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 one eighth to create polygon A′B′C′D′. If the dilation is centered at the origin, determine the vertices of polygon A′B′C′D′.

Question

A′(3.5, −5.25), B′(1.75, −1.75), C′(−3.5, 1.75), D′(−3.5, −3.5)
A′(3.2, −4.8), B′(1.6, −1.6), C′(−3.2, 1.6), D′(3.2, 3.2)
A′(−0.5, 0.75), B′(−0.25, 0.25), C′(0.5, −0.25), D′(0.5, 0.5)
A′(−12, 14), B′(−10, 10), C′(12, −14), D′(12, 12)

Answer 1

To find the vertices of polygon \( A'B'C'D' \) after dilating polygon \( ABCD \) centered at the origin by a scale factor of \( \frac{1}{8} \), we can use the formula for dilation:

\[ A' = k \cdot A \]

where \( k \) is the scale factor and \( A \) is the original vertex coordinates.

Given the vertices of polygon \( ABCD \):

• \( A(-4, 6) \)
• \( B(-2, 2) \)
• \( C(4, -2) \)
• \( D(4, 4) \)

Using the scale factor \( k = \frac{1}{8} \), we will calculate the new vertices:

1. Vertex A': \[ A' = \left(-4 \cdot \frac{1}{8}, 6 \cdot \frac{1}{8}\right) = \left(-\frac{4}{8}, \frac{6}{8}\right) = \left(-\frac{1}{2}, \frac{3}{4}\right) = \left(-0.5, 0.75\right) \]

2. Vertex B': \[ B' = \left(-2 \cdot \frac{1}{8}, 2 \cdot \frac{1}{8}\right) = \left(-\frac{2}{8}, \frac{2}{8}\right) = \left(-\frac{1}{4}, \frac{1}{4}\right) = \left(-0.25, 0.25\right) \]

3. Vertex C': \[ C' = \left(4 \cdot \frac{1}{8}, -2 \cdot \frac{1}{8}\right) = \left(\frac{4}{8}, -\frac{2}{8}\right) = \left(\frac{1}{2}, -\frac{1}{4}\right) = \left(0.5, -0.25\right) \]

4. Vertex D': \[ D' = \left(4 \cdot \frac{1}{8}, 4 \cdot \frac{1}{8}\right) = \left(\frac{4}{8}, \frac{4}{8}\right) = \left(\frac{1}{2}, \frac{1}{2}\right) = \left(0.5, 0.5\right) \]

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

• \( A'(-0.5, 0.75) \)
• \( B'(-0.25, 0.25) \)
• \( C'(0.5, -0.25) \)
• \( D'(0.5, 0.5) \)

Thus, the correct choice from the options provided is:

A′(−0.5, 0.75), B′(−0.25, 0.25), C′(0.5, −0.25), D′(0.5, 0.5)