To determine the vertices of the dilated polygon $A'B'C'D'$, we apply the given scale factor of $\frac{4}{3}$ to each vertex of the original polygon $ABCD$.

Given vertices of polygon $ABCD$:

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

To calculate the vertices of polygon $A'B'C'D'$, multiply each coordinate by the scale factor $\frac{4}{3}$.

Calculating the new vertices:

1. Vertex A: $A' = \left(\frac{4}{3} \times -4, \frac{4}{3} \times 6\right) = \left(-\frac{16}{3}, 8\right)$

2. Vertex B: $B' = \left(\frac{4}{3} \times -2, \frac{4}{3} \times 2\right) = \left(-\frac{8}{3}, \frac{8}{3}\right)$

3. Vertex C: $C' = \left(\frac{4}{3} \times 4, \frac{4}{3} \times -2\right) = \left(\frac{16}{3}, -\frac{8}{3}\right)$

4. Vertex D: $D' = \left(\frac{4}{3} \times 4, \frac{4}{3} \times 4\right) = \left(\frac{16}{3}, \frac{16}{3}\right)$

Vertices of polygon $A'B'C'D'$:

• $A' \left(-\frac{16}{3}, 8\right)$
• $B' \left(-\frac{8}{3}, \frac{8}{3}\right)$
• $C' \left(\frac{16}{3}, -\frac{8}{3}\right)$
• $D' \left(\frac{16}{3}, \frac{16}{3}\right)$

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What happens to the perimeter of the polygon after the dilation?
2. How does the area of the polygon change with the dilation?
3. How do you determine the center of dilation in this problem?
4. Can you find the equation of the line passing through points $A$ and $B$?
5. How does a dilation affect the orientation of a polygon?

Tip: The dilation of a polygon multiplies all distances from the center of dilation by the scale factor, preserving the shape but changing the size.