Triangle XYZ is drawn on a coordinate plane with vertices X (0,4), Y (2,0), and Z (4,7). If you dilate the triangle to form triangle X'Y'Z' using a magnitude of 1/2 and a center of dilation at (4,7), what are the coordinate points of new vertices?

To find the new coordinates of the vertices after dilation, we will use the formula for dilation:

\[ X' = C + k(X - C) \]

where:

• \(C\) is the center of dilation,
• \(k\) is the scale factor,
• \(X\) is the original point,
• \(X'\) is the dilated point.

In this case, the center of dilation \(C\) is at (4,7), and the scale factor \(k\) is \( \frac{1}{2} \).

Now let's apply this to each vertex of triangle XYZ.

Vertex X (0,4):

1. Calculate the difference from the center: \[ X - C = (0 - 4, 4 - 7) = (-4, -3) \]
2. Apply the dilation formula: \[ X' = (4, 7) + \frac{1}{2}(-4, -3) = (4, 7) + (-2, -1.5) = (4 - 2, 7 - 1.5) = (2, 5.5) \]

Vertex Y (2,0):

1. Calculate the difference from the center: \[ Y - C = (2 - 4, 0 - 7) = (-2, -7) \]
2. Apply the dilation formula: \[ Y' = (4, 7) + \frac{1}{2}(-2, -7) = (4, 7) + (-1, -3.5) = (4 - 1, 7 - 3.5) = (3, 3.5) \]

Vertex Z (4,7):

1. Since point Z is the center of dilation, the coordinates remain unchanged: \[ Z' = (4, 7) \]

Conclusion:

The new coordinates of triangle X'Y'Z' are:

• \(X' = (2, 5.5)\)
• \(Y' = (3, 3.5)\)
• \(Z' = (4, 7)\)

Reviewing the available choices, the closest set of dilated coordinates is option b: \(X' (2,5.5), Y' (3,3.5), Z' (4,7)\).