Respuesta :
Given data:
The given coordinates of the pentagon are M(-4,1),N(-2,3). O(0, 3), P(4, 3), and Q(2,-7).
The coordinatte after 90 degrees counterclockwise rotation is,
[tex]\begin{gathered} M(-4,1)\rightarrow M^{\prime}(-1,\text{ -4)} \\ N(-2,\text{ 3)}\rightarrow N^{\prime}(-3,\text{ -2)} \\ O(0,\text{ 3)}\rightarrow O^{\prime}(-3,\text{ 0)} \\ P(4,\text{ 3)}\rightarrow P^{\prime}(-3,\text{ 4)} \\ Q(2,\text{ -7)}\rightarrow Q^{\prime}(7,\text{ 2)} \end{gathered}[/tex]The final coordinates after 2/3 dilation factor is,
[tex]\begin{gathered} M^{\doubleprime}\rightarrow\frac{2}{3}(-1,\text{ -4)} \\ \rightarrow(-\frac{2}{3},\text{ - }\frac{8}{3}) \\ N^{\prime\prime}\rightarrow\frac{2}{3}(-3,\text{ -2)} \\ \rightarrow(-2,\text{ -}\frac{4}{3}) \\ O^{\doubleprime}\rightarrow\frac{2}{3}(-3,\text{ 0)} \\ \rightarrow(-2,\text{ 0)} \\ P^{\doubleprime}\rightarrow\frac{2}{3}(-3,\text{ 4)} \\ \rightarrow(-2,\frac{8}{3})^{} \\ Q^{\doubleprime}\rightarrow\frac{2}{3}(7,\text{ 2)} \\ \rightarrow(\frac{14}{3},\text{ }\frac{4}{3}) \end{gathered}[/tex]Thus, the final coordinates after transformation are M''(-2/3, -8/3), N''(-2, -4/3). O''(-2, 0), P''(-2, 8/3), and Q''(14/3, 4/3).
