Metamath Proof Explorer


Theorem line2x

Description: Example for a horizontal line G passing through two different points in "standard form". (Contributed by AV, 3-Feb-2023)

Ref Expression
Hypotheses line2.i I=12
line2.e E=msup
line2.p P=I
line2.l L=LineME
line2.g G=pP|Ap1+Bp2=C
line2x.x X=102M
line2x.y Y=112M
Assertion line2x ABB0CMG=XLYA=0M=CB

Proof

Step Hyp Ref Expression
1 line2.i I=12
2 line2.e E=msup
3 line2.p P=I
4 line2.l L=LineME
5 line2.g G=pP|Ap1+Bp2=C
6 line2x.x X=102M
7 line2x.y Y=112M
8 5 a1i ABB0CMG=pP|Ap1+Bp2=C
9 1ex 1V
10 2ex 2V
11 9 10 pm3.2i 1V2V
12 c0ex 0V
13 12 jctl M0VM
14 1ne2 12
15 14 a1i M12
16 fprg 1V2V0VM12102M:120M
17 0red 1V2V0
18 simpr 0VMM
19 17 18 anim12i 1V2V0VM0M
20 19 3adant3 1V2V0VM120M
21 prssi 0M0M
22 20 21 syl 1V2V0VM120M
23 16 22 fssd 1V2V0VM12102M:12
24 11 13 15 23 mp3an2i M102M:12
25 1 feq2i 102M:I102M:12
26 24 25 sylibr M102M:I
27 reex V
28 prex 12V
29 1 28 eqeltri IV
30 27 29 elmap 102MI102M:I
31 26 30 sylibr M102MI
32 31 6 3 3eltr4g MXP
33 9 jctl M1VM
34 fprg 1V2V1VM12112M:121M
35 11 33 15 34 mp3an2i M112M:121M
36 1re 1
37 prssi 1M1M
38 36 37 mpan M1M
39 35 38 fssd M112M:12
40 1 feq2i 112M:I112M:12
41 39 40 sylibr M112M:I
42 27 29 pm3.2i VIV
43 elmapg VIV112MI112M:I
44 42 43 mp1i M112MI112M:I
45 41 44 mpbird M112MI
46 45 7 3 3eltr4g MYP
47 opex 10V
48 opex 2MV
49 47 48 pm3.2i 10V2MV
50 opex 11V
51 50 48 pm3.2i 11V2MV
52 49 51 pm3.2i 10V2MV11V2MV
53 14 orci 120M
54 9 12 opthne 102M120M
55 53 54 mpbir 102M
56 55 a1i M102M
57 0ne1 01
58 57 olci 1101
59 9 12 opthne 10111101
60 58 59 mpbir 1011
61 56 60 jctil M1011102M
62 61 orcd M1011102M2M112M2M
63 prneimg 10V2MV11V2MV1011102M2M112M2M102M112M
64 52 62 63 mpsyl M102M112M
65 64 6 7 3netr4g MXY
66 32 46 65 3jca MXPYPXY
67 66 adantl ABB0CMXPYPXY
68 eqid Y1X1=Y1X1
69 eqid Y2X2=Y2X2
70 eqid X2Y1X1Y2=X2Y1X1Y2
71 1 2 3 4 68 69 70 rrx2linest XPYPXYXLY=pP|Y1X1p2=Y2X2p1+X2Y1-X1Y2
72 67 71 syl ABB0CMXLY=pP|Y1X1p2=Y2X2p1+X2Y1-X1Y2
73 8 72 eqeq12d ABB0CMG=XLYpP|Ap1+Bp2=C=pP|Y1X1p2=Y2X2p1+X2Y1-X1Y2
74 rabbi pPAp1+Bp2=CY1X1p2=Y2X2p1+X2Y1-X1Y2pP|Ap1+Bp2=C=pP|Y1X1p2=Y2X2p1+X2Y1-X1Y2
75 7 fveq1i Y1=112M1
76 9 9 14 3pm3.2i 1V1V12
77 fvpr1g 1V1V12112M1=1
78 76 77 mp1i M112M1=1
79 75 78 eqtrid MY1=1
80 6 fveq1i X1=102M1
81 9 12 14 3pm3.2i 1V0V12
82 fvpr1g 1V0V12102M1=0
83 81 82 mp1i M102M1=0
84 80 83 eqtrid MX1=0
85 79 84 oveq12d MY1X1=10
86 1m0e1 10=1
87 85 86 eqtrdi MY1X1=1
88 87 oveq1d MY1X1p2=1p2
89 7 fveq1i Y2=112M2
90 fvpr2g 2VM12112M2=M
91 10 14 90 mp3an13 M112M2=M
92 89 91 eqtrid MY2=M
93 6 fveq1i X2=102M2
94 fvpr2g 2VM12102M2=M
95 10 14 94 mp3an13 M102M2=M
96 93 95 eqtrid MX2=M
