Given:
The vertices of a parallelogram are A (-2,3), B (3,3), C (4,6), and D(-1,6).
It is first rotated 90 degrees clockwise and then translated 4 units left and 2 units down to form quadrilateral A'B'C'D.
To find:
The distance between C' and D'.
Solution:
If a figure rotated 90 degrees clockwise, then
[tex](x,y)\to (y,-x)[/tex]
[tex]C(4,6)\to C_1(6,-4)[/tex]
[tex]D(-1,6)\to D_1(6,-(-1))=D_1(6,1)[/tex]
If figure translated 4 units left and 2 units down, then
[tex](x,y)\to (x-4,y-2)[/tex]
[tex]C_1(6,-4)\to C'(6-4,-4-2)=C'(2,-6)[/tex]
[tex]D_1(6,1)\to D'(6-4,1-2)=D'(2,-1)[/tex]
Distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Distance between C'(2,-6) and D'(2,-1) is
[tex]C'D'=\sqrt{(2-2)^2+(-1-(-6))^2}[/tex]
[tex]C'D'=\sqrt{(0)^2+(-1+6)^2}[/tex]
[tex]C'D'=\sqrt{(5)^2}[/tex]
[tex]C'D'=5[/tex]
Therefore, the distance between C'D' is 5 units.
