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ROTATION
1. ROTATIONJ FORMULA OVER O(0,0) .
    Let A(x,y) any point in plane V ang A’(x’,y’)is
image of A over   R 0, , or A’ = R 0, (A).
Let m(XOA)=  .
We have x =OA cos dan y = OA sin and
              x’ = OA’ cos (+)
                   = OA (cos cos - sin sin)
                   = x cos  - y sin 
A’(x’,y’)




                        A(x,y)
        
                

(0,0)
.
•   y’ = OA’ sin (+)
•      = OA(sin  cos  + cos  sin )
•      = x sin  + y cos 
•   so
•    x’ = xcos  - y sin 
•   y’ = x sin  + y cos 
•   or

       x' cos       sin    x 
            
       y ' sin               y
                    cos    
C’(x’,y’)=C’(x*’,y*’)
            y*


                                   C(x,y)=C(x*,y*)
                 (a,b)
                             x*

(0,0)                                 X

        Y
2. ROTATION OVER P(a,b)

• Let we have coordinate system with
  centre P(a,b)and has two axis X*
  and Y*, X//X* and Y//Y*.
• If C(x*,y*) and C’=RP,(C), then C’
  (x*’,y*’) , we have a relation :




     x*'  cos     sin    x *
           
     y*' sin 
                  cos    y *
                              
In coordinate of X , Y axis , we have :


 x'a  cos   sin    x  a 
        
 y 'b sin  cos    y  b
                             
 x' cos   sin    x   p 
      
 y ' sin  cos    y  q   
                        
         p  a cos   b sin   a
        q  a sin   b cos   b
THEOREM
 • Rotation RP, can represent in
   composition of two lines
   reflection over s and t with P is
   (s,t) and m(<(s,t))=½ .

 • Rotation is an isometry

 • composition         of    two   lines
   reflection :



          SAB            ,if s//t
M t Ms  
          R P,θ , if t and s not paralel
A”

              t
    T


             A’

         Q            s
P                 A
Theorem
            1  R
          R P,    P, -



     Theorem

      R P, R P,   R P,  
•If s perpendicular to t and P=(s,t) ,
•then MtMs=HP.

                   s


                             A”
               P                         E
t
     A             D         A’
.
       For every line a,b with
• Teorema
 a//b, then MbMa=SCD with |CD|=2 x
 distance (a,b) and CD a.
                P’
      P                      P’’



                B
            A
                         D



            a
                 P   b
Mb Ma = Mb I Ma
      = Mb (MsMs )Ma
       = Mb MsMs Ma
       = (Mb Ms )(Ms Ma)
       = HBHA
        = SCD with |CD|= 2 |AB|
•Translation SAB can represent as
composition of two reflection Ms dan Mt
with s//t and s  AB, and distance of
(s,t) is ½ |AB|.

                        t
                                   B
             s




        A
• Given three paralel lines a, b dan c.
• Construct an equilateral triangle ABC
  with condition A on a, B on b and C on c.



          a


                    b



                          c
Contoh permasalahan



                a
    A

                    b
            B
                      c



    C
• a. Fixed any point A on a.
• b. Rotated line c, with angle 60o over A, we got c’.
• c. Intersection of line c’ and line b, ( c’,b) is point
  B.
• d. We can construct equilateral triangle ABC.
•
• We can also start with fixed point B on b or C on
  c.

• Can do it ?
•   a. Fixed any point B on b.
•   b. Rotated line a, with angle 60o over B, we got a’.
•   c. Intersect of line a’ and line c, ( a’,c) is point C .
•   d. We can construct equilqteral triangle ABC.
•
•   We can also start with fixed point A on a or C on
    c.

• Can do it ?

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Yes, we can construct an equilateral triangle ABC with the given conditions using rotations. Starting from any vertex on one of the lines and performing 60° rotations about that point onto the intersecting lines will yield the other two vertices of an equilateral triangle

  • 2.
  • 3.
  • 4.
  • 5.
  • 6. 1. ROTATIONJ FORMULA OVER O(0,0) . Let A(x,y) any point in plane V ang A’(x’,y’)is image of A over R 0, , or A’ = R 0, (A). Let m(XOA)=  . We have x =OA cos dan y = OA sin and x’ = OA’ cos (+) = OA (cos cos - sin sin) = x cos  - y sin 
  • 7. A’(x’,y’) A(x,y)   (0,0)
  • 8. . • y’ = OA’ sin (+) • = OA(sin  cos  + cos  sin ) • = x sin  + y cos  • so • x’ = xcos  - y sin  • y’ = x sin  + y cos  • or  x' cos   sin    x    y ' sin   y    cos    
  • 9. C’(x’,y’)=C’(x*’,y*’) y* C(x,y)=C(x*,y*) (a,b) x* (0,0) X Y
  • 10. 2. ROTATION OVER P(a,b) • Let we have coordinate system with centre P(a,b)and has two axis X* and Y*, X//X* and Y//Y*. • If C(x*,y*) and C’=RP,(C), then C’ (x*’,y*’) , we have a relation :  x*'  cos   sin    x *   y*' sin     cos    y *  
  • 11. In coordinate of X , Y axis , we have :  x'a  cos   sin    x  a    y 'b sin  cos    y  b       x' cos   sin    x   p    y ' sin  cos    y  q          p  a cos   b sin   a q  a sin   b cos   b
  • 12. THEOREM • Rotation RP, can represent in composition of two lines reflection over s and t with P is (s,t) and m(<(s,t))=½ . • Rotation is an isometry • composition of two lines reflection :  SAB ,if s//t M t Ms    R P,θ , if t and s not paralel
  • 13.
  • 14. A” t T A’ Q s P A
  • 15. Theorem 1  R R P, P, - Theorem R P, R P,   R P,  
  • 16.
  • 17. •If s perpendicular to t and P=(s,t) , •then MtMs=HP. s A” P E t A D A’
  • 18. . For every line a,b with • Teorema a//b, then MbMa=SCD with |CD|=2 x distance (a,b) and CD a. P’ P P’’ B A D a P b
  • 19. Mb Ma = Mb I Ma = Mb (MsMs )Ma = Mb MsMs Ma = (Mb Ms )(Ms Ma) = HBHA = SCD with |CD|= 2 |AB|
  • 20. •Translation SAB can represent as composition of two reflection Ms dan Mt with s//t and s  AB, and distance of (s,t) is ½ |AB|. t B s A
  • 21. • Given three paralel lines a, b dan c. • Construct an equilateral triangle ABC with condition A on a, B on b and C on c. a b c
  • 22. Contoh permasalahan a A b B c C
  • 23. • a. Fixed any point A on a. • b. Rotated line c, with angle 60o over A, we got c’. • c. Intersection of line c’ and line b, ( c’,b) is point B. • d. We can construct equilateral triangle ABC. • • We can also start with fixed point B on b or C on c. • Can do it ?
  • 24. a. Fixed any point B on b. • b. Rotated line a, with angle 60o over B, we got a’. • c. Intersect of line a’ and line c, ( a’,c) is point C . • d. We can construct equilqteral triangle ABC. • • We can also start with fixed point A on a or C on c. • Can do it ?