Question

Polygon ABCD with vertices at A(−4, 6), B(−2, 2), C(4, −2), D(4, 4) is dilated using a scale factor of one fourth to create polygon A′B′C′D′. Determine the vertices of polygon A′B′C′D′.

A′(−0.8, 1.2), B′(−0.4, 0.4), C′(0.8, −0.4), D′(0.8, 0.8)
A′(−1, 1.5), B′(−0.5, 0.5), C′(1, −0.5), D′(1, 1)
A′(−2, 3), B′(−1, 1), C′(2, −1), D′(2, 2)
A′(−3, 4.5), B′(−1.5, 1.5), C′(3, −1.5), D′(3, 3)

Answer 1

To find the vertices of polygon A'B'C'D' created by dilating polygon ABCD with a scale factor of \( \frac{1}{4} \), we need to apply the dilation transformation to each vertex of the original polygon.

The formula for dilation from the origin (0, 0) with a scale factor \( k \) is:

\[
(x', y') = (kx, ky)
\]

In our case, the scale factor \( k \) is \( \frac{1}{4} \).

Let's calculate the new vertices:

1. **Vertex A(−4, 6)**:
\[
A' = \left(-4 \cdot \frac{1}{4}, 6 \cdot \frac{1}{4}\right) = \left(-1, 1.5\right)
\]

2. **Vertex B(−2, 2)**:
\[
B' = \left(-2 \cdot \frac{1}{4}, 2 \cdot \frac{1}{4}\right) = \left(-0.5, 0.5\right)
\]

3. **Vertex C(4, −2)**:
\[
C' = \left(4 \cdot \frac{1}{4}, -2 \cdot \frac{1}{4}\right) = \left(1, -0.5\right)
\]

4. **Vertex D(4, 4)**:
\[
D' = \left(4 \cdot \frac{1}{4}, 4 \cdot \frac{1}{4}\right) = \left(1, 1\right)
\]

Now, we can summarize the new vertices:

- \( A' = (-1, 1.5) \)
- \( B' = (-0.5, 0.5) \)
- \( C' = (1, -0.5) \)
- \( D' = (1, 1) \)

Thus, the vertices of polygon A'B'C'D' are:

\[
A'(-1, 1.5), B'(-0.5, 0.5), C'(1, -0.5), D'(1, 1)
\]

Matching this with the provided options, we see that it corresponds to:

**A′(−1, 1.5), B′(−0.5, 0.5), C′(1, −0.5), D′(1, 1)**.