Originally Posted by

**Deveno** what i have written as T(x,y) would be t(0,-1). and what i have written as R would be q0.

so the entire transformation is f = t(0,1) o q0 o t(0,-1).

t(0,-1) moves the line y = 1 "down 1" so that now we have a line going through the origin. q0 reflects the plane about the x-axis (y = 0),

which keeps all the x's the same, but exchanges y with -y. then t(0,1) moves the reflected image "back up 1". on an arbitrary point (x,y), we have:

f(x,y) = t(0,1)(q0(t(0,-1)(x,y))) =

[...([x]....[ 0])].....[0]

[q0([y] + [-1])] + [1] =

[1 .0][..x.].....[0]

[0 -1][y-1] + [1] =

[1 .0][x]...[1 .0][..0]....[0]

[0 -1][y] - [0 -1][-1] + [1] =

[1 .0][x]....[0]

[0 -1][y] + [2], so that our matrix A is just the matrix for q0 and the vector a is (0,2).

as a check, let's verify that f leaves points on the line y = 1 unchanged.

f(x,1) = (x,-1) + (0,2) = (x,1). see? now the point (x,1+k) should go to (x,1-k), we can verify this, too:

f(x,1+k) = (x,-1-k) + (0,2) = (x,1-k), as expected.