{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-point-line-plane-01",
  "mko_id": "mko-point-line-plane",
  "title": {
    "zh-Hant": "點、線、面：在座標模型中判定點是否在直線上",
    "en": "Points, lines and planes: deciding incidence in the coordinate model"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-point-line-plane-01",
    "source": {
      "mko_id": "mko-point-line-plane",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "abf9d0f665f79eaddd4998e3b6080630ebd88fba8faec86475aa135d364f1efd",
      "english_sha256": "aa345ddc765beb7dfffbbb0ddbd644b696872c309ee1222406688ea8469a69cf"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "cartesian_product_def",
      "image_preimage_def",
      "eq_congruence",
      "mul_inverse_nonzero",
      "numeric_evaluation"
    ],
    "goal": {
      "zh-Hant": "在座標平面模型 ℝ² 中，設 P=(1,2)、Q=(3,6)。判定 R=(4,8) 與 S=(2,5) 是否在通過 P、Q 的直線 L 上，並據此判定 P、Q、R 是否共線。",
      "en": "In the coordinate-plane model ℝ², let P=(1,2) and Q=(3,6). Decide whether R=(4,8) and S=(2,5) lie on the line L through P and Q, and hence whether P, Q, R are collinear."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "模型：點是 ℝ² 中的有序實數對；加法、減法與實數倍數逐坐標進行。",
          "en": "Model: points are ordered pairs of reals in ℝ²; addition, subtraction and real multiples act coordinatewise."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "本模型的直線定義：對相異點 P≠Q，通過 P、Q 的直線是 L(P,Q)={P+t(Q−P) ∣ t∈ℝ}。",
          "en": "Lines in this model: for distinct points P≠Q, the line through P and Q is L(P,Q)={P+t(Q−P) ∣ t∈ℝ}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "P=(1,2)、Q=(3,6)、R=(4,8)、S=(2,5)。",
          "en": "P=(1,2), Q=(3,6), R=(4,8), S=(2,5)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "\\text{points and lines of the chosen model}",
        "after_tex": "\\text{point}\\in\\mathbb{R}^2,\\ L(P,Q)=\\{P+t(Q-P)\\mid t\\in\\mathbb{R}\\}\\ (P\\neq Q)",
        "relation": "definition",
        "rule": "mko_model_dependence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 指出線的精確性質由所選模型決定，所以先固定模型：點是 ℝ² 的有序對，直線由 G2 的參數式定義。之後每一步的「在線上」都是指這個模型中的集合歸屬，而不是示意圖上的外觀。",
          "en": "The MKO says the precise properties of lines are fixed by the chosen model, so the model is fixed first: points are ordered pairs in ℝ², and lines are given by the parametric form of G2. From here on, 'on the line' means membership in a set of this model, not how a sketch looks."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_model_dependence"
          ],
          "explanation": {
            "zh-Hant": "此步只選定並寫出模型，不推出新事實；之後所有結論只在這個模型中成立。",
            "en": "This step only chooses and states the model and derives no new fact; every later conclusion holds in this model only."
          }
        },
        "leaf_basis": "mko_model_dependence"
      },
      {
        "id": "S02",
        "uses": [
          "G3",
          "cartesian_product_def"
        ],
        "before_tex": "P=(1,2),\\ Q=(3,6)",
        "after_tex": "P\\neq Q",
        "relation": "implies",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個有序對相等必須兩個坐標都相等；P 與 Q 的第一坐標 1≠3，所以 P≠Q。這是 G2 的直線定義所要求的前提，L(P,Q) 因此確實是一條直線。",
          "en": "Two ordered pairs are equal only if both coordinates agree; the first coordinates of P and Q are 1≠3, so P≠Q. This is the hypothesis the line definition in G2 requires, so L(P,Q) really is a line."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "一個坐標不同就足以推出兩點不等；反向（兩點不等推出某個指定坐標不等）並未宣稱。",
            "en": "One differing coordinate suffices for the points to differ; the converse (distinct points differ in a particular coordinate) is not claimed."
