math - Drawing a Rotated Rectangle

Question

I realize this might be more of a math problem.

To draw the lines for my rectangles I need to solve for their corners. I have a rectangle center at (x,y) With a defined Width and Height.

To find the blue points on a non rotated rectangle on top (angle = 0) It is

UL = (x-Width/2),(y+height/2)
UR = (x+Width/2),(y+height/2)
LR = (x+Width/2),(y-height/2)
LL = (x-Width/2),(y-height/2)

How do I find the points if the angle isn't 0?

Thanks in advance.

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Update: although I have (0,0) in my picture as the center point most likely the center point won't be at that location.

Answer 1

First transform the centre point to 0,0

X' = X-x

Y' = Y-y

Then rotate for an angle of A

X'' = (X-x) * cos A - (Y-y) * sin A

Y'' = (Y-y) * cos A + (X-x) * sin A

Again transform back the centre point to x,y

X''' = (X-x) * cos A - (Y-y) * sin A + x

Y''' = (Y-y) * cos A + (X-x) * sin A + y

Hence compute for all 4 points of (X,Y) with following transformation

X''' = (X-x) * cos A - (Y-y) * sin A + x

Y''' = (Y-y) * cos A + (X-x) * sin A + y

where x, y are the centre points of rectangle and X,Y are the corner points You have n't defined correctly even the corner points when Angle is 0 as I have given in the comments.

After substituting you will get

UL  =  x + ( Width / 2 ) * cos A - ( Height / 2 ) * sin A ,  y + ( Height / 2 ) * cos A  + ( Width / 2 ) * sin A
UR  =  x - ( Width / 2 ) * cos A - ( Height / 2 ) * sin A ,  y + ( Height / 2 ) * cos A  - ( Width / 2 ) * sin A
BL =   x + ( Width / 2 ) * cos A + ( Height / 2 ) * sin A ,  y - ( Height / 2 ) * cos A  + ( Width / 2 ) * sin A
BR  =  x - ( Width / 2 ) * cos A + ( Height / 2 ) * sin A ,  y - ( Height / 2 ) * cos A  - ( Width / 2 ) * sin A

I think this suits your solution.