To dilate triangle XYZ to form triangle X'Y'Z' using a scale factor of 2 and a center of dilation at the point (2,0), you can use the following formula for dilation:
\[ X' = (1 - k) \cdot C_x + k \cdot X \] \[ Y' = (1 - k) \cdot C_y + k \cdot Y \]
Where:
- \( (C_x, C_y) \) is the center of dilation,
- \( (X, Y) \) are the original coordinates of the point,
- \( k \) is the scale factor.
Here, the center of dilation is \( C = (2, 0) \) and the scale factor \( k = 2 \).
For each vertex of triangle XYZ, we will apply this formula. Let's denote the vertices of triangle XYZ as \( X(a, b) \), \( Y(c, d) \), and \( Z(e, f) \).
Assuming you have specific coordinates for points X, Y, and Z, you can substitute those coordinates into the formulas provided above.
Example
Let's say:
- \( X = (1, 1) \)
- \( Y = (3, 2) \)
- \( Z = (4, 5) \)
Now, we will perform the dilation on each vertex.
-
Vertex X (1, 1): \[ X' = (1 - 2) \cdot 2 + 2 \cdot 1 = -2 + 2 = 0 \] \[ Y' = (1 - 2) \cdot 0 + 2 \cdot 1 = 0 + 2 = 2 \] \[ X' = (0, 2) \]
-
Vertex Y (3, 2): \[ X' = (1 - 2) \cdot 2 + 2 \cdot 3 = -2 + 6 = 4 \] \[ Y' = (1 - 2) \cdot 0 + 2 \cdot 2 = 0 + 4 = 4 \] \[ Y' = (4, 4) \]
-
Vertex Z (4, 5): \[ X' = (1 - 2) \cdot 2 + 2 \cdot 4 = -2 + 8 = 6 \] \[ Y' = (1 - 2) \cdot 0 + 2 \cdot 5 = 0 + 10 = 10 \] \[ Z' = (6, 10) \]
Resulting Coordinates
Thus, after performing the dilation on all vertices, the coordinates of the dilated triangle \( X'Y'Z' \) would be:
- \( X' = (0, 2) \)
- \( Y' = (4, 4) \)
- \( Z' = (6, 10) \)
If you provide the actual coordinates of triangle XYZ, I can perform the dilation with those specific values to get the resulting coordinates for triangle X'Y'Z'.
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