suppose we knew how to transform e'-coordinates into e-coordinates, by using some matrix A. it stands to reason that A^-1 would translate e-coordinates to e'-coordinates.

now the matrix that sends e'-coordinates to e-coordinates is very easy to write down. in e'-coordinates, e1' is (1,0). and for ANY 2x2 matrix A, A(1,0) is the first column of the matrix. similarly, in e'-coordinates, e2' is (0,1), and A(0,1) is the second column of the matrix A. so the first column of A is A(e1') = (5/13, 12/13) (in e-coordinates). and the second column of A is e2' in e-coordinates, or (-12/13,5/13) so A =

[5/13 -12/13]

[12/13 ..5/13]

by a standard computation, A^-1 =

[ 5/13 12/13]

[-12/13 5/13]

so in e' coordinates, e1 = (5/13,-12/13) = (5/13)e1' - (12/13)e2', while e2 = (12/13,5/13) = (12/13)e1' + (5/13)e2'.

in e'-coordinates the points (x1,x2) that lie on the line now have coordinates ((5/13)x1+(12/13)x2, (5/13)x2 - (12/13)x1)), so the equation for our line becomes:

2(5/13)x1 + 2(12/13)x2 + 3(5/13)x2 - 3(12/13)x1 = 0, or upon simplfying:

-2x1 + 3x2 = 0