          }
        },
        "leaf_basis": "cartesian_product_def + numeric_evaluation"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "G3"
        ],
        "before_tex": "Q-P",
        "after_tex": "(2,4)",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐坐標相減：Q−P=(3−1, 6−2)=(2,4)。左右兩邊是同一個有序對，這是等值代換。",
          "en": "Subtract coordinatewise: Q−P=(3−1, 6−2)=(2,4). Both sides name the same ordered pair, so this is a replacement of equals."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "coord_ops",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：P+(2,4)=(1+2, 2+4)=(3,6)=Q，還原了 Q−P=(2,4)；兩式表示同一個點。",
            "en": "Backward: P+(2,4)=(1+2, 2+4)=(3,6)=Q, which restores Q−P=(2,4); both expressions name the same point."
          }
        },
        "leaf_basis": "coord_ops + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S02",
          "S03",
          "G3"
        ],
        "before_tex": "L(P,Q)=\\{P+t(Q-P)\\mid t\\in\\mathbb{R}\\}",
        "after_tex": "L=\\{(1+2t,\\,2+4t)\\mid t\\in\\mathbb{R}\\}",
        "relation": "equivalent",
        "rule": "coord_ops",
        "conditions": [
          "P\\neq Q"
        ],
        "explanation": {
          "zh-Hant": "代入 P=(1,2) 與 Q−P=(2,4)：P+t(Q−P)=(1,2)+(2t,4t)=(1+2t, 2+4t)。對每個 t，兩種寫法是同一點，所以兩個集合相同。條件 P≠Q 已在 S02 確認。",
          "en": "Substitute P=(1,2) and Q−P=(2,4): P+t(Q−P)=(1,2)+(2t,4t)=(1+2t, 2+4t). For every t the two expressions are the same point, so the two sets are equal. The condition P≠Q was checked in S02."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S02",
            "S03",
            "coord_ops"
          ],
          "explanation": {
            "zh-Hant": "每個 t 在兩種寫法中給出同一點，集合逐元素相同，雙向成立。",
            "en": "Each t gives the same point in both forms, so the sets agree element by element and the step holds both ways."
          }
        },
        "leaf_basis": "coord_ops + eq_replace"
      },
      {
        "id": "S05",
        "uses": [
          "S04",
          "G3"
        ],
        "before_tex": "R\\in L",
        "after_tex": "\\exists t\\in\\mathbb{R}:\\ (4,8)=(1+2t,\\,2+4t)",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L 是參數 t 走遍 ℝ 時得到的點的集合（函數 t↦(1+2t,2+4t) 的像）。所以 R 屬於 L，恰好表示有某個實數 t 使 R 等於 (1+2t, 2+4t)。",
          "en": "L is the set of points obtained as t runs through ℝ (the image of t↦(1+2t,2+4t)). So R belongs to L exactly when some real t makes R equal to (1+2t, 2+4t)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "這是像集合的展開：找到這樣的 t 就推出 R∈L，R∈L 也必有這樣的 t，雙向成立。",
            "en": "This unfolds membership in an image: such a t gives R∈L, and R∈L supplies such a t, so it holds both ways."
          }
        },
        "leaf_basis": "image_preimage_def + def_unfold"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "cartesian_product_def"
        ],
        "before_tex": "(4,8)=(1+2t,\\,2+4t)",
        "after_tex": "4=1+2t\\ \\land\\ 8=2+4t",
        "relation": "equivalent",
        "rule": "cartesian_product_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "有序對相等當且僅當兩個坐標分別相等，所以一個點的方程拆成兩個實數方程，而且兩個方程用的是同一個 t。",
          "en": "Ordered pairs are equal exactly when both coordinates agree, so the point equation splits into two real equations, and both use the same t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "有序對的相等條件本身是雙向的；拆開後兩個方程不能各自使用不同的 t。",
            "en": "The equality condition for ordered pairs is itself two-way; after splitting, the two equations may not use different values of t."
