posted by Anonymous .
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?
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?
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
ok, then keeping B as it is, I found
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
Yes, but how would I represent this in a matrix?