Theorem 2.11: (The parametric representation of a plane) Let   A,   B,   and   C   be three noncolinear points. Let   D   be any point in the plane. Then there are real numbers   q,   r,   and   s   with

q + r + s = 1

such that

D = qA + rB + sC

Proof: Consider the line through   D   parallel to   BC.   There are two cases to consider. The first case is where this line contains   A,   and the other case is where it doesn't.

If the line through   D   which is parallel to   BC   contains the point   A,   then   D   is on the line through   A   which is parallel to   BC,   so, by Theorem 2.10, there exists a real number   s   such that

D = A + s(C - B)

so

D = qA + rB + sC

where

q = 1 and r = -s

so

q + r + s = 1 - s + s = 1

So, we can assume that the line through   D   which is parallel to   BC   does not contain the point   A. Since   AB   and   BC   intersect at point   B,   by Theorem 1.6 they are not parallel, so any line which is parallel to   BC   is not parallel to   AB,   by Theorem 1.13, so there will be a point where the line through   D,   which is parallel to   BC   will meet the line determined by   A   and   B   by Theorem 1.6. Call this point   E.   Similarly, the line through   D   which is parallel to   BC   will meet   AC   at a point which we will call   F.

In the illustration,   D   is being portrayed as being between   E   and   F,   but it does not have to be in the following proof.

If   E   and   F   were the same point, then since that point would be on the intersection of   AB   and   AC,   and by Theorem 1.6, the point of intersection is unique, that point would have to be   A.   Since we are considering the case where the line through   D,   parallel to   BC   does not contain   A,   this doesn't happen, so we can assume that   E   and   F   are different points.

Let

A = (x0, y0)

B = (x1, y1)

C = (x2, y2)

and

D = (x, y)

Since   EF   is parallel to   BC,   by Theorem 2.9, there will exist a real number   t   such that

E = (1 - t)A + tB

and

F = (1 - t)A + tC

Since   D   is on the line determined by   E   and   F,   by Theorem 2.1, there will exist a real number   u   such that

D = (1 - u)E + uF

If we substitute in the formulas for   E   and   F   we get

D = (1 - u)((1 - t)A + tB) + u((1 - t)A + tC)

Distribute

= (1 - u)(1 - t)A + (1 - u)tB + u(1 - t)A + utC

The   A   terms can be combined to

= (1 - t)A + (1 - u)tB + utC

Let

q = 1 - t,

r = (1 - u)t,

and

s = ut.

Then

D = qA + rB + sC

and

q + r + s

= 1 - t + (1 - u)t + ut

= 1 - t + t - ut + ut

= 1

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