          }
        },
        "leaf_basis": "cartesian_product_def"
      },
      {
        "id": "S07",
        "uses": [
          "S06"
        ],
        "before_tex": "4=1+2t",
        "after_tex": "t=\\tfrac{3}{2}",
        "relation": "equivalent",
        "rule": "L_solve_linear",
        "conditions": [
          "2\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "依引理 L_solve_linear（a=1、b=2、c=4）：4=1+2t ⇔ t=(4−1)/2=3/2。除以 2 需要 2≠0。",
          "en": "By Lemma L_solve_linear (a=1, b=2, c=4): 4=1+2t ⇔ t=(4−1)/2=3/2. Dividing by 2 needs 2≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve_linear"
          ],
          "explanation": {
            "zh-Hant": "代回 1+2·(3/2)=4；兩個方向都由引理保證。",
            "en": "Substituting back, 1+2·(3/2)=4; both directions are given by the lemma."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + two_nonzero"
      },
      {
        "id": "S07.1",
        "uses": [
          "eq_congruence"
        ],
        "before_tex": "c=a+bt",
        "after_tex": "c-a=bt",
        "relation": "equivalent",
        "rule": "eq_congruence",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩邊同減 a：(a+bt)−a=bt（加法結合律、交換律與反元素）。反向是兩邊同加 a。",
          "en": "Subtract a from both sides: (a+bt)−a=bt (associativity, commutativity and inverses for addition). The backward move is to add a to both sides."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "eq_congruence",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "減 a 與加 a 互為反運算，對任何實數 a 都可做，沒有附加條件。",
            "en": "Subtracting a and adding a undo each other and are possible for every real a, with no side condition."
          }
        },
        "leaf_basis": "eq_congruence + add_assoc + add_comm + add_inverse + add_zero"
      },
      {
        "id": "S07.2",
        "uses": [
          "S07.1",
          "mul_inverse_nonzero"
        ],
        "before_tex": "c-a=bt",
        "after_tex": "t=\\frac{c-a}{b}",
        "relation": "equivalent",
        "rule": "mul_inverse_nonzero",
        "conditions": [
          "b\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "因 b≠0，1/b 存在；兩邊同乘 1/b 得 (c−a)/b=t。反向是兩邊同乘 b。若 b=0，1/b 不存在，這一步不能做。S07.1 與 S07.2 合起來證得引理 L_solve_linear。",
          "en": "Since b≠0, 1/b exists; multiplying both sides by 1/b gives (c−a)/b=t. The backward move is to multiply both sides by b. If b=0 there is no 1/b and the step cannot be taken. S07.1 and S07.2 together prove Lemma L_solve_linear."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mul_inverse_nonzero",
            "eq_congruence"
          ],
          "explanation": {
            "zh-Hant": "乘 1/b 與乘 b 互為反運算，前提 b≠0；這正是引理的條件。",
            "en": "Multiplying by 1/b and by b undo each other, given b≠0; that is exactly the lemma's hypothesis."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one"
      },
      {
        "id": "S08",
        "uses": [
          "S06"
        ],
        "before_tex": "8=2+4t",
        "after_tex": "t=\\tfrac{3}{2}",
        "relation": "equivalent",
        "rule": "L_solve_linear",
        "conditions": [
          "4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "同一引理（a=2、b=4、c=8）：8=2+4t ⇔ t=(8−2)/4=6/4=3/2；這裡 4≠0。",
          "en": "The same lemma (a=2, b=4, c=8): 8=2+4t ⇔ t=(8−2)/4=6/4=3/2; here 4≠0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve_linear"
          ],
          "explanation": {
            "zh-Hant": "代回 2+4·(3/2)=8；兩個方向都由引理保證。",
            "en": "Substituting back, 2+4·(3/2)=8; both directions are given by the lemma."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08"
        ],
        "before_tex": "4=1+2t\\ \\land\\ 8=2+4t",
        "after_tex": "R\\in L",
        "relation": "implies",
        "rule": "exists_intro",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩個方程都恰好由 t=3/2 滿足，所以同一個 t=3/2 同時滿足兩者。依 S05、S06 的等價，R∈L，而且 R=P+(3/2)(Q−P)。",
          "en": "Each equation is satisfied exactly by t=3/2, so the single value t=3/2 satisfies both. By the equivalences S05 and S06, R∈L, and indeed R=P+(3/2)(Q−P)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S07",
            "S08",
            "exists_intro"
          ],
          "explanation": {
            "zh-Hant": "需要的是「同一個 t」：S07 與 S08 給出相同的值，存在性才成立。只有一個方程有解並不夠。",
            "en": "What is needed is one common t: S07 and S08 give the same value, which is what makes the existence claim true. A solution of one equation alone is not enough."
