Definition D1
For m > 0, n > 0, the point P dividing segment AB internally in the ratio m : n is the point on segment AB with AP : PB = m : n.
Proof.
(1)Set A = (x₁, y₁), B = (x₂, y₂), P = (x, y).convention
(2)AP : PB = m : n means n·AP = m·PB.convention
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Lemma L1
If x₁ ≠ x₂, then n(x − x₁) = m(x₂ − x)
Proof.
(1)Set A′ = (x₁, 0), P′ = (x, 0), B′ = (x₂, 0).convention
(2)AA′, PP′, BB′ are all vertical lines, hence parallel to one another.(1), A2
(3)The three parallels cut line AB and the x-axis, so A′P′ : P′B′ = AP : PB = m : n(2), A1, D1
(4)That is, n·A′P′ = m·P′B′(3), D1
(5)A′P′ = |x − x₁| and P′B′ = |x₂ − x|(1), A3
(6)P lies on segment AB, so x lies between x₁ and x₂, and x − x₁ and x₂ − x have the same sign.D1, A4
(7)Hence there is a single ε with |x − x₁| = ε(x − x₁), |x₂ − x| = ε(x₂ − x).(6), A5
(8)nε(x − x₁) = mε(x₂ − x) and ε ≠ 0, so n(x − x₁) = m(x₂ − x)(4), (5), (7), A15, A12, A10
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Lemma L2
If n(x − x₁) = m(x₂ − x), then x = (nx₁ + mx₂)/(m + n)
Proof.
(1)nx − nx₁ = mx₂ − mxassumption, A6, A9
(2)Adding mx + nx₁ to both sides gives nx + mx = nx₁ + mx₂(1), E, A7, A8, A9
(3)nx + mx = (n + m)x = (m + n)xA6, A7, A10
(4)Since m > 0 and n > 0, we have m + n > 0, and in particular m + n ≠ 0.D1, A14
(5)Multiplying both sides by 1/(m+n) gives x = (nx₁ + mx₂)/(m + n)(2), (3), (4), E, A13, A10, A11
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Theorem T1
The point dividing segment AB internally in the ratio m : n is P = ((nx₁ + mx₂)/(m+n), (ny₁ + my₂)/(m+n)).
Proof.
(1)If x₁ ≠ x₂, then x = (nx₁ + mx₂)/(m + n).L1, L2
(2)If x₁ = x₂, then P lies on segment AB, so x = x₁ = x₂.D1, A4
(3)In that case (nx₁ + mx₂)/(m + n) = (n + m)x₁/(m + n) = x₁, so the formula holds.(2), A6, A7, A10, A11, A13
(4)For the y-coordinate, drop to the y-axis instead of the x-axis and repeat the same argument.L1, L2
(5)Hence P = ((nx₁ + mx₂)/(m+n), (ny₁ + my₂)/(m+n)).(1), (2), (3), (4)
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Corollary C
If m = n = 1, then P = ((x₁+x₂)/2, (y₁+y₂)/2), and that point is the midpoint.
Proof.
(1)If m = n = 1, then m + n = 1 + 1 = 2.convention
(2)1×x₁ + 1×x₂ = x₁ + x₂, so x = (x₁ + x₂)/2, and likewise for y.(1), T1, A11
(3)AP : PB = 1 : 1, so AP = PB, and since P lies on the segment, it is the midpoint.D1, A11
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