97 92 96 oveq12d MY2X2=MM
98 recn MM
99 98 subidd MMM=0
100 97 99 eqtrd MY2X2=0
101 100 oveq1d MY2X2p1=0p1
102 9 9 15 77 mp3an12i M112M1=1
103 75 102 eqtrid MY1=1
104 96 103 oveq12d MX2Y1= M1
105 ax-1rid M M1=M
106 104 105 eqtrd MX2Y1=M
107 9 12 15 82 mp3an12i M102M1=0
108 80 107 eqtrid MX1=0
109 108 92 oveq12d MX1Y2=0 M
110 98 mul02d M0 M=0
111 109 110 eqtrd MX1Y2=0
112 106 111 oveq12d MX2Y1X1Y2=M0
113 98 subid1d MM0=M
114 112 113 eqtrd MX2Y1X1Y2=M
115 101 114 oveq12d MY2X2p1+X2Y1-X1Y2=0p1+M
116 88 115 eqeq12d MY1X1p2=Y2X2p1+X2Y1-X1Y21p2=0p1+M
117 116 adantl ABB0CMY1X1p2=Y2X2p1+X2Y1-X1Y21p2=0p1+M
118 1 3 rrx2pyel pPp2
119 118 recnd pPp2
120 119 mullidd pP1p2=p2
121 1 3 rrx2pxel pPp1
122 121 recnd pPp1
123 122 mul02d pP0p1=0
124 123 oveq1d pP0p1+M=0+M
125 120 124 eqeq12d pP1p2=0p1+Mp2=0+M
126 117 125 sylan9bb ABB0CMpPY1X1p2=Y2X2p1+X2Y1-X1Y2p2=0+M
127 126 bibi2d ABB0CMpPAp1+Bp2=CY1X1p2=Y2X2p1+X2Y1-X1Y2Ap1+Bp2=Cp2=0+M
128 127 ralbidva ABB0CMpPAp1+Bp2=CY1X1p2=Y2X2p1+X2Y1-X1Y2pPAp1+Bp2=Cp2=0+M
129 98 addlidd M0+M=M
130 129 adantr MpP0+M=M
131 130 eqeq2d MpPp2=0+Mp2=M
132 131 bibi2d MpPAp1+Bp2=Cp2=0+MAp1+Bp2=Cp2=M
133 132 ralbidva MpPAp1+Bp2=Cp2=0+MpPAp1+Bp2=Cp2=M
134 133 adantl ABB0CMpPAp1+Bp2=Cp2=0+MpPAp1+Bp2=Cp2=M
135 1 2 3 4 5 6 7 line2xlem ABB0CMpPAp1+Bp2=Cp2=MA=0M=CB
136 oveq1 A=0Ap1=0p1
137 136 adantr A=0M=CBAp1=0p1
138 137 ad2antlr ABB0CMA=0M=CBpPAp1=0p1
139 123 adantl ABB0CMA=0M=CBpP0p1=0
140 138 139 eqtrd ABB0CMA=0M=CBpPAp1=0
141 140 oveq1d ABB0CMA=0M=CBpPAp1+Bp2=0+Bp2
142 recn BB
143 142 adantr BB0B
144 143 3ad2ant2 ABB0CB
145 144 ad3antrrr ABB0CMA=0M=CBpPB
146 119 adantl ABB0CMA=0M=CBpPp2
147 145 146 mulcld ABB0CMA=0M=CBpPBp2
148 147 addlidd ABB0CMA=0M=CBpP0+Bp2=Bp2
149 141 148 eqtrd ABB0CMA=0M=CBpPAp1+Bp2=Bp2
150 149 eqeq1d ABB0CMA=0M=CBpPAp1+Bp2=CBp2=C
151 simp3 ABB0CC
152 151 recnd ABB0CC
153 152 ad3antrrr ABB0CMA=0M=CBpPC
154 simpl BB0B
155 154 recnd BB0B
156 155 3ad2ant2 ABB0CB
157 156 ad3antrrr ABB0CMA=0M=CBpPB
158 simp2r ABB0CB0
159 158 ad3antrrr ABB0CMA=0M=CBpPB0
160 153 157 146 159 divmuld ABB0CMA=0M=CBpPCB=p2Bp2=C
161 simpr A=0M=CBM=CB
162 161 eqcomd A=0M=CBCB=M
163 162 ad2antlr ABB0CMA=0M=CBpPCB=M
164 163 eqeq1d ABB0CMA=0M=CBpPCB=p2M=p2
165 150 160 164 3bitr2d ABB0CMA=0M=CBpPAp1+Bp2=CM=p2
166 eqcom M=p2p2=M
167 165 166 bitrdi ABB0CMA=0M=CBpPAp1+Bp2=Cp2=M
168 167 ralrimiva ABB0CMA=0M=CBpPAp1+Bp2=Cp2=M
169 168 ex ABB0CMA=0M=CBpPAp1+Bp2=Cp2=M
170 135 169 impbid ABB0CMpPAp1+Bp2=Cp2=MA=0M=CB
171 128 134 170 3bitrd ABB0CMpPAp1+Bp2=CY1X1p2=Y2X2p1+X2Y1-X1Y2A=0M=CB
172 74 171 bitr3id ABB0CMpP|Ap1+Bp2=C=pP|Y1X1p2=Y2X2p1+X2Y1-X1Y2A=0M=CB
173 73 172 bitrd ABB0CMG=XLYA=0M=CB