Definition D1
The distance d(P, ℓ) between a point P and a line ℓ is defined as PH, where H ∈ ℓ is a point with PH ≤ PQ for every point Q on ℓ.
Proof.
That H is not called the foot of the perpendicular. Whether it is the foot of the perpendicular is verified in L2.convention
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Definition D2
ℓ is the set of all points satisfying ax + by + c = 0, where a and b are not both 0. For a point P(x₀, y₀), let k = ax₀ + by₀ + c, s = a² + b², t = k/s, H = (x₀ − ta, y₀ − tb).
Proof.
(1)s ≠ 0, so t is defined and ts = k((1/s)s) = k × 1 = kA14, A15, A17, A18
(2)If b ≠ 0, ℓ is the graph of y = (−a/b)x + (−c/b), a line with slope −a/b.assumption, A15, A16, A19, A20, A21, A6
(3)If b = 0, then a ≠ 0 and ℓ is the set of all points with x = −c/a, that is, a vertical line.assumption, A14, A15, A19, A20, A21, A22
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Lemma L1
H lies on ℓ.
Proof.
(1)a(x₀ − ta) + b(y₀ − tb) + c = (ax₀ + by₀ + c) − (a(ta) + b(tb))D2, A16, A19, A21
(2)a(ta) + b(tb) = ta² + tb² = t(a² + b²) = tsD2, A16, A17
(3)Therefore that value is k − ts = k − k = 0(1), (2), D2, A20
(4)H satisfies ax + by + c = 0, so H ∈ ℓ(3), D2
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Lemma L2
If P ≠ H, the segment PH is perpendicular to the line ℓ.
Proof.
(1)The coordinate differences between P and H are ta in the x-direction and tb in the y-direction.D2, A19, A20
(2)P ≠ H, so t ≠ 0(1), A22
(3)If b = 0, then tb = 0, so PH is a horizontal segment, and ℓ is a vertical line, so they are perpendicular.(1), (2), D2, A22, A4
(4)If a = 0, then ta = 0, so PH is a vertical segment, and ℓ is a horizontal line of slope 0, so they are perpendicular.(1), (2), D2, A22, A4
(5)If a ≠ 0 and b ≠ 0, the slope of PH is (tb)/(ta) = b/a(1), (2), A5, A15, A17
(6)The slope of ℓ is −a/b, and (−a/b)(b/a) = −1, so PH ⊥ ℓ(5), D2, A15, A17, A18, A21, A3
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Lemma L3
For every point Q on ℓ, PH ≤ PQ
Proof.
(1)If P = H, then PH = 0 and PQ ≥ 0, so the claim holds.A7
(2)If P ≠ H and Q = H, then PH = PQassumption
(3)Now suppose P ≠ H, Q ∈ ℓ, Q ≠ H.assumption
(4)H and Q both lie on ℓ and PH is perpendicular to ℓ, so ∠PHQ is a right angle.(3), L1, L2, A23
(5)The hypotenuse of triangle PHQ is PQ, so PQ² = PH² + HQ²(4), A1
(6)HQ² ≥ 0, so PH² ≤ PQ²(5), A9, A24
(7)Lengths are nonnegative, so PH ≤ PQ(6), A7, A8
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Theorem T
d(P, ℓ) = |ax₀ + by₀ + c| / √(a² + b²)
Proof.
(1)H ∈ ℓ and PH ≤ PQ for every Q, so d(P, ℓ) = PHL1, L3, D1
(2)PH² = (ta)² + (tb)² = t²(a² + b²) = t²sD2, A2, A16, A17
(3)PH ≥ 0, so PH = √(t²s) = √(t²)·√s = |t|√s(2), A7, A10, A11, A13
(4)ts = k, so |t|s = |t||s| = |ts| = |k|D2, A12, A13, A14
(5)√s·√s = √(s²) = s and √s > 0, so |t|√s = |k|/√s(4), A10, A11, A13, A15, A17
(6)Therefore d(P, ℓ) = |k|/√s = |ax₀ + by₀ + c| / √(a² + b²)(1), (3), (5), D2
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