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
![(x,y)\to (y,-x)](/tpl/images/0924/0185/a8dc4.png)
![C(4,6)\to C_1(6,-4)](/tpl/images/0924/0185/25d87.png)
![D(-1,6)\to D_1(6,-(-1))=D_1(6,1)](/tpl/images/0924/0185/a3d2a.png)
If figure translated 4 units left and 2 units down, then
![(x,y)\to (x-4,y-2)](/tpl/images/0924/0185/23710.png)
![C_1(6,-4)\to C'(6-4,-4-2)=C'(2,-6)](/tpl/images/0924/0185/2217e.png)
![D_1(6,1)\to D'(6-4,1-2)=D'(2,-1)](/tpl/images/0924/0185/2ba42.png)
Distance formula:
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](/tpl/images/0924/0185/c9678.png)
Distance between C'(2,-6) and D'(2,-1) is
![C'D'=\sqrt{(2-2)^2+(-1-(-6))^2}](/tpl/images/0924/0185/a64f2.png)
![C'D'=\sqrt{(0)^2+(-1+6)^2}](/tpl/images/0924/0185/4236e.png)
![C'D'=\sqrt{(5)^2}](/tpl/images/0924/0185/7da83.png)
![C'D'=5](/tpl/images/0924/0185/c085c.png)
Therefore, the distance between C'D' is 5 units.