### Video Transcript

Point π is the midpoint of the
line π΄πΆ. Does π lie on the line π΅π·? You must show your working.

The first piece of information
weβre given is that point π is the midpoint of the line π΄πΆ. If we begin by finding the vector
that describes π΄ to πΆ, we can then form a vector that describes π΄ to π. We donβt currently know how to get
from π΄ to πΆ. Instead, weβre going to need to
find an alternative route.

We could begin by going from π΄ to
π΅. Then we could travel from π΅ to
πΆ. In vector terms, we can say that
the vector π΄πΆ is equal to the vector π΄π΅ plus the vector π΅πΆ. The vector π΄π΅ is given on the
diagram as two π, and the vector π΅πΆ is given as four π. So we can say that the vector π΄πΆ
is equal to two π plus four π.

Using the fact that π is the
midpoint of this line, we can find the vector π΄π. Itβs half of the vector π΄πΆ. That means itβs a half of two π
plus four π. This might look unnecessarily
complicated. But we can expand it just as we
would expand any normal brackets.

Weβll multiply everything on the
inside of the brackets by the number on the outside. One-half of two π or a half
multiplied by two π is simply π. And a half multiplied by four is
two. So a half multiplied by four π is
two π. And we can see that the vector π΄π
is given by π plus two π.

If the point π does indeed lie on
the line π΅π·, then we will know that the line π΅π· is some multiple of the line
ππ·. Or the line ππ· or the vector ππ·
is a portion or a fraction of the vector π΅π·. We donβt currently have a vector
that describes the line from π to π·, that is assuming π lies on this line.

We do, however, have a vector that
describes the journey from π to πΆ, since we just worked out the vector π΄ to
πΆ. And we have a vector that describes
the journey from πΆ to π·. So we can say that the vector ππ·
is equal to the vector ππΆ plus the vector πΆπ·.

Alternatively, we couldβve used the
vector to get from π to π΄ and then used the vector to get from π΄ to π·. However, since we already know the
vectors ππΆ and πΆπ·, this route is much easier. The vector ππΆ must be equal to
the vector π΄π, since π is the midpoint of that line. So itβs π plus two π. And the vector πΆπ· is given on the
diagram as π minus two π.

We can add together the like
terms. π minus two π is negative π, and
two π plus π is three π. So the vector ππ· is either
negative π plus three π or this is the same as saying three π minus π. All thatβs left is to find the
vector to get from π΅ to π·. This time, weβre going to go from
π΅ to πΆ. And then weβre going to travel from
πΆ to π·. The vector π΅πΆ is given as four
π. And once again, weβre using the
vector πΆπ· as π minus two π. So the vector π΅π· is given as five
π minus two π.

Remember, we said that π lies on
the line π΅π· if π΅π· is some multiple of ππ·. In this case, five π minus two π
is some multiple of three π minus π. Well, letβs see if we can work out
what that multiple would need to be. To get from negative π to negative
two π, this multiple would need to be two. However, two multiplied by three π
would be six π. So thereβs no number that we can
multiply the vector three π minus π by to get five π minus two π. This means that π does not lie on
the line π΅π·.