+129899 A) coordinates of D = [ 5 + (3 - 1) , 1 + (4 - 3) ] = ( 7 , 2 )
B) Area = AD = sqrt [ (7 - 3)^2 + (4 - 2)^2 ] = sqrt [ 4^2 + 2^2 ] = sqrt 20 = 2sqrt 5
Next we need to find the equation of the the line containing segment AD
Slope of AD =( 2-4)/ (7-3) = -1/2
Equation of line through AD = y = -(1/2)(x - 3) + 4
Put this into standard form.....multiply through by -2
-2y = x - 3 - 8
-x - 2y = -11
x + 2y = 11
x + 2y - 11 = 0
Using point C, we can find the distance from C to AD thusly
l 5 + 2(1) - 11 l l - 4 l
_____________ = ______ = 4 / sqrt 5 = the height of the parallelogram
sqrt ( 1^2 + 2^2) sqrt 5
Area = AD * height = 2sqrt 5 * 4 /sqrt 5 = 8
