The image of the resulting figure (quadrilateral A'B'C'D') is presented below as attachment.
How to graph a dilated quadrilateral
According to geometry and linear algebra, a dilation of a point [tex]P(x,y)[/tex] with respect to another point [tex]G(x,y)[/tex] is determined by the following expression:
[tex]P'(x,y) = P(x,y) +k\cdot [P(x,y) - G(x,y)][/tex] (1)
Where:
Now we proceed to make the corresponding operations ([tex]A(x,y) = (-1, 3)[/tex], [tex]B(x,y) = (2,0)[/tex], [tex]C(x,y) = (2, -1)[/tex], [tex]D(x,y) = (-3, -1)[/tex]):
[tex]A'(x,y) = (-1,3)+(-0.5)\cdot [(-1,3)-(2,1)][/tex]
[tex]A'(x,y) = (-1, 3) + (-0.5)\cdot (-3, 2)[/tex]
[tex]A'(x,y) = (-1, 3) + (1.5, -1)[/tex]
[tex]A'(x,y) = (0.5, 2)[/tex]
[tex]B'(x,y) = (2,0) + (-0.5)\cdot [(2,0)-(2,1)][/tex]
[tex]B'(x,y) = (2,0) + (-0.5)\cdot (0, -1)[/tex]
[tex]B'(x,y) = (2,0) + (0, 0.5)[/tex]
[tex]B'(x,y) = (2, 0.5)[/tex]
[tex]C'(x,y) = (2, -1) + (-0.5)\cdot [(2,-1)-(2,1)][/tex]
[tex]C'(x,y) = (2, -1) + (-0.5)\cdot (0,-2)[/tex]
[tex]C'(x,y) = (2,-1) + (0, 1)[/tex]
[tex]C'(x,y) = (2, 0)[/tex]
[tex]D'(x,y) = (-3,-1) + (-0.5)\cdot [(-3, -1)-(2,1)][/tex]
[tex]D'(x,y) = (-3,-1) + (-0.5)\cdot (-5, -2)[/tex]
[tex]D'(x,y) = (-3, -1) + (2.5, 1)[/tex]
[tex]D'(x,y) = (-0.5, 0)[/tex]
Lastly, we proceed to graph both the original and resulting figure. [tex]\blacksquare[/tex]
The statement is incomplete. Complete description is shown below:
Draw the image of quadrilateral ABCD ([tex]A(x,y) = (-1, 3)[/tex], [tex]B(x,y) = (2,0)[/tex], [tex]C(x,y) = (2, -1)[/tex], [tex]D(x,y) = (-3, -1)[/tex]) under the dilation with scale factor -0.5 and center of dilation [tex]G(x,y) = (2, 1)[/tex].
To learn more on quadrilaterals, we kindly invite to check this verified question: https://brainly.com/question/25240753
