Given the points A(-2, -1) and B(3,2), graph {P|dT(P,A) + dT(P,B) = dT(A,B)}.

My work so far:
I know the distance metric in taxicab geometry is:
dT(A,B) = |a1 - b1| + |a2 - b2|.

So dT(P,A) = |p1 - (-2)| + |p2 - (-1)| = |p1 +2| + |p2 + 1|, and
dT(P,B) = |p1 - 3| + |p2 - 2|.

I computed dT(A,B) = |-2 - 3| + |-1 - 2| = |-5| + |-3| = 8

If I add dT(P,A) + dT(P,B), do I need to have two cases since the quantity within the absolute value signs can be (+) or (-)? This is where I got confused.

I tried to compute P|dT(P,A) + dT(P,B) = dT(A,B) to see if I could come up with a graphable equation. I got this far:

|p1 + 2| + |p2 + 1| + |p1 - 3| + |p2 - 2| = 8
p1 + 2 + p2 + 1 + p1 - 3 + p2 - 2 = 8
2p1 + 2p2 = 10
p1 + p2 = 5
So I get the line: p2 = -p1 + 5

Is that even close to logical/right?
Any help, corrections, suggestions, hints, and/or tips are greatly appreciated! Thank you for your time.