Answer:
The coordinates of P" is (m,n+e-3).
Step-by-step explanation:
From the figure it is noticed that the line s is vertical transformation of line r, e units up. Then the x coordinates are same as in line r, but e units subtracted from y-coordinates.
If a graph vertically translated by m units up, then
[tex](x,y)\rightarrow (x,y+m)[/tex]
If a graph vertically translated by m units down, then
[tex](x,y)\rightarrow (x,y-m)[/tex]
Since the line Q is a vertical translation of line S, 3 units down, therefore x-coordinate remains same but 3 units are subtracted from y-coordinate.
[tex](x,y)\rightarrow (x,y-3)[/tex]
It is given that P" is the image of P' and P'(m,n+e).
[tex](m,n+e)\rightarrow (m,n+e-3)[/tex]
Therefore the coordinates of P" is (m,n+e-3).
