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Question 33017: A(-1;4):B(2;3): C(0;-3): D(-3;-2)
these points on a graph make a rectangle...specify...
what is the equation for the diagonales d1 and d2 of the rectangle?
what is the equation of the medianes m1 and m2 of the rectangle?
what is the area of this rectangle?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website!
A(-1;4):B(2;3): C(0;-3): D(-3;-2)
Vector AB = B-A = (3, -1) = 3i - j
DC = C - D = (-3, -1) = 3 i - j,
BC = C - B = (-2, -6) = -2 i - 6j
DA = D -A = (-2, -6) = -2 i - 6j
We see that AB = CD & BC = DA so AB // CD and BC // DA
Also, the dot product AB. DA = (3i-j).(2i+6j) = 6-6 =0
Hence, ABCD is a rectangle
Use the basic two points formula (don't ask me!) to get the equation for
diagonals AC and BD. (d1 & d2)
Also the four midpoitns are (1/2, 1/2), (1,0), (-1/2, -1/2) and (-1,1)
so you can get the eqatioins for m1 & m2.
The area of ABCD is equal to the magnitude of the direct product
AB x AD = (3i-j) x (2i - 6j) = 2 k - 18 k = -16 k.
So area = 16.
Kenny
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