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parallelogram

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by goyalsau » Thu Dec 02, 2010 8:47 am
In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
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by pellucide » Thu Dec 02, 2010 8:57 am
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
My answer (C)

These three points make a triangle. Each side of the triangle can be the diagonal of the parallelogram. Since there are three sides of this triangle, there could be three distinct parallelograms.
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by Rahul@gurome » Thu Dec 02, 2010 9:09 am
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
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by rishab1988 » Thu Dec 02, 2010 9:22 am
Great question guys.

rahul it would be great if you could shed some light on whether combinatorics can be used to solve this question,because what you showed was quite intuitive and might not have been thought of others.
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by goyalsau » Thu Dec 02, 2010 9:26 am
Rahul@gurome wrote:
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
Saurabh Goyal
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by rishab1988 » Thu Dec 02, 2010 9:37 am
goyalsau wrote:
Rahul@gurome wrote:
goyalsau wrote:In the xy-plane, if points (5, 2), (2, 5), and (-2, -5) are the vertex of a parallelogram, how many such parallelograms are possible?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6

Image

See the above image.
For three non collinear points in a plane always only three parallelograms can be drawn with those points as vertex.

The correct answer is C.
Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
Goyal try drawing the xy coordinate plan and then plot those points.Join those points and form a triangle.Rahul's concept is based on keeping each side once as a diagonal and the other 2 as one of the 2 parallel sides.

But this is very intuitive method.I got to agree.

I was hoping that this question could be solved using combinatorics.
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by Rahul@gurome » Thu Dec 02, 2010 9:51 am
goyalsau wrote:Thanks Rahul for the figure but , I am not able to understand it completely Please explain with some detail so i can understand it.
In the figure, points A, B and C are any three given points in a plane which are non-collinear. Points O, P and Q are possible points each of which with A, B and C gives rise to a parallelogram. Thus the possible parallelograms are OBCA, ABCQ and ABPC.
rishab1988 wrote:rahul it would be great if you could shed some light on whether combinatorics can be used to solve this question,because what you showed was quite intuitive and might not have been thought of others.
The question is based on simple knowledge of geometry (to be precise concept of parallel lines) and can be answered applying the concept. If you insist then combinatorics can be applied as follows,
  • The three given points can be joined by three lines. Now three of them simultaneously cannot be sides of a parallelogram, but any two of them can. Number of ways to select two lines out of three = 3C2 = 3
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by rishab1988 » Thu Dec 02, 2010 9:59 am
got it.

thanks for the amazing explanation.

I know that it could be solved the first way,but I always like to know as many methods of solving as question as possible.You never know which one might be more useful...
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