Expert: Sherman D. - 12/17/2004

Question
Quadrilateral JAME has vertices J(2,-2),A(8,-1),M(9,3), and E(3,2). Prove that quadrilateral Jame is a parallelogram and find its area.

Answer
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
p^2 + q^2 = 2(a^2 + b^2)


J(2,-2), A(8,-1), M(9,3), E(3,2)

J(2,-2) and A(8,-1)
JA = sqrt((8 - (2))^2 + ((-1) - (-2))^2)))
JA = sqrt((8 - 2)^2 + (1)^2)
JA = sqrt(6^2 + 1)
JA = sqrt(36 + 1)
JA = sqrt(37)

A(8,-1) and M(9,3)
AM = sqrt((9 - 8)^2 + (3 - (-1))^2)
AM = sqrt(1^2 + (3 + 1)^2)
AM = sqrt(1 + (4)^2)
AM = sqrt(1 + 16)
AM = sqrt(17)

M(9,3) and E(3,2)
ME = sqrt((3 - 9)^2 + (2 - 3)^2)
ME = sqrt((-6)^2 + (-1)^2)
ME = sqrt(36 + 1)
ME = sqrt(37)

J(2,-2) and E(3,2)
JE = sqrt((3 - 2)^2 + (2 - (-2))^2)
JE = sqrt((1)^2 + (2 + 2)^2)
JE = sqrt(1 + (4)^2)
JE = sqrt(1 + 16)
JE = sqrt(17)

So we know
JA = ME and AM = JE

Now we have to proof that its a parallelogram

J(2,-2) and M(9,3)
JM = sqrt((9 - 2)^2 + (3 - (-2))^2)
JM = sqrt((7)^2 + (3 + 2)^2)
JM = sqrt(49 + (5)^2)
JM = sqrt(49 + 25)
JM = sqrt(74)

A(8,-1) and E(3,2)
AE = sqrt((3 - 8)^2 + (2 - (-1))^2)
AE = sqrt((-5)^2 + (2 + 1)^2)
AE = sqrt(25 + (3)^2)
AE = sqrt(25 + 9)
AE = sqrt(34)

so if

(JM)^2 + (AE)^2 = 2((AM)^2 + (JA)^2)
so
(sqrt(74))^2 + (sqrt(34))^2 = 2((sqrt(17))^2 + (sqrt(37))^2)
74 + 34 = 2(17 + 37)
108 = 2(54)
108 = 108

Not certain if that definately proofs that the Quadrilateral JAME is a parallelogram or not, and if not, here is where i got the info from

http://mathforum.org/dr.math/faq/formulas/faq.quad.html#parallelogram
and
http://math.about.com/library/bldistance.htm