Question

If you dilate triangle abc to form xyz using a magnitude of 2 and a canter of dilation at (2,0), what are the coordinates of the dilated triangle

Answer 1

To dilate triangle ABC to form triangle XYZ using a magnitude of 2 and a center of dilation at the point (2, 0), you'll follow these steps for each vertex of triangle ABC.

### Dilation Formula

The formula for dilating a point \( P(x, y) \) from a center of dilation \( C(cx, cy) \) with a scale factor \( k \) is given by:

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

Where:
- \( P' \) is the new point after dilation.
- \( C \) is the center of dilation.
- \( k \) is the scale factor.
- \( P \) is the original point.

### Steps

1. **Identify the vertices of triangle ABC**: Let's denote the vertices of triangle ABC as \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \).
2. **Apply the dilation formula** to each vertex.

### Example Calculation

Let's assume the vertices of triangle ABC are:
- \( A(1, 1) \)
- \( B(3, 1) \)
- \( C(2, 2) \)

The center of dilation is \( C(2, 0) \) and the scale factor \( k = 2 \).

**For point A(1, 1)**:
\[
P' = (2, 0) + 2((1, 1) - (2, 0))
\]
\[
= (2, 0) + 2(-1, 1)
\]
\[
= (2, 0) + (-2, 2) = (0, 2)
\]

**For point B(3, 1)**:
\[
P' = (2, 0) + 2((3, 1) - (2, 0))
\]
\[
= (2, 0) + 2(1, 1)
\]
\[
= (2, 0) + (2, 2) = (4, 2)
\]

**For point C(2, 2)**:
\[
P' = (2, 0) + 2((2, 2) - (2, 0))
\]
\[
= (2, 0) + 2(0, 2)
\]
\[
= (2, 0) + (0, 4) = (2, 4)
\]

### Coordinates of Triangle XYZ

After applying the dilation to all vertices, the coordinates of triangle XYZ will be:
- \( X(0, 2) \)
- \( Y(4, 2) \)
- \( Z(2, 4) \)

You can apply this method to any points of triangle ABC to get the corresponding dilated triangle.