Explanation
Graphing the polygon
We plot and join the given ordered pairs.
Graphing the image of the polygon after the dilation
The formula of dilation when it is not centred at the origin is:
[tex]\begin{gathered} (x,y)\rightarrow(k(x-a)+a,k(y-b)+b) \\ \text{ Where} \\ k\text{ is the scale factor} \\ (a,b)\text{ is the center of the dilation} \end{gathered}[/tex]
Then, we can find the coordinates of the image:
[tex]\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(7-(-2))-2,\frac{2}{3}(1-4)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(7+2)-2,\frac{2}{3}(-3)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(\frac{2}{3}(9)-2,\frac{2}{3}(-3)+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(6-2,-2+4) \\ T(7,1)\operatorname{\rightarrow}T^{\prime}(4,2) \end{gathered}[/tex][tex]\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(4-(-2))-2,\frac{2}{3}(4-4)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(4+2)-2,\frac{2}{3}(4-4)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(\frac{2}{3}(6)-2,\frac{2}{3}(0)+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(4-2,0+4) \\ U(4,4)\operatorname{\rightarrow}U^{\prime}(2,4) \end{gathered}[/tex]
[tex]\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1-(-2))-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1+2)-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(3)-2,\frac{2}{3}(9)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(2-2,6+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(0,10) \end{gathered}[/tex][tex][/tex]
