To find the scale factor let's use the fact that
[tex]k\cdot\overline{SU}=\overline{S'U'}[/tex]
Therefore let's find SU and S'U' length
[tex]\begin{gathered} \overline{SU}=\sqrt[]{4^2+8^2} \\ \\ \overline{SU}=\sqrt[]{80} \end{gathered}[/tex]
And S'U'
[tex]\begin{gathered} \overline{S^{\prime}U^{\prime}}=\sqrt[]{2^2+4^2} \\ \\ \overline{S^{\prime}U^{\prime}}=\sqrt[]{20} \end{gathered}[/tex]
Using our first equation:
[tex]\begin{gathered} k\cdot\overline{SU}=\overline{S^{\prime}U^{\prime}}\Rightarrow k=\frac{\overline{S^{\prime}U^{\prime}}}{\overline{SU}} \\ \\ k=\frac{\sqrt[]{20}}{\overline{\sqrt[]{80}}} \\ \\ k=\sqrt[]{\frac{1}{4}} \\ \\ k=\frac{1}{2} \end{gathered}[/tex]
And to find the center of dilatation we can use the equation:
[tex]\begin{gathered} T=(x_1,y_1) \\ T^{\prime}=(x_2,y_2) \\ \\ x_0=\frac{kx_1-x_2}{k-1} \\ \\ y_0=\frac{ky_1-y_2}{k-1} \end{gathered}[/tex]
Where (x₀, y₀) is the center of dilatation, using the point T a reference we get
[tex]\begin{gathered} T=(9_{},0) \\ T^{\prime}=(1_{},4) \\ \\ x_0=\frac{\frac{1}{2}\cdot9_{}-1}{\frac{1}{2}-1}=-7 \\ \\ y_0=\frac{\frac{1}{2}\cdot0-4}{\frac{1}{2}-1}=8 \end{gathered}[/tex]
Then, the center of dilatation is
[tex](-7,8)[/tex]
