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# A right triangle ABC has to be constructed in the xy-plane

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A right triangle ABC has to be constructed in the xy-plane [#permalink]

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22 May 2010, 15:05
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A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER
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22 May 2010, 21:54
shekar123 wrote:

A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER

You are going in the right direction. For A there are 90 possibilities. However, for B we are bound by A, so we have only 8 possibilities because B has to have the same x-coordinate as A. Similarly C has 9 possibilities.
Total: 90*8*10= 7200
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23 May 2010, 05:48
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Friends,
I am trying to provide my thinking below:
For A, there are 90 possibilities on the plane.
For each location of A, as the triangle has to have same properties, there are 8 possibilities on the x axis and 9 possibilities on y axis.
So the answer is 90*8*9 = 6480

Consider giving me KUDOS if you find my post useful.
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23 May 2010, 06:06
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shekar123 wrote:

A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

my analysis

A(x,y) - 9,10- 90 possible values
B(x,y) - 8,10 =80 possible values
C(X,Y) -9,9 =81 POBBILE VALUES

i DON'T KNOW HOW TO PROCEED LATER

amitjash's solution is correct.

We have the rectangle with dimensions 9*10 (9 horizontal dots and 10 vertical). AB is parallel to x-axis and AC is parallel to y-axis.

Choose the (x,y) coordinates for vertex A: 9C1*10C1=90;
Choose the x coordinate for vertex B (as y coordinate is fixed by A): 8C1, (9-1=8 as 1 horizontal dot is already occupied by A);
Choose the y coordinate for vertex C (as x coordinate is fixed by A): 9C1, (10-1=9 as 1 vertical dot is already occupied by A).

9C1*10C1*8C1*9C1=6480.

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21 Dec 2010, 09:10
A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

100
6480
2320
1500
9000
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21 Dec 2010, 09:15
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16 Jun 2011, 00:41
A 9*10 for x and y.

B 8* 1

C 1* 9
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Re: A right triangle ABC has to be constructed in the xy-plane [#permalink]

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11 Jan 2014, 04:43
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A right triangle ABC has to be constructed in the xy-plane [#permalink]

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30 Jul 2014, 06:55
[quote="shekar123"]A right triangle ABC has to be constructed in the xy-plane so that the right angle is at A and AB is parallel to x axis. The coordinates of A, B and C are to satisfy the inequalities -3 ≤ x ≤ 5 and 2 ≤ y ≤11 and x and y are integers. The number of different triangles that can be constructed with these properties are?

A. 100
B. 6480
C. 2320
D. 1500
E. 9000

Sol:

let's find point A first:
X- axis_ point A :: point A can be any where between -3 and 5 on X- axis (total point = 5-(-3)+1 = 9)= 9c1
Y- axis_ point A :: for each x-coordinate the point A can take any value from 2 to 11 (11-2+1) in Y -coordinate= 10c1

A : 9c1*10c1

Lets Now find point B :

X- axis _ Point B:: from remaining X cordinates( A has already taken one), B can take any value so 8C1
Y - axis_point B :: No bother because it will be on same co-ordinate as A

B: 8C1

Now point C:
X: It has to be alias with A so no bother
Y: It can take any value from remaining 9 (one already taken by A) , 9C1

total: 9c1*10c1*8c1*9c1 = 6480 !!!!
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Re: A right triangle ABC has to be constructed in the xy-plane [#permalink]

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19 Sep 2015, 23:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A right triangle ABC has to be constructed in the xy-plane   [#permalink] 19 Sep 2015, 23:06
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