1. The problem states that a figure is dilated by a factor of $\frac{3}{2}$ centered at the origin. We need to find the coordinates of the resulting image after dilation. 2. The formula for dilation centered at the origin is: $$ (x', y') = \left(kx, ky\right) $$ where $k$ is the dilation factor, and $(x, y)$ are the original coordinates. 3. The original points are: - $G(-4, 0)$ - $J(0, 6)$ - $I(4, 0)$ - $H(0, -4)$ 4. Applying the dilation factor $k = \frac{3}{2}$ to each point: - $G': \left(\frac{3}{2} \times -4, \frac{3}{2} \times 0\right) = \left(-6, 0\right)$ - $J': \left(\frac{3}{2} \times 0, \frac{3}{2} \times 6\right) = \left(0, 9\right)$ - $I': \left(\frac{3}{2} \times 4, \frac{3}{2} \times 0\right) = \left(6, 0\right)$ - $H': \left(\frac{3}{2} \times 0, \frac{3}{2} \times -4\right) = \left(0, -6\right)$ 5. The resulting image is a diamond-shaped figure with vertices at $(-6, 0)$, $(0, 9)$, $(6, 0)$, and $(0, -6)$. Final answer: The dilated figure has vertices $G'(-6, 0)$, $J'(0, 9)$, $I'(6, 0)$, and $H'(0, -6)$.