General

[quote=bloodisshed]Direct Proof
Let P1 (x1, y1, z1) and P2 (x2,y2,z2) be points in S

P ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)
= P(x1/2, y1/2, z1/2) + P(x2/2, y2/2, z2/2) by linearity
= 1/2 P(x1, y1, z1) + 1/2 P(x2, y2, z2) by linearity
= 1/2 P1 + 1/2 P2 by definition of P1 and P2

thus, P is a linear combination of P1 and P2
since P1 and P2 are defined to be in surface S,
P must also be in S since it's just a linear combination of P1 and P2.[/quote]

Thanks a lot!

Direct Proof
Let P1 (x1, y1, z1) and P2 (x2,y2,z2) be points in S

P ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)
= P(x1/2, y1/2, z1/2) + P(x2/2, y2/2, z2/2) by linearity
= 1/2 P(x1, y1, z1) + 1/2 P(x2, y2, z2) by linearity
= 1/2 P1 + 1/2 P2 by definition of P1 and P2

thus, P is a linear combination of P1 and P2
since P1 and P2 are defined to be in surface S,
P must also be in S since it's just a linear combination of P1 and P2.

That's literally just the midpoint between the two points on the surface.
I don't know of a real formal way to prove that, but you could use the definition of distance and then use those provided points ((x_1+x_2)/2, etc.) to prove that the distance from the original point to the provided points is 1/2 the distance between the two original points. It's not a direct proof tho so I'm not sure if your professor would accept it.

kkkkkkkkkk

I wish I could meet the guy or gal who came up with this question

Seriously, just to see what they're like--I'd buy them some beers and just ask, "why?"

OT: I never made it past pre-calc so I'm of no use here but still, "why"