Theorem 2.10: Let A, B, and C be three noncolinear points. If D is on the line through A which is parallel to BC then there is a real number s such that

D = A + s(C - B)

Proof: If A = D, then we can take s = 0, and we're done. So we can assume that D is a different point than A.

Let

A = (x0, y0)

B = (x1, y1)

C = (x2, y2)

and

D = (x, y)

Here we are considering the case where the line is vertical.

If BC is vertical, then x1 = x2 but since B and C are distinct points, their y-coordinates must be different. Define

Then

A + s(C - B)

= A - sB + sC

= (x0, y0 ) - s(x1, y1) + s(x2, y2)

= (x0 - sx1 + sx2, y0 - sy1 + sy2)

= (x0 + s(x2 - x1), y0 + s(y2 - y1))

as before, but in this case, x1 = x2, and since AD is also vertical, x0 = x. With these substitutions, we get

= (x, y0 + s(y2 - y1))

= (x, y) = D

top

horizontal case

the case where the line is neither horizontal or vertical

next theorem (2.11)