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| Mirrors > Home > MPE Home > Th. List > Mathboxes > i2linesd | Structured version Visualization version GIF version | ||
| Description: Solve for the intersection of two lines expressed in Y = MX+B form (note that the lines cannot be vertical). Here we use deduction form. We just solve for X, since Y can be trivially found by using X. This is an example of how to use the algebra helpers. Notice that because this proof uses algebra helpers, the main steps of the proof are higher level and easier to follow by a human reader. (Contributed by David A. Wheeler, 15-Oct-2018.) |
| Ref | Expression |
|---|---|
| i2linesd.1 | ⊢ (𝜑 → 𝐴 ∈ ℂ) |
| i2linesd.2 | ⊢ (𝜑 → 𝐵 ∈ ℂ) |
| i2linesd.3 | ⊢ (𝜑 → 𝐶 ∈ ℂ) |
| i2linesd.4 | ⊢ (𝜑 → 𝐷 ∈ ℂ) |
| i2linesd.5 | ⊢ (𝜑 → 𝑋 ∈ ℂ) |
| i2linesd.6 | ⊢ (𝜑 → 𝑌 = ((𝐴 · 𝑋) + 𝐵)) |
| i2linesd.7 | ⊢ (𝜑 → 𝑌 = ((𝐶 · 𝑋) + 𝐷)) |
| i2linesd.8 | ⊢ (𝜑 → (𝐴 − 𝐶) ≠ 0) |
| Ref | Expression |
|---|---|
| i2linesd | ⊢ (𝜑 → 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | i2linesd.1 | . . 3 ⊢ (𝜑 → 𝐴 ∈ ℂ) | |
| 2 | i2linesd.3 | . . 3 ⊢ (𝜑 → 𝐶 ∈ ℂ) | |
| 3 | 1, 2 | subcld 11469 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ∈ ℂ) |
| 4 | i2linesd.5 | . 2 ⊢ (𝜑 → 𝑋 ∈ ℂ) | |
| 5 | i2linesd.8 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ≠ 0) | |
| 6 | 2, 4 | mulcld 11129 | . . . 4 ⊢ (𝜑 → (𝐶 · 𝑋) ∈ ℂ) |
| 7 | i2linesd.4 | . . . . 5 ⊢ (𝜑 → 𝐷 ∈ ℂ) | |
| 8 | i2linesd.2 | . . . . 5 ⊢ (𝜑 → 𝐵 ∈ ℂ) | |
| 9 | 7, 8 | subcld 11469 | . . . 4 ⊢ (𝜑 → (𝐷 − 𝐵) ∈ ℂ) |
| 10 | 1, 4 | mulcld 11129 | . . . . . 6 ⊢ (𝜑 → (𝐴 · 𝑋) ∈ ℂ) |
| 11 | i2linesd.6 | . . . . . . 7 ⊢ (𝜑 → 𝑌 = ((𝐴 · 𝑋) + 𝐵)) | |
| 12 | i2linesd.7 | . . . . . . 7 ⊢ (𝜑 → 𝑌 = ((𝐶 · 𝑋) + 𝐷)) | |
| 13 | 11, 12 | eqtr3d 2768 | . . . . . 6 ⊢ (𝜑 → ((𝐴 · 𝑋) + 𝐵) = ((𝐶 · 𝑋) + 𝐷)) |
| 14 | 10, 8, 13 | mvlraddd 11524 | . . . . 5 ⊢ (𝜑 → (𝐴 · 𝑋) = (((𝐶 · 𝑋) + 𝐷) − 𝐵)) |
| 15 | 6, 7, 8, 14 | assraddsubd 11528 | . . . 4 ⊢ (𝜑 → (𝐴 · 𝑋) = ((𝐶 · 𝑋) + (𝐷 − 𝐵))) |
| 16 | 6, 9, 15 | mvrladdd 11527 | . . 3 ⊢ (𝜑 → ((𝐴 · 𝑋) − (𝐶 · 𝑋)) = (𝐷 − 𝐵)) |
| 17 | 1, 4, 2, 16 | joinlmulsubmuld 49805 | . 2 ⊢ (𝜑 → ((𝐴 − 𝐶) · 𝑋) = (𝐷 − 𝐵)) |
| 18 | 3, 4, 5, 17 | mvllmuld 11950 | 1 ⊢ (𝜑 → 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 = wceq 1541 ∈ wcel 2111 ≠ wne 2928 (class class class)co 7346 ℂcc 11001 0cc0 11003 + caddc 11006 · cmul 11008 − cmin 11341 / cdiv 11771 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1968 ax-7 2009 ax-8 2113 ax-9 2121 ax-10 2144 ax-11 2160 ax-12 2180 ax-ext 2703 ax-sep 5234 ax-nul 5244 ax-pow 5303 ax-pr 5370 ax-un 7668 ax-resscn 11060 ax-1cn 11061 ax-icn 11062 ax-addcl 11063 ax-addrcl 11064 ax-mulcl 11065 ax-mulrcl 11066 ax-mulcom 11067 ax-addass 11068 ax-mulass 11069 ax-distr 11070 ax-i2m1 11071 ax-1ne0 11072 ax-1rid 11073 ax-rnegex 11074 ax-rrecex 11075 ax-cnre 11076 ax-pre-lttri 11077 ax-pre-lttrn 11078 ax-pre-ltadd 11079 ax-pre-mulgt0 11080 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 848 df-3or 1087 df-3an 1088 df-tru 1544 df-fal 1554 df-ex 1781 df-nf 1785 df-sb 2068 df-mo 2535 df-eu 2564 df-clab 2710 df-cleq 2723 df-clel 2806 df-nfc 2881 df-ne 2929 df-nel 3033 df-ral 3048 df-rex 3057 df-rmo 3346 df-reu 3347 df-rab 3396 df-v 3438 df-sbc 3742 df-csb 3851 df-dif 3905 df-un 3907 df-in 3909 df-ss 3919 df-nul 4284 df-if 4476 df-pw 4552 df-sn 4577 df-pr 4579 df-op 4583 df-uni 4860 df-br 5092 df-opab 5154 df-mpt 5173 df-id 5511 df-po 5524 df-so 5525 df-xp 5622 df-rel 5623 df-cnv 5624 df-co 5625 df-dm 5626 df-rn 5627 df-res 5628 df-ima 5629 df-iota 6437 df-fun 6483 df-fn 6484 df-f 6485 df-f1 6486 df-fo 6487 df-f1o 6488 df-fv 6489 df-riota 7303 df-ov 7349 df-oprab 7350 df-mpo 7351 df-er 8622 df-en 8870 df-dom 8871 df-sdom 8872 df-pnf 11145 df-mnf 11146 df-xr 11147 df-ltxr 11148 df-le 11149 df-sub 11343 df-neg 11344 df-div 11772 |
| This theorem is referenced by: (None) |
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