          }
        },
        "leaf_basis": "exists_intro + and_intro_elim + logic_chain"
      },
      {
        "id": "S10",
        "uses": [
          "S04",
          "G3"
        ],
        "before_tex": "S\\in L",
        "after_tex": "\\exists t\\in\\mathbb{R}:\\ 2=1+2t\\ \\land\\ 5=2+4t",
        "relation": "equivalent",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對 S 重複 S05、S06 的兩個展開：S∈L 當且僅當有某個 t 同時滿足兩個坐標方程 2=1+2t 與 5=2+4t。",
          "en": "Repeat the two unfoldings of S05 and S06 for S: S∈L exactly when some t satisfies both coordinate equations 2=1+2t and 5=2+4t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "image_preimage_def",
            "cartesian_product_def"
          ],
          "explanation": {
            "zh-Hant": "與 S05、S06 同理，兩個展開都是雙向的。",
            "en": "As in S05 and S06, both unfoldings are two-way."
          }
        },
        "leaf_basis": "image_preimage_def + cartesian_product_def"
      },
      {
        "id": "S11",
        "uses": [
          "S10"
        ],
        "before_tex": "2=1+2t\\ \\land\\ 5=2+4t",
        "after_tex": "t=\\tfrac{1}{2}\\ \\land\\ t=\\tfrac{3}{4}",
        "relation": "equivalent",
        "rule": "L_solve_linear",
        "conditions": [
          "2\\neq 0",
          "4\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "分別解兩個方程：2=1+2t ⇔ t=1/2；5=2+4t ⇔ t=3/4。每個方程各自等價變形，所以合取前後同真同假。",
          "en": "Solve each equation: 2=1+2t ⇔ t=1/2 and 5=2+4t ⇔ t=3/4. Each equation is transformed by an equivalence, so the conjunction before and after is true or false together."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "L_solve_linear"
          ],
          "explanation": {
            "zh-Hant": "代回 1+2·(1/2)=2、2+4·(3/4)=5；兩個等價各自雙向。",
            "en": "Substituting back, 1+2·(1/2)=2 and 2+4·(3/4)=5; each equivalence is two-way."
          }
        },
        "leaf_basis": "eq_congruence + mul_inverse_nonzero + two_nonzero + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S10",
          "S11"
        ],
        "before_tex": "\\exists t:\\ t=\\tfrac{1}{2}\\ \\land\\ t=\\tfrac{3}{4}",
        "after_tex": "S\\notin L",
        "relation": "implies",
        "rule": "contradiction",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若有 t 同時等於 1/2 與 3/4，依等號的對稱與傳遞得 1/2=3/4，但 1/2≠3/4。所以不存在這樣的 t；依 S10 的等價，S∉L。",
          "en": "If some t equalled both 1/2 and 3/4, symmetry and transitivity of equality would give 1/2=3/4, but 1/2≠3/4. So no such t exists, and by the equivalence S10, S∉L."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10",
            "S11",
            "contradiction"
          ],
          "explanation": {
            "zh-Hant": "矛盾來自「同一個 t 要同時滿足兩個方程」；每個方程單獨都有解，所以不能只看其中一個坐標。",
            "en": "The contradiction comes from one t having to satisfy both equations; each equation alone has a solution, so one coordinate alone cannot decide membership."
