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    This example is given in the Pure Mathematics book of Year 1/AS Edexcel.

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    OABC is a parallelogram. P is the point where the diagonals OB and AC intersect. The vectors a and c are equal to OA and OC respectively.

    Prove that the diagonals biscect each other.

    Working:
    OB = OC + CB = c + a
    and AC = AO + C
    = -OA +OC = -a+c
    P lies on OB --> OP = λ(c+a)
    P lies on AC --> OP = OA + AP
    = a + μ(-a+c)

    ---> λ(c+a) = a + μ(-a+c)
    ---> λ = 1-μ and λ=μ
    ---> λ=μ=1/2, so P is the midpoint of both diagonals, so the diagonals bisect each other.

    I understand everything up the point where:
    ---> λ = 1-μ and λ=μ
    takes place, they refer to this as "forming and solving a pair of simulataneous equations by equating the coefficients of a and c" in the book.

    Can someone please explain to me how they went from:
    ---> λ(c+a) = a + μ(-a+c)
    to
    λ = 1-μ and λ=μ

    Thank you!
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    If you try expanding the brackets in both, it will become:

    λc+λa = a-μa+μc

    Then you can rearrange:

    λc+λa = a(1-μ) + μc

    Then equating coefficients of a and c:

    λ = (1-μ) and λ = μ

    Thus:

    1-μ = μ

    And μ = 1/2 and λ = 1/2
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    (Original post by Truleyhero)
    If you try expanding the brackets in both, it will become:

    λc+λa = a-μa+μc

    Then you can rearrange:

    λc+λa = a(1-μ) + μc

    Then equating coefficients of a and c:

    λ = (1-μ) and λ = μ

    Thus:

    1-μ = μ

    And μ = 1/2 and λ = 1/2
    OHHHHHHHHHHH! I understand it now.

    Thanks for the help!
 
 
 
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