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    Can anyone give me a little clue on how to do question b)???
    I worked out vector AQX to be -4a+12b but then how do i prove that to be a straight line??

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    Ah, I was doing this just before Christmas.
    With vectors, for 2 lines to be on the same line, they must share 2 things: A common point and the same gradient. Hope this info helps, gimme a few minutes and I'll work it out myself and can therefore help you further.
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    (Original post by RamonDavis)
    Ah, I was doing this just before Christmas.
    With vectors, for 2 lines to be on the same line, they must share 2 things: A common point and the same gradient. Hope this info helps, gimme a few minutes and I'll work it out myself and can therefore help you further.
    ok
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    I've worked it out
    Firstly, to Prove it is a straight line you basically need to prove that the middle point, Q, is on the line.
    So here is a question for you, what 2 lines could you use to show Q is going in the same direction as X?
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    (Original post by RamonDavis)
    I've worked it out
    Firstly, to Prove it is a straight line you basically need to prove that the middle point, Q, is on the line.
    So here is a question for you, what 2 lines could you use to show Q is going in the same direction as X?
    I don't understand....
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    Hmm ok.
    You should have A to Q as 3b-a from the last question.
    So this means that if A to X has the same gradient, it must be on the same line as they share the common point of A.
    So you need to work out A to X
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    Yeah Ramon is right. Find the vector for AQ, and then for AX. You'll find that AX is a multiple of AQ. This means that they have the same direction (or gradient), but different magnitudes (lengths). In other words, AX is parallel to AQ. However because they both go through the same point, A, the must be the same line.
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    (Original post by NothingButWaleed)
    I don't understand....
    To prove that three points A, B and C are in a straight line using vectors, you need to show that any two of the vectors that make up the line are parallel.

    E.g. you could show that \vec{AB} is parallel to \vec{AC} or you could show that \vec{AB} is parallel to \vec{BC} etc.
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    (Original post by Darth_Narwhale)
    Yeah Ramon is right. Find the vector for AQ, and then for AX. You'll find that AX is a multiple of AQ. This means that they have the same direction (or gradient), but different magnitudes (lengths). In other words, AX is parallel to AQ. However because they both go through the same point, A, the must be the same line.
    But in the question it doesn't say anywhere that AQ is a straight line and we can't assume!!??
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    (Original post by NothingButWaleed)
    But the in doesn't say anywhere that AQ is a straight line and we can't assume!!??
    The vector AQ is defined as the straight line between A and Q
 
 
 
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