trapezoid with vertices R(-13,7), S(-1,7), T(-1,-5), U(-13,-9)

The scale factor, k, equal a quarter. The center of dilation, denoted by (a, b), equals (-5, 3). Now we shall arrive at our formula through deduction:

[tex]\begin{aligned}&\left.D_{o, k}(x, y)=(k(x-a)+a), k(y-b)+b\right) \\&=\left(\frac{1}{4}(x-(-5))+(-5), \frac{1}{4}(y-3)+3\right) \\&\left.=\left(\frac{1}{4}(x+5)-5\right), \frac{1}{4}(y-3)+3\right) \\&\left.\left.=\left(\frac{1}{4} x+\frac{5}{4}\right)-5, \frac{1}{4} y-\frac{3}{4}\right)+3\right) \\&=\left(\frac{1}{4} x-4, \frac{1}{4} y+\frac{9}{4}\right)\end{aligned}[/tex]

This is how we arrived at our formula.

Now that we have this information, we can determine the coordinates needed to plot the deflated figure.

In order to determine what $R' is, we shall replace the vertices of R with the values (x, y).

[tex]\begin{aligned}&R^{\prime}=\left(\frac{1}{4}(-13)-4, \frac{1}{4}(7)\right. \\&=\left(\frac{-29}{4}, 4\right)=(-7.25,4)\end{aligned}[/tex]

Similarly,

[tex]$S^{\prime}=\left(\frac{1}{4}(-1)-4, \frac{1}{4}(7)+\frac{9}{4}\right)\\\\$$=\left(\frac{-17}{4}, 4\right)=(-4.25,4)\\\\$$T^{\prime}=\left(\frac{1}{4}(-1)-4, \frac{1}{4}(-5)+\frac{9}{4}\right)\\\\$$=\left(\frac{-17}{4}, 1\right)=(-4.25,1)\\\\$$U^{\prime}=\left(\frac{1}{4}(-13)-4, \frac{1}{4}(-9)+\frac{9}{4}\right)\\\\$[/tex]

In conclusion, R'(-7.25, 4) , S'(-4.25, 4) , T'(-4.25, 1) , U'(-7.25, 0) are the coordinates and hence our figure.