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Given C(2,8),D(6,4),E(0,4),U(1,4),V(3,2),C(2,-8),D(-6,4),E(0,4),U(1,-4),V(-3,2), and \newlineW(0,2)W(0,2), and that CDE\angle CDE is the preimage of UVW\angle UVW, represent the transformation algebraically.\newline(x,y)(x,y)(x,y)\rightarrow(\square x,\square y)

Full solution

Q. Given C(2,8),D(6,4),E(0,4),U(1,4),V(3,2),C(2,-8),D(-6,4),E(0,4),U(1,-4),V(-3,2), and \newlineW(0,2)W(0,2), and that CDE\angle CDE is the preimage of UVW\angle UVW, represent the transformation algebraically.\newline(x,y)(x,y)(x,y)\rightarrow(\square x,\square y)
  1. Analyze Coordinates: To solve this problem, we need to find the transformation that maps triangle CDECDE to triangle UVWUVW. We will start by comparing the coordinates of the corresponding points to determine the transformation.
  2. X-Coordinate Comparison: Let's first look at the x-coordinates of the corresponding points. We can compare the x-coordinates of CC and UU, DD and VV, EE and WW to find a pattern.\newlineC(2)U(1)C(2) \rightarrow U(1), D(6)V(3)D(-6) \rightarrow V(-3), E(0)W(0)E(0) \rightarrow W(0).\newlineIt seems that the x-coordinate of the image is half the x-coordinate of the preimage. This suggests a scaling transformation with a scale factor of 12\frac{1}{2} in the x-direction.
  3. Y-Coordinate Comparison: Now let's look at the y-coordinates of the corresponding points. We can compare the y-coordinates of CC and UU, DD and VV, EE and WW to find a pattern.\newlineC(8)U(4)C(-8) \rightarrow U(-4), D(4)V(2)D(4) \rightarrow V(2), E(4)W(2)E(4) \rightarrow W(2).\newlineIt seems that the y-coordinate of the image is half the y-coordinate of the preimage. This suggests a scaling transformation with a scale factor of 12\frac{1}{2} in the y-direction as well.
  4. Algebraic Representation: Since both the xx and yy-coordinates are scaled by a factor of 12\frac{1}{2}, the transformation can be represented algebraically as:\newline(x,y)(12×x,12×y)(x, y) \rightarrow (\frac{1}{2} \times x, \frac{1}{2} \times y).\newlineThis transformation scales any point by a factor of 12\frac{1}{2} in both the xx and yy directions.
  5. Transformation Verification: To check for any math errors, we can apply the transformation to one of the points in triangle CDECDE and see if we get the corresponding point in triangle UVWUVW. Let's apply it to point C(2,8)C(2, -8): (12×2,12×8)=(1,4)(\frac{1}{2} \times 2, \frac{1}{2} \times -8) = (1, -4), which is point UU, as expected. No math errors are detected.