# Every Point Is Incident With at Least Two Distinct Lines

(D4M) — Here is the video transcript for this video.

00:00
hi everyone welcome back in this episode we’re going to prove that every point
is incident with at least two distinct lines in other words if i put down any
point i want then i know there’s at least two
distinct lines going through that point there so let’s do some math
so we’re going to begin by briefly recapping what the incident axioms are
so we’re going to have a point line and an incidence
are going to be the undefined terms action one says if we’re given two
distinct points we have a unique line passing through them
axiom two says for every line we have two points on it
and a three uh says that there exists three distinct points with the property
that no line passes through all of them so we’re using the word incident
sometimes i’ll use the word passes through

00:01
instead of incident or i’ll say p lies on a line but those are all just
shortcuts for the word incident all right and we have three definitions
here uh points are called collinear if they all lie on a
the same line and concurrent uh lines um they all have a point in common
and in parallel means they have no point in common
all right and so let’s get started but you know
actually before we get started though i want to mention though that if you’re
not clear on all this right here uh make sure and check out the link
below for the full uh series uh in the first video in this episode in the first
episode in the series we uh i went through these axioms very
carefully very slowly and so i recommend watching that episode
also we proved each of these theorems so this is our
episode on theorem one episode on theorem two and so on and right now

00:02
we’re on theorem six so we’ve already proved theorems one through five
in the series uh in previous episodes and so now it’s time for theorem six
every point is incident with at least two lines
and so let’s get started here here we go and i’m going to start by doing this in
a column format so i’ll make a list of statements
and a list of justifications and then after we’re done with that we’ll get on
with the uh paragraph proof all right so here we go um the first
statement i’m going to make is number one is
p is a point so i’m saying every point right so i’m going to name my point it’s
just an arbitrary point and p is a point it could be any point at all
but i want to pick a name for that point so that i can refer to it
but again it’s any point all right justification so that’s just hypothesis and
number two now i’m going to say that a b and c are

00:03
distinct collinear points a b and c are distinct collinear points
and that’s by a3 so a and b are different b and c are
different a and c are different we have three points
and uh they’re non-collinear points sorry meant to say non-collinear points
a3 says they’re non uh non-clean your points there that’s important
all right step three so i know a and b are different so i
know there’s a unique line passing through them and b and c and a and c so
i’m going to write all that out real quick in my proof so there exist lines
there exist lines and i’m going to say a b and then i’m going to say bc

00:04
and i’m going to say ac and so and this this one is going to be incident with
and if you don’t like the word incident here you could say a b passes through
uh a and b and line bc is incident with b and c and incident with a and c
so all three of these statements right here which i’m combining into one
statement is true by axiom a1 so there’s actually a one there
all right so we got these three lines and a and b are on this line and b and c
are on this line and a and c on this line and we can say all these because a
b and c are distinct points they’re all distinct they’re also non-clean year
we’ll get to that here in a second so step four is
i’m going to say what happens with a b line a b and b c

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i know there’s only one line going through a b and i know there’s only one
line going through b c but how do we know about line a b and b
c well i’m going to do an r a hypothesis here i’m going to say line a b is equal
to line ac and that’s going to be our a hypothesis here now
that can’t happen though because uh a b and c are non-collinear if these
three if these two points are equal to each other then i got a line going
through all three points in other words a b and c are collinear
and that’s just definition of collinear and that’s a contradiction
and that would be steps um two and five two says nonconlinear steps
that’s five it says collinear so steps two and five are a direct contradiction

00:06
and so a b and ac i assumed they were equal and we got a
contradiction so they’re not equal so a b is not equal to ac
in other words there we have two distinct lines here a b and a c
now we’re trying to get two distinct lines through p
and we haven’t shown that yet and so right now i’m kind of ignoring p
and what i’m working on are the a b and c and i’m just getting together facts
about them i got these three points here and i got these three lines here
but because we’re trying we’re interested in lines over here so i’m
going to try to get some of these lines going through p and
make sure that they’re different lines so that’s kind of the idea here
now now that i showed that a b line a b and line bc are not the same line
what about you know the b c and ac or what about a b and a e c

00:07
so let’s work on those cases as well so uh let’s go to um
step eight now if i can squeeze that in so step eight will be uh a b and b c
oh i skipped to a b and a c let’s do a b and b c now
so what if these are equal a b and b c and so that these two equal so we’re
going to kind of repeat these steps here four through seven here
um but instead of a b and a c now i’m going to use a b and b c so what if
there is equal to each other so there’s my r a hypothesis
and so now if i get these are equal if these two lines are equal then like i
did right here i’ll get a b and c or collinear so i’ll write that out right
over here i’ll say about right here nine step nine a b and c are collinear are

