Video: Pack 4 β€’ Paper 2 β€’ Question 17

Pack 4 β€’ Paper 2 β€’ Question 17

07:04

Video Transcript

𝑂𝑋𝐴, 𝑂𝑀𝐡, and π΄π΅π‘Œ are straight lines. Four 𝑂𝑋 equals 𝑂𝐴. π΄π‘Œ equals three over two 𝐴𝐡. Vector 𝑂𝐴 equals π‘Ž. Vector 𝑂𝐡 equals 𝑏. π‘‹π‘€π‘Œ is a straight line. Prove that 𝑀 is the midpoint of 𝑂𝐡.

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

Now, let’s move on and actually solve the problem. So to start off with, what we can look at is that four 𝑂𝑋 is equal to 𝑂𝐴. So therefore, this will tell us that the vector 𝑂𝑋 is going to equal to a quarter of the vector 𝑂𝐴. So therefore, this is gonna be equal to a quarter π‘Ž cause we’ve already been told that the vector 𝑂𝐴 is equal to π‘Ž.

So next, what we’re gonna do is to take a look at the vector 𝐴𝐡. So there’s going from 𝐴 to 𝐡. Well, to go from 𝐴 to 𝐡, using the vectors that we know, we go from 𝐴 to 𝑂 and then 𝑂 to 𝐡. So vector 𝐴𝐡 is equal to 𝐴𝑂 plus 𝑂𝐡, which is gonna be equal to vector 𝑂𝐡 minus vector 𝑂𝐴. And that’s because we actually went in the opposite direction of our way. So we actually went from 𝐴 to 𝑂, which can also be written as negative 𝑂𝐴. So therefore, this is gonna be equal to 𝑏 minus π‘Ž. And that’s because we’re told that vector 𝑂𝐴 is equal to π‘Ž and vector 𝑂𝐡 is equal to 𝑏.

So now, what we can do is actually use a bit of information that tells us that π΄π‘Œ is equal three over two 𝐴𝐡 because, therefore, vector π΄π‘Œ is gonna be equal to three over two vector 𝐴𝐡. And we’ve actually just found vectors 𝐴𝐡. So therefore, we can say that vector π΄π‘Œ is gonna be equal to three over two 𝑏 minus π‘Ž. 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 𝑀 because what we can say about the vector 𝑂𝑀 is this is gonna be equal to π‘˜, which is just a constant, multiplied by the vector 𝑂𝐡. And we know that because we’re told at the beginning that 𝑂𝑀𝐡 is actually a straight line. So therefore, as we know that vector 𝑂𝐡 is equal to 𝑏, we can actually say that vector 𝑂𝑀 is gonna be equal to π‘˜π‘. And it’s this π‘˜ that we want to find because if we find this π‘˜, it’s actually gonna help us to prove that 𝑀 is the midpoint of 𝑂𝐡.

Well, now, let’s actually consider vector π‘‚π‘Œ. Well, vector π‘‚π‘Œ is gonna be equal to vector 𝑂𝐴 plus vector π΄π‘Œ. Well, these are actually both vectors that we know. So therefore, we can say that vector π‘‚π‘Œ is gonna be equal to π‘Ž plus three over two 𝑏 minus π‘Ž.

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

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

Well, next, what we can actually do is move on and start to think about 𝑋𝑀 because what that 𝑋𝑀 can do is actually help us to think about the π‘˜ that we’re looking for earlier. So if we actually travel from 𝑋 to 𝑀, so we take the vector 𝑋𝑀, then we’re actually gonna go from 𝑋 to 𝑂 and then 𝑂 to 𝑀. So therefore, it’s gonna be equal to vector 𝑂𝑀 minus vector 𝑂𝑋 cause like we said we went 𝑋 to 𝑂, which is actually the negative of 𝑂 to 𝑋. So therefore, this is gonna give us that the vector 𝑋𝑀 is equal to π‘˜π‘ minus a quarter π‘Ž. And that’s cause our vector 𝑂𝑀 was π‘˜π‘ and vector 𝑂𝑋 is actually a quarter π‘Ž.

So therefore, what we can actually say is that because π‘‹π‘€π‘Œ is a straight line, then π‘‹π‘Œ is gonna be a multiple of 𝑋𝑀. 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 𝑀 is the midpoint of 𝑂𝐡. So therefore, as we already said, the vector π‘‹π‘Œ is gonna be equal to 𝑀, the vector 𝑋𝑀, where 𝑀 is actually a multiple. So we’re saying it’s gonna be a multiple of the vector 𝑋𝑀.

So now, we need to consider what possibly that could be. Well, if we take a look at the coefficients of our π‘Žs, then, we’ve 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 π‘‹π‘Œ is three times the vector 𝑋𝑀. So we can actually say that vector π‘‹π‘Œ equals three vector 𝑋𝑀.

But as we’re trying to find π‘˜, there’s actually a more useful way that we can think of this. And that’s that a third of vector π‘‹π‘Œ is equal to the vector 𝑋𝑀. And we’ve already proved this with the π‘Ž coefficient because actually a quarter or negative quarter is actually a third of negative three-quarters. So therefore, if it’s actually a third with the coefficients of π‘Ž, it must be the same with the coefficients of 𝑏.

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 that’s gonna help us? Well, using this bit of information, so π‘˜ being equal to a half, we can say that therefore vector 𝑂𝑀 is equal to a half vector 𝑂𝐡. And that’s because we found out earlier that vector 𝑂𝑀 was equal to π‘˜ vector 𝑂𝐡.

So therefore, what we can actually say is we’ve actually proved that 𝑀 is the midpoint of 𝑂𝐡.

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