Given:
Quadrilateral ABCD has vertices A(–1, –2), B(–1, 3), C(4, 3) and D(4, –2).
It's dilated by a factor of 2 with the center of dilation at the origin.
To find:
The coordinates of the resulting quadrilateral A'B'C'D'.
Solution:
The figure dilated by a factor of 2 with the center of dilation at the origin. So, the rule of dilation is
[tex](x,y)\to (2x,2y)[/tex]
Using the above rule, we get
[tex]A(-1,-2)\to A'(2(-1),2(-2))=A'(-2,-4)[/tex]
[tex]B(-1,3)\to B'(2(-1),2(3))=B'(-2,6)[/tex]
[tex]C(4,3)\to C'(2(4),2(3))=C'(8,6)[/tex]
[tex]D(4,-2)\to D'(2(4),2(-2))=D'(8,-4)[/tex]
Therefore, the coordinates of the resulting quadrilateral A'B'C'D' are A'(-2,-4), B'(-2,6), C'(8,6) and D'(8,-4).
