Theorem i2linesi 51042

Description: Solve for the intersection of two lines expressed in Y = MX+B form (note that the lines cannot be vertical). Here we use inference 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, 11-Oct-2018.)

Hypotheses

Ref	Expression
i2linesi.1	⊢ 𝐴 ∈ ℂ
i2linesi.2	⊢ 𝐵 ∈ ℂ
i2linesi.3	⊢ 𝐶 ∈ ℂ
i2linesi.4	⊢ 𝐷 ∈ ℂ
i2linesi.5	⊢ 𝑋 ∈ ℂ
i2linesi.6	⊢ 𝑌 = ((𝐴 · 𝑋) + 𝐵)
i2linesi.7	⊢ 𝑌 = ((𝐶 · 𝑋) + 𝐷)
i2linesi.8	⊢ (𝐴 − 𝐶) ≠ 0

Assertion

Ref	Expression
i2linesi	⊢ 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))

Proof of Theorem i2linesi

Step	Hyp	Ref	Expression
1		i2linesi.1	. . 3 ⊢ 𝐴 ∈ ℂ
2		i2linesi.3	. . 3 ⊢ 𝐶 ∈ ℂ
3	1, 2	subcli 11637	. 2 ⊢ (𝐴 − 𝐶) ∈ ℂ
4		i2linesi.5	. 2 ⊢ 𝑋 ∈ ℂ
5		i2linesi.8	. 2 ⊢ (𝐴 − 𝐶) ≠ 0
6	2, 4	mulcli 11319	. . . 4 ⊢ (𝐶 · 𝑋) ∈ ℂ
7		i2linesi.4	. . . . 5 ⊢ 𝐷 ∈ ℂ
8		i2linesi.2	. . . . 5 ⊢ 𝐵 ∈ ℂ
9	7, 8	subcli 11637	. . . 4 ⊢ (𝐷 − 𝐵) ∈ ℂ
10	1, 4	mulcli 11319	. . . . . 6 ⊢ (𝐴 · 𝑋) ∈ ℂ
11		i2linesi.6	. . . . . . 7 ⊢ 𝑌 = ((𝐴 · 𝑋) + 𝐵)
12		i2linesi.7	. . . . . . 7 ⊢ 𝑌 = ((𝐶 · 𝑋) + 𝐷)
13	11, 12	eqtr3i 2785	. . . . . 6 ⊢ ((𝐴 · 𝑋) + 𝐵) = ((𝐶 · 𝑋) + 𝐷)
14	10, 8, 13	mvlraddi 51035	. . . . 5 ⊢ (𝐴 · 𝑋) = (((𝐶 · 𝑋) + 𝐷) − 𝐵)
15	6, 7, 8, 14	assraddsubi 51036	. . . 4 ⊢ (𝐴 · 𝑋) = ((𝐶 · 𝑋) + (𝐷 − 𝐵))
16	6, 9, 15	mvrladdi 11578	. . 3 ⊢ ((𝐴 · 𝑋) − (𝐶 · 𝑋)) = (𝐷 − 𝐵)
17	1, 4, 2, 16	joinlmulsubmuli 51039	. 2 ⊢ ((𝐴 − 𝐶) · 𝑋) = (𝐷 − 𝐵)
18	3, 4, 5, 17	mvllmuli 12153	1 ⊢ 𝑋 = ((𝐷 − 𝐵) / (𝐴 − 𝐶))

This proof depends on syntax axioms:   = wceq 1569   ∈ wcel 2144   ≠ wne 2955  (class class class)co 7420  ℂcc 11201  0cc0 11203   + caddc 11206   · cmul 11208   − cmin 11544   / cdiv 11976

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pow 5326  ax-pr 5390  ax-un 7751  ax-resscn 11260  ax-1cn 11261  ax-icn 11262  ax-addcl 11263  ax-addrcl 11264  ax-mulcl 11265  ax-mulrcl 11266  ax-mulcom 11267  ax-addass 11268  ax-mulass 11269  ax-distr 11270  ax-i2m1 11271  ax-1ne0 11272  ax-1rid 11273  ax-rnegex 11274  ax-rrecex 11275  ax-cnre 11276  ax-pre-lttri 11277  ax-pre-lttrn 11278  ax-pre-ltadd 11279  ax-pre-mulgt0 11280

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-nel 3062  df-ral 3077  df-rex 3087  df-rmo 3365  df-reu 3366  df-rab 3413  df-v 3452  df-sbc 3739  df-csb 3847  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-pw 4558  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5545  df-po 5558  df-so 5559  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-iota 6494  df-fun 6540  df-fn 6541  df-f 6542  df-f1 6543  df-fo 6544  df-f1o 6545  df-fv 6546  df-riota 7377  df-ov 7423  df-oprab 7424  df-mpo 7425  df-er 8720  df-en 8977  df-dom 8978  df-sdom 8979  df-pnf 11348  df-mnf 11349  df-xr 11350  df-ltxr 11351  df-le 11352  df-sub 11546  df-neg 11547  df-div 11977

This theorem is used by: (None)