# Video: AQA GCSE Mathematics Higher Tier Pack 3 β’ Paper 3 β’ Question 26

Point π is the midpoint of the line π΄πΆ. Does π lie on the line π΅π·? You must show your working.

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### Video Transcript

Point π is the midpoint of the line π΄πΆ. Does π lie on the line π΅π·? You must show your working.

The first piece of information weβre given is that point π is the midpoint of the line π΄πΆ. If we begin by finding the vector that describes π΄ to πΆ, we can then form a vector that describes π΄ to π. We donβt currently know how to get from π΄ to πΆ. Instead, weβre going to need to find an alternative route.

We could begin by going from π΄ to π΅. Then we could travel from π΅ to πΆ. In vector terms, we can say that the vector π΄πΆ is equal to the vector π΄π΅ plus the vector π΅πΆ. The vector π΄π΅ is given on the diagram as two π, and the vector π΅πΆ is given as four π. So we can say that the vector π΄πΆ 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. Itβs half of the vector π΄πΆ. That means itβs 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.

Weβll 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 ππ·. Or the line ππ· or the vector ππ· is a portion or a fraction of the vector π΅π·. We donβt 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 πΆ, since we just worked out the vector π΄ to πΆ. And we have a vector that describes the journey from πΆ to π·. So we can say that the vector ππ· is equal to the vector ππΆ plus the vector πΆπ·.

Alternatively, we couldβve 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 ππΆ and πΆπ·, this route is much easier. The vector ππΆ must be equal to the vector π΄π, since π is the midpoint of that line. So itβs π plus two π. And the vector πΆπ· 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 ππ· is either negative π plus three π or this is the same as saying three π minus π. All thatβs left is to find the vector to get from π΅ to π·. This time, weβre going to go from π΅ to πΆ. And then weβre going to travel from πΆ to π·. The vector π΅πΆ is given as four π. And once again, weβre using the vector πΆπ· 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 ππ·. In this case, five π minus two π is some multiple of three π minus π. Well, letβs 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 thereβs 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·.