i dont quite get the question but...
i guess this is how it is.
Take the mirror image of∆ABC Through the a line through the point y=3.
The new ∆ABC would have point C=(4,2)
B=(3,-6) A=(1,-3)
Now shifting the ∆ABC one unit (i.e. 2 acc. to the graph as scale is 1 unit =2) towards right ( or adding 2 to the x coordinates of ∆ABC)
We get the Coordinates of triangle ABC as A=(3,-3) B=(5,-6) C=(6,2).
This coordinate is the same coordinates of ∆A"B"C".
Hope it helps...
Regards;
Leukonov/Olegion.