Definition D1
The line through the origin with slope m is y = mx, and it passes through the point (1, m).
Proof.
(1)Perpendicularity of two lines is unchanged by translation, so assume both pass through the origin.convention
(2)Substituting x = 1 gives y = m.(1), A7
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Lemma L1
If O = (0, 0) and P = (1, m), then OP² = 1 + m²
Proof.
(1)Let H = (1, 0).convention
(2)If m = 0, then P = H and OP = 1, so OP² = 1 = 1 + 0²(1), A3, A15
(3)Now suppose m ≠ 0.assumption
(4)OH is a horizontal segment with OH = 1; HP is a vertical segment with HP = |m|(1), A3
(5)Triangle OHP has a right angle at H and its hypotenuse is OP.(4), A2
(6)OP² = OH² + HP² = 1 + |m|²(5), (4), A1
(7)OP² = 1 + m²(6), A4
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Lemma L2
For all u, v, (u − v)² = u² − 2uv + v²
Proof.
(1)(u − v)² = (u + (−v))(u + (−v))convention
(2) = u(u + (−v)) + (−v)(u + (−v))(1), A5, A6
(3)u(u + (−v)) = u² + u(−v) = u² + (−(uv))A5, A8
(4)(−v)(u + (−v)) = (−v)u + (−v)(−v) = (−(uv)) + v²A5, A6, A8, A9
(5)(−(uv)) + (−(uv)) = (1 + 1)(−(uv)) = 2(−(uv))A5, A6, A7
(6)2(−(uv)) = −(2(uv)), which is written −2uv.(5), A8, convention
(7)(u − v)² = u² − 2uv + v²(2), (3), (4), (6), A10, A11
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Theorem T
If two lines with slopes m₁, m₂ are perpendicular, then m₁m₂ = −1
Proof.
(1)Take the lines to be y = m₁x, y = m₂x, perpendicular at the origin.assumption, D1
(2)A = (1, m₁), B = (1, m₂) are points on the two lines respectively.D1
(3)Triangle OAB has a right angle at O and its hypotenuse is AB.(1), (2)
(4)OA² + OB² = AB²(3), A1
(5)OA² = 1 + m₁², OB² = 1 + m₂²L1
(6)A and B have the same x-coordinate, so AB is a vertical segment and AB = |m₁ − m₂|(2), A3
(7)AB² = (m₁ − m₂)²(6), A4
(8)(1 + m₁²) + (1 + m₂²) = (m₁ − m₂)²(4), (5), (7)
(9)(m₁ − m₂)² = m₁² + m₂² + (−(2m₁m₂))L2, A10, A11
(10)(m₁² + m₂²) + 2 = (m₁² + m₂²) + (−(2m₁m₂))(8), (9), A10, A11
(11)2 = −(2m₁m₂)(10), A12
(12)2 × 1 = 2 and −(2m₁m₂) = 2(−(m₁m₂))A7, A8
(13)0 ≠ 2, so 1 = −(m₁m₂)(11), (12), A15, A13
(14)m₁m₂ + (−(m₁m₂)) = 0, so m₁m₂ + 1 = 0A14, (13), E
(15)The inverse is unique, so m₁m₂ = −1(14), A14
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