-15, -15)
- B'(-9, -3)
- C'(-3, -15)
To visualize, you can plot these points on a coordinate plane, connecting A', B', and C' to form the image of the dilated triangle.
If your original triangle has different coordinates, please provide them, and I can assist you with the calculation.