We present the resulting figure in the image attached.
The original figure is formed by the following vertices: [tex]A(x,y) = (6, 9)[/tex], [tex]B(x,y) = (7.5, 9)[/tex], [tex]C(x,y) = (7.5, 6)[/tex], [tex]D(x,y) = (9,6)[/tex], [tex]E(x,y) = (9, 4.5)[/tex] y [tex]F(x,y) = (6, 4.5)[/tex]. The image is found by the following transformation:
[tex]P'(x,y) = O(x,y)+k\cdot [P(x,y)-O(x,y)][/tex] (1)
Where:
- [tex]O(x,y)[/tex] - Centre of dilation.
- [tex]P(x,y)[/tex] - Original point.
- [tex]P'(x,y)[/tex] - Resulting point.
- [tex]k[/tex] - Dilation factor.
If we know that [tex]O(x,y) = (0,3)[/tex] and [tex]k = \frac{1}{3}[/tex], then the resulting points of the figure are, respectively:
[tex]A'(x,y) = (0,3) + \frac{1}{3}\cdot [(6,9)-(0,3)][/tex]
[tex]A'(x,y) = (0,3) + \frac{1}{3}\cdot (6,6)[/tex]
[tex]A'(x,y) = (0,3) +(2,2)[/tex]
[tex]A'(x,y) = (2,5)[/tex]
[tex]B'(x,y) = (0,3) + \frac{1}{3}\cdot [(7.5, 9)-(0,3)][/tex]
[tex]B'(x,y) = (0,3) +\frac{1}{3}\cdot (7.5, 6)[/tex]
[tex]B'(x,y) = (0,3) + (2.5, 2)[/tex]
[tex]B'(x,y) = (2.5, 5)[/tex]
[tex]C'(x,y) = (0,3) + \frac{1}{3}\cdot [(7.5,6)-(0,3)][/tex]
[tex]C'(x,y) = (0,3) +\frac{1}{3}\cdot (7.5, 3)[/tex]
[tex]C'(x,y) = (0,3) +(2.5, 1)[/tex]
[tex]C'(x,y) = (2.5, 4)[/tex]
[tex]D'(x,y) = (0,3) + \frac{1}{3}\cdot [(9,6)-(0,3)][/tex]
[tex]D'(x,y) = (0,3)+\frac{1}{3}\cdot (9, 3)[/tex]
[tex]D'(x,y) = (0,3) +(3, 1)[/tex]
[tex]D'(x,y) = (3, 4)[/tex]
[tex]E'(x,y) = (0,3)+\frac{1}{3}\cdot [(9, 4.5)-(0,3)][/tex]
[tex]E'(x,y) = (0,3) + \frac{1}{3} \cdot (9,1.5)[/tex]
[tex]E'(x,y) = (0,3) + (3, 0.5)[/tex]
[tex]E'(x,y) = (3, 3.5)[/tex]
[tex]F'(x,y) = (0,3) + \frac{1}{3}\cdot [(6, 4.5)-(0,3)][/tex]
[tex]F'(x,y) = (0,3) + \frac{1}{3} \cdot (6, 1.5)[/tex]
[tex]F'(x,y) = (0,3) +(2, 0.5)[/tex]
[tex]F'(x,y) = (2, 3.5)[/tex]
Finally, we present the resulting figure in the image attached.
We kindly invite to check this question on dilations: https://brainly.com/question/13176891
