To dilate a triangle using a center of dilation and a scale factor, you use the following formula for each vertex \((x, y)\):
\[ (x', y') = (kx + (1-k)x_c, ky + (1-k)y_c) \]
where:
- \((x', y')\) are the coordinates of the new point after dilation,
- \(k\) is the scale factor (in this case, 12),
- \((x_c, y_c)\) is the center of dilation (in this case, \((2, 2)\)),
- \((x, y)\) are the coordinates of the original point.
Let’s denote the original triangle's vertices as \(A (x_1, y_1)\), \(B (x_2, y_2)\), and \(C (x_3, y_3)\).
I will assume that the vertices of the original triangle are given. You can apply the dilation process to each vertex using the above formula.
As an example, let's say the vertices of triangle \(ABC\) are:
- \(A(1, 1)\)
- \(B(1, 3)\)
- \(C(3, 1)\)
Using the center of dilation \((2,2)\) and a scale factor of \(12\):
-
A (1, 1): \[ (x', y') = (12 \cdot 1 + (1 - 12) \cdot 2, 12 \cdot 1 + (1 - 12) \cdot 2) \] \[ x' = 12 + (-11) \cdot 2 = 12 - 22 = -10 \] \[ y' = 12 + (-11) \cdot 2 = 12 - 22 = -10 \] So, \(A'(-10, -10)\).
-
B (1, 3): \[ (x', y') = (12 \cdot 1 + (1 - 12) \cdot 2, 12 \cdot 3 + (1 - 12) \cdot 2) \] \[ x' = 12 - 22 = -10 \] \[ y' = 36 - 22 = 14 \] So, \(B'(-10, 14)\).
-
C (3, 1): \[ (x', y') = (12 \cdot 3 + (1 - 12) \cdot 2, 12 \cdot 1 + (1 - 12) \cdot 2) \] \[ x' = 36 - 22 = 14 \] \[ y' = 12 - 22 = -10 \] So, \(C'(14, -10)\).
The coordinates of the dilated triangle \(A'B'C'\) are:
- \(A'(-10, -10)\)
- \(B'(-10, 14)\)
- \(C'(14, -10)\)
Please provide the original coordinates of the triangle if different to ensure accurate results.
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