Definition D1
The distance d(P, Q) between two points P and Q is the length of the segment PQ joining them.
Proof.
(1)Let P = (x₁, y₁), Q = (x₂, y₂).convention
(2)Distance is not defined by coordinates. Coordinates are used to measure that length.convention
∎
Lemma L1
For every real number t, √(t²) = |t|
Proof.
(1)0 ≤ t² + 0² = t² + 0 = t²A6, A8
(2)There exists a unique s ≥ 0 with s² = t², and it is written √(t²).(1), A7
(3)|t| ≥ 0 and |t|² = t²A4
(4)|t| satisfies the condition of (2), so √(t²) = |t|(2), (3)
∎
Theorem T1
If x₁ ≠ x₂ and y₁ ≠ y₂, then d(P, Q) = √((x₂−x₁)² + (y₂−y₁)²)
Proof.
(1)Let R = (x₂, y₁).convention
(2)P and R have the same y-coordinate, so PR is a horizontal segment and PR = |x₂−x₁|A3
(3)R and Q have the same x-coordinate, so RQ is a vertical segment and RQ = |y₂−y₁|A3
(4)The two segments are perpendicular, so triangle PRQ has a right angle at R and its hypotenuse is PQ.(2), (3), A2
(5)PQ² = PR² + RQ²(4), A1
(6)PQ² = |x₂−x₁|² + |y₂−y₁|² = (x₂−x₁)² + (y₂−y₁)²(5), (2), (3), A4
(7)0 ≤ (x₂−x₁)² + (y₂−y₁)²A6
(8)The s ≥ 0 with s² = (x₂−x₁)² + (y₂−y₁)² is unique.(7), A7
(9)PQ ≥ 0, so PQ is that s.(6), (8), A5
(10)d(P, Q) = PQ = √((x₂−x₁)² + (y₂−y₁)²)(9), D1
∎
Corollary C
For all points P, Q, d(P, Q) = √((x₂−x₁)² + (y₂−y₁)²)
Proof.
(1)The case x₁ ≠ x₂ and y₁ ≠ y₂ is T1.T1
(2)Suppose y₁ = y₂.assumption
(3)PQ is a horizontal segment, so its length is |x₂−x₁|.(2), A3
(4)y₂ − y₁ = 0 and 0² = 0(2), A8
(5)√((x₂−x₁)² + (y₂−y₁)²) = √((x₂−x₁)²) = |x₂−x₁|(4), A8, L1
(6)d(P, Q) = |x₂−x₁|, so the two values are equal.(3), (5), D1
(7)The case x₁ = x₂ holds by the same argument with a vertical segment.A3, A8, L1, D1
(8)The formula holds in every case.(1), (6), (7)
∎