00:08
linear so if these two lines are equal right so
a is on this line that’s why i wrote it up here a is on this line and b is on
this line and c is on this line and the lines are equal so that means c is also
on this line you know so they’re all on one line
so i’ll say definition of collinear definition of collinear
and for step 10 i’m going to have a contradiction just like i had here and
this will be steps 2 which says it’s not clean year and
step 9 which says it is colleen year all right and then step 11 here
um so i got a contradiction so my r a hypothesis has been refuted
so these are not equal so line a b is not equal to line bc r a a conclusion
and then i’m going to repeat this one more time because we have one more

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possibility so a b is line a b is not equal to line ac
line a b is not equal to line b c which is [Music]
uh yeah we did that right there and so now the last one to check is is line bc
equal to the line ac so number 12 here bc is equal to ac
and that’s my r a hypothesis again and 13.
if these two lines are equal a b and c are collinear and step 14 oops arcelinia
are collinear and that’s the definition of collinear

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i’ll just abbreviate that there step 14 here is contradiction
and this will be steps 2 and 13. 13 says they are collinear
step 2 says they’re not so contradiction and then i have my r a hypothesis here
that these are not equal so bc and a ac these two right here are not equal
and this will be the raa conclusion there
so we have refuted this hypothesis here all right so we’re good to go so far
so what we have is these three lines existing
and we just showed that they’re off uh all three distinct lines we have three
distinct lines there um and so a quick diagram of our proof so far
i’ll erase this real quick and so what we have going on here is

00:11
we haven’t picked a p just any point p and we have points a b and c
and what we know about them is they’re non-collinear
and i drew them as separate from p but we don’t know actually one of them might
be p we’ll deal with that in a second but what we’ve shown so far
is for these three points here we got these lines here
and they’re distinct lines they’re not the same lines
and that’s what we’ve shown so far is we’ve got three distinct lines now we
need to get some lines going through p though right all right so here we go so
step 16 here is going to be exactly one must hold exactly one must hold

00:12
and p p is is one of a b or c or not in other words p is equal to one of
these three points or it’s not equal to any of those three points right so this
is just the law of excluded middle now by the way uh when i was writing up
write a proof no one proof has to be exactly the same this is just the way i
chose to do to do this um so now i got case one p is equal to a i’ll just call
this one here right here case one um actually let me let me bump that out
because i have an or not here so p so k step 17 here is p is one of the three so
this will be case one so case one is it’s equal to one of them
i don’t know which one but it’s equal to one of them

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and now step eighteen is well what if it’s actually equal to the a so i’m
gonna call this here case one a so now we know definitively we’re we’re
in the definitive case that p is equal to a so the diagram i had to go a minute
ago with p here and then we have a b and a c
and we got three distinct lines going through them because we’re in the case
where p is actually equal to the a okay and so then we have this this line
here we’ve already shown it and we have this line here we’ve already shown it
and we have two distinct lines so that that’s what we’ll say here so lines
lines a b and a c are distinct and pass through p and pass through point p
so why is this well we’ve already shown these lines are distinct

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and p is equal to a so it passes through a right a is on this one a is on this
one a is equal to p right so this is just steps of 3 and 11.
i’ll just say previous steps there um previous steps
right we’ve already shown those lines are distinct
and we’re in the case p is equal to a and a is on both of them we’ve already
said all that all right so now we’re on the case where p is equal to b
and we’re going to call this case 1 b 21 so now p is equal to b now p is p is
right here and so this is a a over here now we have the lines

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uh you know a b and b c so the lines a b and b c are distinct and pass through
p pass through p so lines a b and b c are distinct and pass through p
and again this is previous steps here just kind of summarizing our
previous steps there all right and then the last case for case one will be p is
equal to c and so this will be case one c and then 23 step 23 says lines so
if if p is equal to c then we’re going to have a line uh ac and bc so lines ac

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and bc are distinct and pass through p pass through p
and the justification for that is the previous steps because we’ve
already said everything in that uh were in that word and we’re
very shown that they’re distinct and in this case right here when p is
equal to c they pass through c because we said that one passes
through c and we already said that one passes through c all right so in
case one we’re finished and in case one is three sub cases and in each of the
sub cases we came up with two distinct lines passing through p
so let’s go on now to uh case two so i’ll uh get rid of this now and now

00:17
we’re on case 2. so case 2 starts off with um
so what line number are we on so we’re on line number 22 20 23
and that says p is not equal to a p is not equal to b and p is not equal to c
so we’re going to call this here case 2.
all right there we go sorry i got to get that focused all right so
p is not any of them p is not equal to a
it’s not going to be it’s not equal to c
that’s case 2. so case 1 was p is one of them and we went through all the sub
cases all right now case two so now for case two now these are
different points here a is not p p is not a not b right so we can use

