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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 11567 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ∈ ℂ) |
4 | i2linesd.5 | . 2 ⊢ (𝜑 → 𝑋 ∈ ℂ) | |
5 | i2linesd.8 | . 2 ⊢ (𝜑 → (𝐴 − 𝐶) ≠ 0) | |
6 | 2, 4 | mulcld 11230 | . . . 4 ⊢ (𝜑 → (𝐶 · 𝑋) ∈ ℂ) |
7 | i2linesd.4 | . . . . 5 ⊢ (𝜑 → 𝐷 ∈ ℂ) | |
8 | i2linesd.2 | . . . . 5 ⊢ (𝜑 → 𝐵 ∈ ℂ) | |
9 | 7, 8 | subcld 11567 | . . . 4 ⊢ (𝜑 → (𝐷 − 𝐵) ∈ ℂ) |
10 | 1, 4 | mulcld 11230 | . . . . . 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 2775 | . . . . . 6 ⊢ (𝜑 → ((𝐴 · 𝑋) + 𝐵) = ((𝐶 · 𝑋) + 𝐷)) |
14 | 10, 8, 13 | mvlraddd 11620 | . . . . 5 ⊢ (𝜑 → (𝐴 · 𝑋) = (((𝐶 · 𝑋) + 𝐷) − 𝐵)) |
15 | 6, 7, 8, 14 | assraddsubd 11624 | . . . 4 ⊢ (𝜑 → (𝐴 · 𝑋) = ((𝐶 · 𝑋) + (𝐷 − 𝐵))) |
16 | 6, 9, 15 | mvrladdd 11623 | . . 3 ⊢ (𝜑 → ((𝐴 · 𝑋) − (𝐶 · 𝑋)) = (𝐷 − 𝐵)) |
17 | 1, 4, 2, 16 | joinlmulsubmuld 47723 | . 2 ⊢ (𝜑 → ((𝐴 − 𝐶) · 𝑋) = (𝐷 − 𝐵)) |
18 | 3, 4, 5, 17 | mvllmuld 12042 | 1 ⊢ (𝜑 → 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 = wceq 1542 ∈ wcel 2107 ≠ wne 2941 (class class class)co 7404 ℂcc 11104 0cc0 11106 + caddc 11109 · cmul 11111 − cmin 11440 / cdiv 11867 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1914 ax-6 1972 ax-7 2012 ax-8 2109 ax-9 2117 ax-10 2138 ax-11 2155 ax-12 2172 ax-ext 2704 ax-sep 5298 ax-nul 5305 ax-pow 5362 ax-pr 5426 ax-un 7720 ax-resscn 11163 ax-1cn 11164 ax-icn 11165 ax-addcl 11166 ax-addrcl 11167 ax-mulcl 11168 ax-mulrcl 11169 ax-mulcom 11170 ax-addass 11171 ax-mulass 11172 ax-distr 11173 ax-i2m1 11174 ax-1ne0 11175 ax-1rid 11176 ax-rnegex 11177 ax-rrecex 11178 ax-cnre 11179 ax-pre-lttri 11180 ax-pre-lttrn 11181 ax-pre-ltadd 11182 ax-pre-mulgt0 11183 |
This theorem depends on definitions: df-bi 206 df-an 398 df-or 847 df-3or 1089 df-3an 1090 df-tru 1545 df-fal 1555 df-ex 1783 df-nf 1787 df-sb 2069 df-mo 2535 df-eu 2564 df-clab 2711 df-cleq 2725 df-clel 2811 df-nfc 2886 df-ne 2942 df-nel 3048 df-ral 3063 df-rex 3072 df-rmo 3377 df-reu 3378 df-rab 3434 df-v 3477 df-sbc 3777 df-csb 3893 df-dif 3950 df-un 3952 df-in 3954 df-ss 3964 df-nul 4322 df-if 4528 df-pw 4603 df-sn 4628 df-pr 4630 df-op 4634 df-uni 4908 df-br 5148 df-opab 5210 df-mpt 5231 df-id 5573 df-po 5587 df-so 5588 df-xp 5681 df-rel 5682 df-cnv 5683 df-co 5684 df-dm 5685 df-rn 5686 df-res 5687 df-ima 5688 df-iota 6492 df-fun 6542 df-fn 6543 df-f 6544 df-f1 6545 df-fo 6546 df-f1o 6547 df-fv 6548 df-riota 7360 df-ov 7407 df-oprab 7408 df-mpo 7409 df-er 8699 df-en 8936 df-dom 8937 df-sdom 8938 df-pnf 11246 df-mnf 11247 df-xr 11248 df-ltxr 11249 df-le 11250 df-sub 11442 df-neg 11443 df-div 11868 |
This theorem is referenced by: (None) |
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