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Math Help - sum of coordinates

  1. #1
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    sum of coordinates

    Points A = ( 3, 9) B = (1 , 1) c = ( 5, 3 ) and D = (a , b) lie in the first quadrant and are the vertices of quadrilateral ABCD. The quadrilateral formed by joining the midpoints of ab, bc, cd, and da is a square. What is the sum of the coordinates of point D?

    I solved the problem but someone said that I could've solved it easier and faster by using vectors. I looked up vectors on google and I dont understand it that well. Could someone take a look at my work and show me where i could apply vectors.

    Thanks

    Vicky.
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  2. #2
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    Hi

    Your work is correct
    Let I be the midpoint of [AB], J the midpoint of [BC], K the midpoint of [CD] and L the midpoint of [DA]
    From the data we have I(2,5) ; J(3,2) ; K((a+5)/2,(b+3)/2) ; L((a+3)/2,(b+9)/2)
    Using vectors
    \vec{IJ} (1,-3)

    \vec{JK} \left(\frac{a-1}{2},\frac{b-1}{2} \right)

    [IJ] and [JK] being two sides of a square, they are perpendicular
    The dot product of \vec{IJ} and \vec{JK} is equal to 0

    \frac{a-1}{2}-3\:\frac{b-1}{2} = 0

    \frac{a-1}{2}=3\:\frac{b-1}{2} [1]

    The lengths of [IJ] and [JK] are equal

    10 = \left(\frac{a-1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2

    Using [1]
    10 = 10\:\frac{(b-1)^2}{4}

    (b-1)^2 = 4

    b-1=2 => b=3 => a=7 => D(7,3)
    b-1=-2 => b=-1 => a=-5 => D'(-5,-1)

    There are 2 solutions depending if you consider ABCD' as a quadrilateral (probably not)
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    Your work is correct
    Let I be the midpoint of [AB], J the midpoint of [BC], K the midpoint of [CD] and L the midpoint of [DA]
    From the data we have I(2,5) ; J(3,2) ; K((a+5)/2,(b+3)/2) ; L((a+3)/2,(b+9)/2)
    Using vectors
    \vec{IJ} (1,-3)

    \vec{JK} \left(\frac{a-1}{2},\frac{b-1}{2} \right)

    [IJ] and [JK] being two sides of a square, they are perpendicular
    The dot product of \vec{IJ} and \vec{JK} is equal to 0

    \frac{a-1}{2}-3\:\frac{b-1}{2} = 0

    \frac{a-1}{2}=3\:\frac{b-1}{2} [1]

    The lengths of [IJ] and [JK] are equal

    10 = \left(\frac{a-1}{2}\right)^2+\left(\frac{b-1}{2}\right)^2

    Using [1]
    10 = 10\:\frac{(b-1)^2}{4}

    (b-1)^2 = 4

    b-1=2 => b=3 => a=7 => D(7,3)
    b-1=-2 => b=-1 => a=-5 => D'(-5,-1)

    There are 2 solutions depending if you consider ABCD' as a quadrilateral (probably not)
    Thanks for your help.
    But I think I need to study vectors some more.
    Vicky.
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