          }
        },
        "leaf_basis": "contradiction + eq_refl_symm_trans + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S04",
          "G3"
        ],
        "before_tex": "t=0,\\ t=1",
        "after_tex": "P\\in L,\\ Q\\in L",
        "relation": "implies",
        "rule": "image_preimage_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "取 t=0 得 (1+0, 2+0)=(1,2)=P；取 t=1 得 (1+2, 2+4)=(3,6)=Q。所以 L 確實通過兩個給定點。",
          "en": "Taking t=0 gives (1+0, 2+0)=(1,2)=P, and t=1 gives (1+2, 2+4)=(3,6)=Q. So L does pass through the two given points."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "image_preimage_def"
          ],
          "explanation": {
            "zh-Hant": "每個點只需找到一個參數值；這一步不宣稱參數值唯一。",
            "en": "Each point needs only one parameter value; the step does not claim the value is unique."
          }
        },
        "leaf_basis": "image_preimage_def + coord_ops + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S09",
          "S12",
          "S13"
        ],
        "before_tex": "P,Q,R\\in L,\\ S\\notin L",
        "after_tex": "P,Q,R\\ \\text{collinear};\\ S\\notin L(P,Q)",
        "relation": "implies",
        "rule": "collinear_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "L 同時包含 P、Q、R，依共線定義三點共線；S 不在 L 上。兩個結論都只在本模型中成立。",
          "en": "L contains P, Q and R, so by the definition the three points are collinear; S is not on L. Both conclusions hold in this model."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S09",
            "S12",
            "S13",
            "collinear_def"
          ],
          "explanation": {
            "zh-Hant": "目標 G：P、Q、R 共線且 S∉L(P,Q)。足夠條件 H 是兩件事的合取：L 同時含 P、Q、R，且 S∉L。前一項由 S09、S13 得到，後一項由 S12 得到；共線定義加合取引入給出 H⇒G。沒有把 G 當作前提。",
            "en": "Goal G: P, Q, R are collinear and S∉L(P,Q). The sufficient condition H is the conjunction that L contains P, Q and R and that S∉L. S09 and S13 establish the first conjunct; S12 establishes the second. The collinearity definition and conjunction introduction give H⇒G, without using G as a premise."
          }
        },
        "leaf_basis": "collinear_def + and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在 ℝ² 座標模型中，R=(4,8) 在通過 P、Q 的直線上（t=3/2），所以 P、Q、R 共線；S=(2,5) 不在這條直線上，因為兩個坐標方程要求 t=1/2 與 t=3/4，互相矛盾。",
      "en": "In the coordinate model of ℝ², R=(4,8) lies on the line through P and Q (at t=3/2), so P, Q, R are collinear; S=(2,5) does not lie on this line, because its two coordinate equations demand t=1/2 and t=3/4, which contradict each other."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "S 在圖上看起來很接近直線，所以 S 在 L 上",
          "en": "S looks very close to the line in a sketch, so S is on L"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "入射由方程決定；S 的兩個坐標方程要求 t=1/2 與 t=3/4，沒有共同解。",
          "en": "Incidence is decided by equations; the coordinate equations of S demand t=1/2 and t=3/4, with no common solution."
        }
      },
      {
        "claim": {
          "zh-Hant": "由 2=1+2t 得 t=1/2，所以 S∈L",
          "en": "From 2=1+2t we get t=1/2, so S∈L"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "第二坐標不成立：2+4·(1/2)=4≠5。",
          "en": "The second coordinate fails: 2+4·(1/2)=4≠5."
        }
      },
      {
        "claim": {
          "zh-Hant": "任意兩點 P、Q（包括 P=Q）都決定一條直線 {P+t(Q−P)}",
          "en": "Any two points P, Q (including P=Q) determine a line {P+t(Q−P)}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P=Q 時這個集合只有一點 P。",
          "en": "When P=Q this set is the single point P."
