Question 33017

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