Here is the graph of the line 2x + 3y = 6 and the point (-1,1):



2x + 3y = 6

is equivalent to

2x + 3y - 6 = 0.

Every line parallel to the line 

2x + 3y - 6 = 0 

is of the form

2x + 3y + K = 0.

The distance from the point (p,q) to the line 
Ax + By + C = 0 is given by the formula:



So since the distance from (-1,1) to the line 2x + 3y + K = 0

is to be , we have







Cross multiply and solve for K, 

using the + we get K = 4 and using th - we get -6 

For K = 4 the solution is

2x + 3y + 4 = 0

which has this graph (in green):



using K = -6. we have the solution

2x + 3y + (-6) = 0

2x + 3y - 6 = 0

Wow! That's the GIVEN LINE!!!  As it turns out, the given line 
was itself  units from the point (-1,1).  
However we wanted a line PARALLEL to 2x + 3y - 6 = 0, not that 
line itself.  So the only solution is

2x + 3y + 4 = 0

Edwin