        }
      },
      {
        "claim": {
          "zh-Hant": "在任何幾何模型中，三點是否共線都可用這組座標方程判定",
          "en": "In every geometric model, collinearity of three points can be decided by these coordinate equations"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這組方程屬於 ℝ² 的座標模型；MKO 指出線的性質由所選模型決定，換了模型（例如球面幾何）直線的定義也不同。",
          "en": "These equations belong to the coordinate model of ℝ²; the MKO says the properties of lines depend on the chosen model, and in another model (for example spherical geometry) lines are defined differently."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "cartesian_product_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。",
          "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d."
        }
      },
      {
        "id": "image_preimage_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。",
          "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse."
        }
      },
      {
        "id": "eq_congruence",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。",
          "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "two_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。",
          "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0."
        }
      },
      {
        "id": "add_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。",
          "en": "Associativity of addition: (u+v)+w=u+(v+w)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法單位元：u+0=u。",
          "en": "Additive identity: u+0=u."
        }
      },
      {
        "id": "mul_assoc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法結合律：(uv)w=u(vw)。",
          "en": "Associativity of multiplication: (uv)w=u(vw)."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "mul_one",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法單位元：1·u=u。",
          "en": "Multiplicative identity: 1·u=u."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "logic_chain",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。",
          "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔."
        }
      },
      {
        "id": "contradiction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。",
          "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds."
        }
      },
      {
        "id": "eq_refl_symm_trans",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。",
          "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w."
        }
      },
      {
        "id": "mko_model_dependence",
        "kind": "cited",
        "source": {
          "mko_id": "mko-point-line-plane",
          "sha256": "abf9d0f665f79eaddd4998e3b6080630ebd88fba8faec86475aa135d364f1efd",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在指定的幾何模型中，點表示位置；線與面是承載點並具有相應維度的幾何對象，其精確性質由該模型的公理與結構決定。",
          "en": "In a specified geometric model, a point represents a position. Lines and planes are geometric objects carrying points with their respective dimensions, and their precise properties are determined by the axioms and structure of that model."
        }
      },
      {
        "id": "coord_ops",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "逐坐標運算：(a,b)+(c,d)=(a+c,b+d)，(a,b)−(c,d)=(a−c,b−d)，t(a,b)=(ta,tb)。",
          "en": "Coordinatewise operations: (a,b)+(c,d)=(a+c,b+d), (a,b)−(c,d)=(a−c,b−d), t(a,b)=(ta,tb)."
        }
      },
      {
        "id": "collinear_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "共線：若存在一條直線同時包含若干點，這些點稱為共線。",
          "en": "Collinear: points are collinear if some line contains all of them."
        }
      },
      {
        "id": "L_solve_linear",
        "kind": "lemma",
        "proved_in": [
          "S07.1",
          "S07.2"
        ],
        "statement": {
          "zh-Hant": "對實數 a、c 與 b≠0：c=a+bt ⇔ t=(c−a)/b。",
          "en": "For reals a, c and b≠0: c=a+bt ⇔ t=(c−a)/b."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "3-1",
        "rhs": "2"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "6-2",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S07",
