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    The points A(1,0), B(7,2) and C(13,7) are the vertices of a triangle.
    The midpoints of the sides BC, CA and AB are L, M and N.
    Find the vector equation of the lines AL

    I got an answer of (1,0) + λ(3,2.5) by doing:
    AL = OA + AL, however the book has an answer of (1,0) + λ(2,1), and i'm not sure how they got that answer.
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    (Original post by znx)
    The points A(1,0), B(7,2) and C(13,7) are the vertices of a triangle.
    The midpoints of the sides BC, CA and AB are L, M and N.
    Find the vector equation of the lines AL

    I got an answer of (1,0) + λ(3,2.5) by doing:
    AL = OA + AL, however the book has an answer of (1,0) + λ(2,1), and i'm not sure how they got that answer.
    L is the midpoint of BC. So its coordinate is (10,4.5). The vector AL is thus OL-OA = (9,4.5) = 4.5(2,1) which has direction (2,1).
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    (Original post by RDKGames)
    L is the midpoint of BC. So its coordinate is (10,4.5). The vector AL is thus OL-OA = (9,4.5) = 4.5(2,1) which has direction (2,1).
    Thank you mate for the help
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    (Original post by RDKGames)
    L is the midpoint of BC. So its coordinate is (10,4.5). The vector AL is thus OL-OA = (9,4.5) = 4.5(2,1) which has direction (2,1).
    For the vector equation for BM:
    I did the same thing as for AL, which is BM = OM - OB, and i got an answer of (0,1.5) However the book says it should be (0,1)?

    And for CN:
    I got it as (-3,-2), but the book got it as (3,2) - does the negative make a difference?
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    (Original post by znx)
    For the vector equation for BM:
    I did the same thing as for AL, which is BM = OM - OB, and i got an answer of (0,1.5) However the book says it should be (0,1)?

    And for CN:
    I got it as (-3,-2), but the book got it as (3,2) - does the negative make a difference?
    You can take out a factor of 1.5 and have the direction as (0,1)

    The -ve doesnt make a difference as that's just going the other way. You can factor it out too.


    In case you're wondering why we can do these, it's because if you end up with some vector equation \mathbf{r} = \mathbf{a} + t (5\mathbf{b}) then you can just move the common factor of 5 from the vector  \mathbf{b} to combine it with the t to give 5t. But time is generally an unrestricted parameter so 5t will take all the values anyway just as t would on its own, so you may as well replace 5t back with t as it makes no difference.
 
 
 
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