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

Point π‘š is the midpoint of the line 𝐴𝐢. Does π‘š lie on the line 𝐡𝐷? You must show your working.

04:09

Video Transcript

Point π‘š is the midpoint of the line 𝐴𝐢. Does π‘š lie on the line 𝐡𝐷? You must show your working.

The first piece of information we’re given is that point π‘š is the midpoint of the line 𝐴𝐢. If we begin by finding the vector that describes 𝐴 to 𝐢, we can then form a vector that describes 𝐴 to π‘š. We don’t currently know how to get from 𝐴 to 𝐢. Instead, we’re going to need to find an alternative route.

We could begin by going from 𝐴 to 𝐡. Then we could travel from 𝐡 to 𝐢. In vector terms, we can say that the vector 𝐴𝐢 is equal to the vector 𝐴𝐡 plus the vector 𝐡𝐢. The vector 𝐴𝐡 is given on the diagram as two π‘Ž, and the vector 𝐡𝐢 is given as four 𝑏. So we can say that the vector 𝐴𝐢 is equal to two π‘Ž plus four 𝑏.

Using the fact that π‘š is the midpoint of this line, we can find the vector π΄π‘š. It’s half of the vector 𝐴𝐢. That means it’s a half of two π‘Ž plus four 𝑏. This might look unnecessarily complicated. But we can expand it just as we would expand any normal brackets.

We’ll multiply everything on the inside of the brackets by the number on the outside. One-half of two π‘Ž or a half multiplied by two π‘Ž is simply π‘Ž. And a half multiplied by four is two. So a half multiplied by four 𝑏 is two 𝑏. And we can see that the vector π΄π‘š is given by π‘Ž plus two 𝑏.

If the point π‘š does indeed lie on the line 𝐡𝐷, then we will know that the line 𝐡𝐷 is some multiple of the line π‘šπ·. Or the line π‘šπ· or the vector π‘šπ· is a portion or a fraction of the vector 𝐡𝐷. We don’t currently have a vector that describes the line from π‘š to 𝐷, that is assuming π‘š lies on this line.

We do, however, have a vector that describes the journey from π‘š to 𝐢, since we just worked out the vector 𝐴 to 𝐢. And we have a vector that describes the journey from 𝐢 to 𝐷. So we can say that the vector π‘šπ· is equal to the vector π‘šπΆ plus the vector 𝐢𝐷.

Alternatively, we could’ve used the vector to get from π‘š to 𝐴 and then used the vector to get from 𝐴 to 𝐷. However, since we already know the vectors π‘šπΆ and 𝐢𝐷, this route is much easier. The vector π‘šπΆ must be equal to the vector π΄π‘š, since π‘š is the midpoint of that line. So it’s π‘Ž plus two 𝑏. And the vector 𝐢𝐷 is given on the diagram as 𝑏 minus two π‘Ž.

We can add together the like terms. π‘Ž minus two π‘Ž is negative π‘Ž, and two 𝑏 plus 𝑏 is three 𝑏. So the vector π‘šπ· is either negative π‘Ž plus three 𝑏 or this is the same as saying three 𝑏 minus π‘Ž. All that’s left is to find the vector to get from 𝐡 to 𝐷. This time, we’re going to go from 𝐡 to 𝐢. And then we’re going to travel from 𝐢 to 𝐷. The vector 𝐡𝐢 is given as four 𝑏. And once again, we’re using the vector 𝐢𝐷 as 𝑏 minus two π‘Ž. So the vector 𝐡𝐷 is given as five 𝑏 minus two π‘Ž.

Remember, we said that π‘š lies on the line 𝐡𝐷 if 𝐡𝐷 is some multiple of π‘šπ·. In this case, five 𝑏 minus two π‘Ž is some multiple of three 𝑏 minus π‘Ž. Well, let’s see if we can work out what that multiple would need to be. To get from negative π‘Ž to negative two π‘Ž, this multiple would need to be two. However, two multiplied by three 𝑏 would be six 𝑏. So there’s no number that we can multiply the vector three 𝑏 minus π‘Ž by to get five 𝑏 minus two π‘Ž. This means that π‘š does not lie on the line 𝐡𝐷.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.