        "lhs": "(4-1)/2",
        "rhs": "3/2"
      },
      {
        "id": "N4",
        "step": "S08",
        "lhs": "(8-2)/4",
        "rhs": "3/2"
      },
      {
        "id": "N5",
        "step": "S11",
        "lhs": "(2-1)/2",
        "rhs": "1/2"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "(5-2)/4",
        "rhs": "3/4"
      },
      {
        "id": "N7",
        "step": "S12",
        "lhs": "1/2",
        "relation": "!=",
        "rhs": "3/4"
      },
      {
        "id": "N8",
        "step": "S09",
        "lhs": "(4-1)*4-(8-2)*2",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S12",
        "lhs": "(2-1)*4-(5-2)*2",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "1+2*1",
        "rhs": "3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只在一個明示的模型（ℝ² 的座標模型，直線以參數式定義）中判定入射關係，示範 MKO「線與面的精確性質由所選模型決定」這一點。不處理一般公理化幾何、平面或高維情形；不改動 canonical MKO 或英文 edition。",
      "en": "Decides incidence only in one stated model (the coordinate model of ℝ², with lines defined parametrically), illustrating the MKO's point that the precise properties of lines and planes are fixed by the chosen model. It does not treat axiomatic geometry in general, planes or higher dimensions, and does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "\\mathbb{R}^2",
        "有序實數對的集合；本稿的點",
        "the set of ordered pairs of reals; the points here"
      ],
      [
        "P,Q,R,S",
        "四個給定點",
        "the four given points"
      ],
      [
        "L(P,Q)",
        "通過相異點 P、Q 的直線",
        "the line through distinct points P and Q"
      ],
      [
        "t",
        "直線的實數參數",
        "the real parameter of the line"
      ],
      [
        "Q-P",
        "方向向量，逐坐標相減",
        "direction vector, coordinatewise difference"
      ]
    ],
    "md_error_controls": [
      {
        "title": "看起來在線上不等於在線上 / Appearing to lie on a line is not the same as lying on it",
        "zh": "在示意圖上 S=(2,5) 很靠近 L：x=2 時 L 上的點是 (2,4)，只差 1。MKO 提醒不要把圖的外觀當成定義。本稿的判定完全由模型中的方程決定：S 的兩個坐標方程沒有共同解，所以 S∉L，不論圖上看起來多接近。",
        "en": "In a sketch S=(2,5) is close to L: at x=2 the point of L is (2,4), only 1 away. The MKO warns against taking a drawing's appearance as a definition. Here membership is decided entirely by the model's equations: the two coordinate equations of S have no common solution, so S∉L however close it looks."
      },
      {
        "title": "只核一個坐標不夠 / One coordinate is not enough",
        "zh": "S 的第一個坐標方程 2=1+2t 有解 t=1/2，單看它會誤以為 S 在線上。但 t=1/2 代入第二坐標得 2+4·(1/2)=4≠5。入射需要同一個 t 同時滿足兩個坐標方程（S06 與 S12）。",
        "en": "The first coordinate equation of S, 2=1+2t, has the solution t=1/2, and on its own suggests S is on the line. But t=1/2 in the second coordinate gives 2+4·(1/2)=4≠5. Incidence needs one t that satisfies both coordinate equations (S06 and S12)."
      },
      {
        "title": "為什麼先確認 P≠Q / Why P≠Q is checked first",
        "zh": "若 P=Q，Q−P=(0,0)，集合 {P+t·(0,0) ∣ t∈ℝ} 只有一個點 P，不是直線。G2 的定義只對相異兩點給出直線，所以 S02 必須先確認 P≠Q，S04 才能把 L 當作直線使用。",
        "en": "If P=Q then Q−P=(0,0) and {P+t·(0,0) ∣ t∈ℝ} is the single point P, not a line. The definition in G2 gives a line only for two distinct points, so S02 must establish P≠Q before S04 can treat L as a line."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不證明「通過兩相異點的直線唯一」，所以只宣稱 S∉L(P,Q)，不宣稱 P、Q、S 不在任何一條直線上。",
        "en": "This guide does not prove that the line through two distinct points is unique, so it claims only S∉L(P,Q), not that P, Q, S lie on no common line."
      },
      {
        "zh-Hant": "N8、N9 是以行列式形式做的獨立交叉核對，只對這組數字有效，本稿的論證不依賴它們。",
        "en": "N8 and N9 are independent cross-checks in determinant form, valid for these numbers only; the argument does not depend on them."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "acf8e1534a6a5a0b508ab339ff6a480eccfed9de9c617ae4b40a60230ec65bfc",
    "original_source_sha256": "fe3365f185ad4359da5ff93616296d2ba284d6ea11f6bb9927b953ceefeb6905",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M001"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}