Question

Graph the polygon and it’s image after a dilation centered C with scale factor k

Answer 1

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:

$\begin{gathered} (x,y)\rightarrow(k(x-a)+a,k(y-b)+b) \\ \text{ Where} \\ k\text{ is the scale factor} \\ (a,b)\text{ is the center of the dilation} \end{gathered}$

Then, we can find the coordinates of the image:

$\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}$$\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}$

$\begin{gathered} k=\frac{2}{3} \\ (a,b)=C(-2,4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1-(-2))-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(1+2)-2,\frac{2}{3}(13-4)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(\frac{2}{3}(3)-2,\frac{2}{3}(9)+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(2-2,6+4) \\ V(1,13)\operatorname{\rightarrow}V^{\prime}(0,10) \end{gathered}$