Question

Polygon ABCD with vertices at A(−4, 6), B(−2, 2), C(4, −2), and D(4, 4) is dilated using a scale factor of three fifths to create polygon A′B′C′D′. If the dilation is centered at the origin, determine the vertices of polygon A′B′C′D′.

A′(5.8, −3), B′(1.6, −1.5), C′(−1.6, 3), D′(2.5, 3)
A′(−12, 18), B′(−6, 6), C′(12, −6), D′(12, 12)
A′(2.4, −3.6), B′(1.2, −1.2), C′(−2.4, 1.26), D′(−2.4, −2.4)
A′(−2.4, 3.6), B′(−1.2, 1.2), C′(2.4, −1.2), D′(2.4, 2.4)

Answer 1

To determine the vertices of the dilated polygon \( A'B'C'D' \) from the original polygon \( ABCD \), we apply the dilation centered at the origin with a scale factor of \( \frac{3}{5} \).

The formula for dilating a point \( (x, y) \) by a factor \( k \) centered at the origin is given by:
\[
(x', y') = (k \cdot x, k \cdot y)
\]

Here, \( k = \frac{3}{5} \). We will apply this formula to each vertex of polygon \( ABCD \).

1. **Vertex A (-4, 6)**:
\[
A' = \left(\frac{3}{5} \cdot (-4), \frac{3}{5} \cdot 6\right) = \left(-\frac{12}{5}, \frac{18}{5}\right) = \left(-2.4, 3.6\right)
\]

2. **Vertex B (-2, 2)**:
\[
B' = \left(\frac{3}{5} \cdot (-2), \frac{3}{5} \cdot 2\right) = \left(-\frac{6}{5}, \frac{6}{5}\right) = \left(-1.2, 1.2\right)
\]

3. **Vertex C (4, -2)**:
\[
C' = \left(\frac{3}{5} \cdot 4, \frac{3}{5} \cdot (-2)\right) = \left(\frac{12}{5}, -\frac{6}{5}\right) = \left(2.4, -1.2\right)
\]

4. **Vertex D (4, 4)**:
\[
D' = \left(\frac{3}{5} \cdot 4, \frac{3}{5} \cdot 4\right) = \left(\frac{12}{5}, \frac{12}{5}\right) = \left(2.4, 2.4\right)
\]

So the vertices of polygon \( A'B'C'D' \) after dilation are:

- \( A'(-2.4, 3.6) \)
- \( B'(-1.2, 1.2) \)
- \( C'(2.4, -1.2) \)
- \( D'(2.4, 2.4) \)

Matching these results with the options available:

- A′(−2.4, 3.6), B′(−1.2, 1.2), C′(2.4, −1.2), D′(2.4, 2.4)

Thus, the correct vertices of polygon \( A'B'C'D' \) are:

**A′(−2.4, 3.6), B′(−1.2, 1.2), C′(2.4, −1.2), D′(2.4, 2.4)**