### Video Transcript

πππ΄, πππ΅, and π΄π΅π are straight lines. Four ππ equals ππ΄. π΄π equals three over two π΄π΅. Vector ππ΄ equals π. Vector ππ΅ equals π. πππ is a straight line. Prove that π is the midpoint of ππ΅.

So the first thing I want to actually draw your attention towards is actually some
notation. So as you can see, Iβve actually written here π with a line underneath and π with a
line underneath. And we can actually write these instead of bold π and bold π. The reason Iβve done this is one: itβs easy for me to write; two: itβs easy for you
to see. And then also when youβre doing an exam, itβs actually easy for you to write and be
clear for an examiner. Okay, it also means the exact same thing.

Now, letβs move on and actually solve the problem. So to start off with, what we can look at is that four ππ is equal to ππ΄. So therefore, this will tell us that the vector ππ is going to equal to a quarter
of the vector ππ΄. So therefore, this is gonna be equal to a quarter π cause weβve already been told
that the vector ππ΄ is equal to π.

So next, what weβre gonna do is to take a look at the vector π΄π΅. So thereβs going from π΄ to π΅. Well, to go from π΄ to π΅, using the vectors that we know, we go from π΄ to π and
then π to π΅. So vector π΄π΅ is equal to π΄π plus ππ΅, which is gonna be equal to vector ππ΅
minus vector ππ΄. And thatβs because we actually went in the opposite direction of our way. So we actually went from π΄ to π, which can also be written as negative ππ΄. So therefore, this is gonna be equal to π minus π. And thatβs because weβre told that vector ππ΄ is equal to π and vector ππ΅ is
equal to π.

So now, what we can do is actually use a bit of information that tells us that π΄π
is equal three over two π΄π΅ because, therefore, vector π΄π is gonna be equal to
three over two vector π΄π΅. And weβve actually just found vectors π΄π΅. So therefore, we can say that vector π΄π is gonna be equal to three over two π
minus π. So great, we found that. And now, what we can do is actually move on to the next step.

Now, the next step is actually to take a look at our π because what we can say about
the vector ππ is this is gonna be equal to π, which is just a constant,
multiplied by the vector ππ΅. And we know that because weβre told at the beginning that πππ΅ is actually a
straight line. So therefore, as we know that vector ππ΅ is equal to π, we can actually say that
vector ππ is gonna be equal to ππ. And itβs this π that we want to find because if we find this π, itβs actually gonna
help us to prove that π is the midpoint of ππ΅.

Well, now, letβs actually consider vector ππ. Well, vector ππ is gonna be equal to vector ππ΄ plus vector π΄π. Well, these are actually both vectors that we know. So therefore, we can say that vector ππ is gonna be equal to π plus three over two
π minus π.

So now, what we can actually do is actually consider the vector ππ. And what Iβve done to highlight this is actually drawing it with a dashed pink
line. Well, the vector ππ is actually gonna be equal to vector ππ minus vector
ππ. And weβve actually chosen this rule because these are both the vectors that we know
cause weβve found them out. So therefore, we can say that this is actually gonna be equal to π plus three over
two multiplied by π minus π then minus a quarter π. And this is because a quarter π was actually ππ, weβve found that earlier on.

So therefore, if we actually expand the bracket, what weβre actually gonna get is π
plus three over two π minus three over two π minus a quarter π. And this is gonna give us the expression three over two π minus three-quarter
π. And we got the three-quarter π because we had π and then we subtracted one and a
half π from it, which gives us negative a half π. And then, we subtracted another quarter π from it, which gave us negative
three-quarter π. So great, weβve actually done that. Now, we found vector ππ.

Well, next, what we can actually do is move on and start to think about ππ because
what that ππ can do is actually help us to think about the π that weβre looking
for earlier. So if we actually travel from π to π, so we take the vector ππ, then weβre
actually gonna go from π to π and then π to π. So therefore, itβs gonna be equal to vector ππ minus vector ππ cause like we said
we went π to π, which is actually the negative of π to π. So therefore, this is gonna give us that the vector ππ is equal to ππ minus a
quarter π. And thatβs cause our vector ππ was ππ and vector ππ is actually a quarter
π.

So therefore, what we can actually say is that because πππ is a straight line,
then ππ is gonna be a multiple of ππ. And we can actually use that to actually help us to find out what our π is, which
therefore will enable us to prove that π is the midpoint of ππ΅. So therefore, as we already said, the vector ππ is gonna be equal to π, the vector
ππ, where π is actually a multiple. So weβre saying itβs gonna be a multiple of the vector ππ.

So now, we need to consider what possibly that could be. Well, if we take a look at the coefficients of our πs, then, weβve got negative
three-quarters and negative a quarter. We actually have to multiply negative a quarter by three to get to negative
three-quarters. So you can see that actually the vector ππ is three times the vector ππ. So we can actually say that vector ππ equals three vector ππ.

But as weβre trying to find π, thereβs actually a more useful way that we can think
of this. And thatβs that a third of vector ππ is equal to the vector ππ. And weβve already proved this with the π coefficient because actually a quarter or
negative quarter is actually a third of negative three-quarters. So therefore, if itβs actually a third with the coefficients of π, it must be the
same with the coefficients of π.

So what we can say that actually π must be a third of three over two. And we can rewrite it like this: a third multiplied by three over two is gonna be
equal to π. So therefore, we can say that three over six is equal to π because you actually have
one multiplied by three is three and three multiplied by two is six. So this actually gives us a value of π of a half.

Okay, great, but how thatβs gonna help us? Well, using this bit of information, so π being equal to a half, we can say that
therefore vector ππ is equal to a half vector ππ΅. And thatβs because we found out earlier that vector ππ was equal to π vector
ππ΅.

So therefore, what we can actually say is weβve actually proved that π is the
midpoint of ππ΅.