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discuss the differences between a coordinate geometry proof and a proof method that does not require coordinate geometry. when would it be appropriate to use a coordinate proof rather than another proof method?

A coordinate geometry proof typically sets up a coordinate system and then rewrites the entire problem in terms of algebra. The proof then becomes an algebraic computation (often rather long and tedious). A good choice of coordinates systems can make things simplier. 

A coordinate-free proof does not make reference to any particular coordinate system. The result is usually simpler and easier to follow, but harder to find in the first place. 

For example: Prove that the centroid of a triangle (the intersection of the medians) divides a median in the ratio of 2:1. Specifically, for any triangle ABC with centroid P and median AM, show that AP:PM=2:1. (Or, equivalently, AM:PM=3:1.) 

For coordinates, set up a coordinate system with origin at A, (0,0), and an x-axis along AB, scaled so that B is at (2,0). Then C can have any coordinates (2p,2q), for any real numbers p and q. The midpoint, M, of BC becomes [(2,0)+(2p,2q)]/2=(1+p,q). To find the centroid, we need another median, so take the midpoint (1,0) of AB. The equation of that median to C is: (2p-1)(y-0)=(2q-0)(x-1) while the equation of AM is (1+p-0)(y-0)=(q-0)(x-0). Solving that system of equations for 'x' and 'y' (tedious arithmetic omitted) gives the centroid as (2/3(1+p),2/3(q)). So AP is 2/3 of AM. leaving 1/3 of AM for PM, hence AM:PM=3:1. 

For a non-coordinate proof, P is the intersection of the medians AM and BN. Note that triangles with equal bases and the same height have equal area, which will imply that the three triangles PAB, PAC, and PBC have equal area (make a diagram for yourself): |ABM|=|ACM| and |PBM|=|PCM| ⇒ |PAB|=|ABM|-|PBM|=|ACM|-|PCM|=|PAC|. Similarly, |BCN|=|BAN| and |PCN|=|PAN| ⇒ |PBC|=|BCN|-|PCN|=|BAN|-|PAN|=|PAB|. 
Hence |PBC|=|PAB|=|PAC| and so |ABC|=3|PBC|. Those two triangles have the same base BC, hence the height of the first, AX, is three times the height of the later, PY. Since AX is parallel PY and XM is parallel YM and AM is parallel PM, the triangles AXM and PYM are similar and so AM:PM = AX:PY = 3:1.