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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 11620 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ∈ ℂ) |
| 4 | i2linesd.5 | . 2 ⊢ (𝜑 → 𝑋 ∈ ℂ) | |
| 5 | i2linesd.8 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ≠ 0) | |
| 6 | 2, 4 | mulcld 11281 | . . . 4 ⊢ (𝜑 → (𝐶 · 𝑋) ∈ ℂ) |
| 7 | i2linesd.4 | . . . . 5 ⊢ (𝜑 → 𝐷 ∈ ℂ) | |
| 8 | i2linesd.2 | . . . . 5 ⊢ (𝜑 → 𝐵 ∈ ℂ) | |
| 9 | 7, 8 | subcld 11620 | . . . 4 ⊢ (𝜑 → (𝐷 − 𝐵) ∈ ℂ) |
| 10 | 1, 4 | mulcld 11281 | . . . . . 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 2779 | . . . . . 6 ⊢ (𝜑 → ((𝐴 · 𝑋) + 𝐵) = ((𝐶 · 𝑋) + 𝐷)) |
| 14 | 10, 8, 13 | mvlraddd 11673 | . . . . 5 ⊢ (𝜑 → (𝐴 · 𝑋) = (((𝐶 · 𝑋) + 𝐷) − 𝐵)) |
| 15 | 6, 7, 8, 14 | assraddsubd 11677 | . . . 4 ⊢ (𝜑 → (𝐴 · 𝑋) = ((𝐶 · 𝑋) + (𝐷 − 𝐵))) |
| 16 | 6, 9, 15 | mvrladdd 11676 | . . 3 ⊢ (𝜑 → ((𝐴 · 𝑋) − (𝐶 · 𝑋)) = (𝐷 − 𝐵)) |
| 17 | 1, 4, 2, 16 | joinlmulsubmuld 49293 | . 2 ⊢ (𝜑 → ((𝐴 − 𝐶) · 𝑋) = (𝐷 − 𝐵)) |
| 18 | 3, 4, 5, 17 | mvllmuld 12099 | 1 ⊢ (𝜑 → 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))) |
| Colors of variables: wff setvar class |
| Syntax hints: → wi 4 = wceq 1540 ∈ wcel 2108 ≠ wne 2940 (class class class)co 7431 ℂcc 11153 0cc0 11155 + caddc 11158 · cmul 11160 − cmin 11492 / cdiv 11920 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-10 2141 ax-11 2157 ax-12 2177 ax-ext 2708 ax-sep 5296 ax-nul 5306 ax-pow 5365 ax-pr 5432 ax-un 7755 ax-resscn 11212 ax-1cn 11213 ax-icn 11214 ax-addcl 11215 ax-addrcl 11216 ax-mulcl 11217 ax-mulrcl 11218 ax-mulcom 11219 ax-addass 11220 ax-mulass 11221 ax-distr 11222 ax-i2m1 11223 ax-1ne0 11224 ax-1rid 11225 ax-rnegex 11226 ax-rrecex 11227 ax-cnre 11228 ax-pre-lttri 11229 ax-pre-lttrn 11230 ax-pre-ltadd 11231 ax-pre-mulgt0 11232 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-or 849 df-3or 1088 df-3an 1089 df-tru 1543 df-fal 1553 df-ex 1780 df-nf 1784 df-sb 2065 df-mo 2540 df-eu 2569 df-clab 2715 df-cleq 2729 df-clel 2816 df-nfc 2892 df-ne 2941 df-nel 3047 df-ral 3062 df-rex 3071 df-rmo 3380 df-reu 3381 df-rab 3437 df-v 3482 df-sbc 3789 df-csb 3900 df-dif 3954 df-un 3956 df-in 3958 df-ss 3968 df-nul 4334 df-if 4526 df-pw 4602 df-sn 4627 df-pr 4629 df-op 4633 df-uni 4908 df-br 5144 df-opab 5206 df-mpt 5226 df-id 5578 df-po 5592 df-so 5593 df-xp 5691 df-rel 5692 df-cnv 5693 df-co 5694 df-dm 5695 df-rn 5696 df-res 5697 df-ima 5698 df-iota 6514 df-fun 6563 df-fn 6564 df-f 6565 df-f1 6566 df-fo 6567 df-f1o 6568 df-fv 6569 df-riota 7388 df-ov 7434 df-oprab 7435 df-mpo 7436 df-er 8745 df-en 8986 df-dom 8987 df-sdom 8988 df-pnf 11297 df-mnf 11298 df-xr 11299 df-ltxr 11300 df-le 11301 df-sub 11494 df-neg 11495 df-div 11921 |
| This theorem is referenced by: (None) |
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