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If the vertices of quadrilateral ABCD are A(1,1), B(-2,3), C(-4,-1), and D(2,-3), and the quadrilateral is dilated so that it's perimeter is 2 times the original perimeter, what would the vertex matrix look like?

  • math -

    Several questions are gnawing at me as I read your question.

    1. What form does your dilation take?

    you could keep one point in its present position and simply double the length of each side.
    e.g. keep B(-2,3) but move A to A1(4,-1) so that A becomes the midpoint of BA1.
    I would be relatively easy to work your way around in the same way for the other sides.
    Of course you could do this again by keeping point A constant, etc.

    Another way would be to find the intersection of the two diagonals, call it F, and then joining F to the original vertices and extending FA,FB,FC,and FD its own length.

    2. Some of this can be done with vector geometry, but do you know how to work with vectors?

    3. I am not sure what you mean by a "vertex matrix".

    4. What course level is this? Is it simply analytical geometry?

  • math -

    1. not sure
    2. Don't remember but the book says to do it with matrices.
    3. My book says a vertex matrix is a matrix into which the coordinates of a polygon's matrices are placed.
    4. Algebra 2 - Transformations with matrices

  • math -

    ok, then keeping B as it is, I found
    A1(4,-1)
    C1(-6,-5)
    D1(6,-9)

    each of my new sides are twice as long as the old ones, but the new lines are parallel to the original
    e.g. AD is parallel to A1D1

  • math -

    Yes, but how would I represent this in a matrix?

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