1. **State the problem:** We are given the original vertices of a triangle and the vertices of a new triangle after a transformation. We need to determine which rule was applied to the original triangle to create the new triangle. 2. **List the original and new vertices:** Original: $(-6, 9), (3, -6), (-9, 12)$ New: $(-4, 6), (2, -4), (-6, 8)$ 3. **Check each transformation rule:** - Rule 1: $(x, y) \to \left(\frac{3}{2}x, \frac{3}{2}y\right)$ - Rule 2: $(x, y) \to \left(\frac{2}{3}x, \frac{2}{3}y\right)$ - Rule 3: $(x, y) \to (x + 2, y - 3)$ - Rule 4: $(x, y) \to (x - 1, y + 2)$ 4. **Test Rule 1 on the first vertex:** $$\left(\frac{3}{2} \times -6, \frac{3}{2} \times 9\right) = (-9, 13.5)$$ This does not match $(-4, 6)$. 5. **Test Rule 2 on the first vertex:** $$\left(\frac{2}{3} \times -6, \frac{2}{3} \times 9\right) = (-4, 6)$$ This matches the new vertex. 6. **Test Rule 2 on the second vertex:** $$\left(\frac{2}{3} \times 3, \frac{2}{3} \times -6\right) = (2, -4)$$ Matches the new vertex. 7. **Test Rule 2 on the third vertex:** $$\left(\frac{2}{3} \times -9, \frac{2}{3} \times 12\right) = (-6, 8)$$ Matches the new vertex. 8. **Conclusion:** The transformation rule applied is scaling by $\frac{2}{3}$ with center of dilation at $(0,0)$. **Final answer:** The rule applied is $(x, y) \to \left(\frac{2}{3}x, \frac{2}{3}y\right)$.