Plane Affine Algebra (II)

See also Part I and Part III.

As noted before, the algebra permits us to calculate proofs without resorting to coordinate manipulations. As a demonstration, I'll present a very simple calculation, and then a theorem that I didn't know (and I would be very hard-pressed to come up with a classical proof).

Given points P, Q, R, the altitude of the triangle PQR through the vertex Q is the segment QH perpendicular to the base PR. Denoting hgt.P.Q.R = |Q−H| (we write |•−•| = |‹•−•›| per abusum linguae), it is:

(17)  hgt.P.Q.R = (‹R−P›⊥⋅‹Q−P›)/|R−P|

Proof:

   hgt.P.Q.R
= { definition }
   |Q−H|
= { definition of the sine line }
   |Q−P|·sin.(θ.‹R−P›.‹Q−P›)
= { (16) with u, v := ‹R−P›, ‹Q−P› }
   (‹R−P›⊥⋅‹Q−P›)/|R−P|

The line through point P with direction vector v is the set of points P⇒v that, in the words of Euclid, "lies equally on the points of itself":

(18.0)  Q ∈ P⇒v ≡ |v|⋅‹Q − P› = |Q − P|⋅v

The following are theorems:

(18.1)  P ∈ P⇒v
(18.2)  ⟨∀ k : k ≠ 0 : P⇒(k⋅v) = P⇒v ⟩
(18.2)  P⇒v = ⟨k ∈ R | P → k⋅v⟩

The parametric line through points P and Q is defined as the real function:

(18)  ray.P.Q.k = P → k·‹Q−P›

From (18.2), it is immediate that ray.P.Q maps R to the line P⇒‹Q−P›.

As a special case, the midpoint of the segment PQ is:

(19)  ray.P.Q.½ = P → ½·‹Q−P›

Given points P, Q, R and S, the intersection of the lines containing the segments PQ and RS is:

(20)  meet.P.Q.R.S = ray.P.Q.((‹S−R›⊥⋅‹R−P›)/(‹S−R›⊥⋅‹Q−P›))

Proof: At the intersection point meet.P.Q.R.S, for some k, l ∈ R:

   ray.P.Q.k = ray.R.S.l
= { (18) }
   P → k·‹Q−P› = R → l·‹S−R›
= { (6) }
   ‹(P → k·‹Q−P›) − (R → l·‹S−R›)› = 0
= { (9) }
   ‹P−R› + k·‹Q−P› − l·‹S−R› = 0
= { heading towards eliminating l; inner product }
   ‹S−R›⊥⋅(‹P−R› + k·‹Q−P› − l·‹S−R›) = 0
= { linearity of inner product }
   ‹S−R›⊥⋅‹P−R› + k·‹S−R›⊥⋅‹Q−P› − l·‹S−R›⊥⋅‹S−R› = 0
= { (14) }
   ‹S−R›⊥⋅‹P−R› + k·‹S−R›⊥⋅‹Q−P› = 0
= { (5); algebra }
   k = (‹S−R›⊥⋅‹R−P›)/(‹S−R›⊥⋅‹Q−P›)

In Part III, the theorem.