In order to find the new coordinates for the dilated figure, we just need to multiply the coordinates of each point by the scale factor. So we have the following:
[tex]\begin{gathered} I(0,4)\to I^{\prime}(0\cdot\frac{3}{2},4\cdot\frac{3}{2})=I^{\prime}(0,6) \\ J(4,-2)\to J^{\prime}(4\cdot\frac{3}{2},-2\cdot\frac{3}{2})=J^{\prime}(6,-3) \\ K(-4,-2)\to K^{\prime}(-4\cdot\frac{3}{2},-2\cdot\frac{3}{2})=K^{\prime}(-6,-3) \end{gathered}[/tex]
So the coordinates of the points in the dilated figure are I'(0, 6), J'(6, -3) and K'(-6, -3).
