Transformation

No 2. A point of K has segment of AB. K isn’t has AB. And A  line of g make  g parallel with AB. And distance between K and AB as twice more than distance between k and g. There is a domain T with range from AB and the range of g. and then that if p has AB then T (P) = p’ KP

Answer :

Firstly we must check it. Is it v to v ?

AB

P

P

   E                                   P                                         F      D

A                                                                                    B                                  

                                          K

                                                                                         g

D’= g(D)  F’                    P’                 E’ = g(E)

a.     What the shape of  set P’ . if  The P’ move on segment of AB ?      

b.     Prove that T is injectif  !

c.      If E and F are two point of AB. Are they distance of E’ F’ if E’ = T (E) and F’ = T(F) ? 

           Look at the triangle

EKF and E’K F’

For example P

Every triangle comparation with the triangle 1 : ½

For ecample the pont of D and E

D = E is

D

The proved of contradiction.

We take DFE get 

That is contradiction when we were using assumption  make a onto function.