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 the new vertices?

Answer 1

To find the coordinates of the new vertices of triangle XYZ after dilating triangle XYZ with a center of dilation at point \( C(4,7) \) and a magnitude of \( \frac{1}{2} \), we can use the formula for dilation:

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

where:

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

Let's apply this to each vertex:

Vertex X (0, 4)

1. Find the vector from C to X: \[ X - C = (0,4) - (4,7) = (0 - 4, 4 - 7) = (-4, -3) \]

2. Scale the vector by \( \frac{1}{2} \): \[ \frac{1}{2}(-4, -3) = \left(-2, -\frac{3}{2}\right) \]

3. Find the new coordinate: \[ X' = C + k(X - C) = (4, 7) + \left(-2, -\frac{3}{2}\right) = (4 - 2, 7 - \frac{3}{2}) = (2, 7 - 1.5) = (2, 5.5) \]

Vertex Y (2, 0)

1. Find the vector from C to Y: \[ Y - C = (2,0) - (4,7) = (2 - 4, 0 - 7) = (-2, -7) \]

2. Scale the vector by \( \frac{1}{2} \): \[ \frac{1}{2}(-2, -7) = \left(-1, -\frac{7}{2}\right) \]

3. Find the new coordinate: \[ Y' = C + k(Y - C) = (4, 7) + \left(-1, -\frac{7}{2}\right) = (4 - 1, 7 - \frac{7}{2}) = (3, 7 - 3.5) = (3, 3.5) \]

Vertex Z (4, 7)

1. Find the vector from C to Z: \[ Z - C = (4,7) - (4,7) = (0,0) \]

2. Scale the vector by \( \frac{1}{2} \): \[ \frac{1}{2}(0,0) = (0,0) \]

3. Find the new coordinate (since it remains the same): \[ Z' = C + k(Z - C) = (4, 7) + (0, 0) = (4, 7) \]

Summary of new vertices

The coordinates of the vertices of triangle \( X'Y'Z' \) after the dilation are:

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

Thus, the final points of the new triangle are:

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