Ok the question is:

Show that A(1,-1) , B(7,3) , C(3,5) and D(-3.1) are the vertices of a parallelogram.

help please :confused: I dont know how to solve this..

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- Mar 13th 2006, 10:09 PMGreeneyedgirlVertices of a parallelogram ???
Ok the question is:

Show that A(1,-1) , B(7,3) , C(3,5) and D(-3.1) are the vertices of a parallelogram.

help please :confused: I dont know how to solve this.. - Mar 13th 2006, 10:57 PMrgep
A parallelogram is defined by having its opposite edges equal and parallel: that is, the same length and direction. So as direction (or displacement) vectors they are equal (vectors are equal if and only if they have the same direction and magnitude). If the position vectors of the vertices are a,b,c,d then the direction vectors representing the edges are b-a = d-c and c-b = d-a. You can check that this is indeed true for your four points.

Incidentally you can see that b-a = d-c => b+c = a+d => c-b = d-a and vice versa. So you only need to check one of these three equalities. From the middle equality here you also derive the condition (b+c)/2 = (a+d)/2: that is, a quadrilateral is a parallelogram if and only if the mid-points of the two diagonals coincide. - Mar 14th 2006, 07:34 AMThePerfectHackerQuote:

Originally Posted by**rgep**

Look it up in your version of the "Elements". You should have it attached to your keychain in minature size like the Bible and carry it wherever your go :D - Mar 15th 2006, 01:35 PMrgep
Well, if I'm mistaken I'm in good company. Euclid Book I Proposition 33 (in Heath's edition, which I do have by me) states "The straight lines joining equal and parallel straight lines in the same direction are themselves equal and parallel". From then on he speaks of "parallelogrammic areas" and Proclus observes that this is the foundation of the theory of parallelograms. So Euclid takes the same property that I stated as fundamental.

In fact we now have 3 possible definitions for ABCD to be a parallelogram:

I. AB equal and parallel to DC and AD equal and parallel to BC (me)

II. AB equal and parallel to DC (Euclid)

III. AB parallel to DC and AD parallel to BC (Perfect Hacker)

In vector notation:

I. b-a = c-d and d-a = c-b

II. b-a = c-d

III. b-a = u(c-d) and (d-a) = v(c-b) for (non-zero) scalars u,v.

Clearly I => II and I => III. Euclid's proposition 33 is that II => I and we can see this immediately from the vector formulation. It's interesting to prove III => I by vectors. If -a + b - uc + ud = 0 and -a +vb -vc + d = 0 then we have two equations giving (v-1)b + (u-v)c + (1-u)d = 0. If say, u is not 1 then b = (u-v)/(1-v) c + (1-u)/(1-v) d and this expresses the fact that B lies on the line from C to D, a contradiction unless ABCD is degenerate. So u=1. Similarly v=1. Hence if the quadrilateral in non-degenerate and satisfies III then it also satisfies I.

Thus all three possible definitions are equivalent and so equally "correct". - Mar 15th 2006, 02:38 PMThePerfectHackerQuote:

Originally Posted by**rgep**

III. This also makes sense.

I. Why, are you making you definiton more than it should, because III assures that the are equal.

You can also give two more definitons:

IV: Diagnols bisect each other.

V:Opposite angles are equal.

In fact all 5 of these defintions are biconditional logical statements of each other.

I just happen to think that III is most elegant definition.