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axiom one again so here we go there exist lines and i’m going to say ap and bp
and cp and this one is going to be incident with a and p and incident with bmp
and this one is incident with c and p and this is of course by axiom one
so i use all those one three separate statements use axiom one
i put them all together in one step there so there we go
so we got these three lines ap bp and cp it has all these points on it
a ap is on ap and so on all right so there we go so now what i want i want

00:19
two lines through p and it looks like we have three but actually it
doesn’t necessarily hold that they’re all different lines for example
if you have a b and c as we’ve been writing before our p who knows where it’s at
p actually might be like right here it might be on that line uh we’re under
the case where p is not a b and c so p is not here not here not here but that
doesn’t mean p isn’t on that line um but but even if p is on this line
we would still have two distinct lines going through p um it would be the line
you know bp or ap or ab they would all be equal to each other because they’re
clean here and the line uh cp so i would have two distinct lines going through p

00:20
um but you know we don’t know that we have three you know just because
we write three down here that doesn’t mean any of them are the same or not all
right so long story short let’s assume that they’re all one line
so i’m going to do an r a hypothesis that all three lines are the same
actually we just need to get rid of this case that they’re all the same line
if they’re not all the same line then at
least there’s two lines here there could
be more they all three might be distinct but um if we get rid of the case that
they’re all the same line and that’s false then we’ll have at least two lines
among these three so let’s go here with our a hypothesis
so a would be on this line and b would be on this line and c would be on this
line and so we would have them that they’re collinear so are collinear

00:21
a b and c are collinear and that’s just by definition of collinear
and then now step 27 well that’s a contradiction because steps two
uh steps two and twenty six are negations of each other this one
says there are collinear step 2 says they’re not go linear
all right and then so step 8 28 is the raa conclusion we need to negate this
now how would we negate this so this is really an and here
it’s line ap is equal to line bp and line bp is equal to line cp and
line ap is equal to line cp and so to negate that
you know we’re just going to have that one of them is

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it’s going to be an or so ap is not equal to bp and then bp is not equal to cp
or ap is not equal to cp we don’t know which one of these three holds
but at least we get two points uh sorry at least we get two lines going through
p and so that’s the r a conclusion and so you know that’s basically the proof
right there either these two are not equal or these
two are not equal or these two are not equal but p is on all of them
so it doesn’t matter which ones are not equal you’re going to get at least two
it’s not the case one whole so they can’t be all all these cannot be one
line right all right so there’s at least two lines there could be more okay
so there are the steps to this theorem six here and now let’s look at

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a write-up of the uh proof here so i’m going to
get rid of this here for us and let’s look at a proof now so the first thing
is i’m going to say p is a point and i’m not going to say anything else about p
um it’s just a point i’m not saying if it’s equal to anything else or
anything else i’m just starting off with a point that’s important there
and then now i’m going to try to get some more points axiom three says we
have three distinct non-collinear points
and and like i said a minute ago i don’t know if p’s any of them are not
so it could be or could be not we have to look at all those cases all right so
since that since a b and c are uh non-collinear uh i put clean you
there so it should be non-clean you’re here non-linear here
that needs to be fixed there so since a b and c are
non-collinear it follows that line a b line ac and line bc are three distinct

00:24
lines now we went through those proofs line by line case by case
and so i’m just kind of summarizing that here each case was very short in itself
and so i’m just summarizing that so uh you know for example it follows these
are three distinct lines so the argument was these two are not the same line if
these two are the same line then a b and c are collinear
but they’re non-collinear axiom a3 says these three lines these three points are
non-collinear and so we made that argument for each pair of lines right there so
there’s the next sentence so now we’re going to suppose
that p is one of these points so we went through the case p is equal
to a p is equal to b p is equal to c we went through each of those cases and
they all ended up exactly the same so i’m going to say without loss of

00:25
generality that p is equal to the a and in that case by axiom one it follows
that these right here are two distinct lines uh that are incident with p so
by axiom one um you know we have um you know p is equal to a
so we know that they’re incident with with p and we know we have
uh those two lines right there so when when
when p is equal to a we’re done we got the two lines that go through p they go
through p because uh p is equal to a in this case you know p is equal to a
um and we have two distinct lines as we’ve already proven up here
all right and so now what if we have four distinct points so by axiom one
there exists line ap incident with a and p and b p incident with b and p and c p

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incident with c and p so we got axiom one giving us those three lines
we don’t know if those three lines are all different
or some of them are the same or in fact the worst case would be that they’re all
equal to each other so let’s look at the
worst case what if all these three lines are equal to each other
so then that would put a b and c all on the same line that would make them
collinear but we know that they’re non-collinear
so hence these cannot be the same line so at least two of these are different
from each other so we have at least two lines passing through p and there we go
there’s the proof and hey you want to do some more math
together let’s do let’s do some right here i’ll see you in that episode