PS: Coordinate Geometry

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PS: Coordinate Geometry

by Arsenal » Tue Sep 27, 2016 1:03 pm
Right triangle PQR is to be constructed in XY plane so that right angle is at P and PR is parallel to x-axis. X and Y cordinates of P,Q & R are to be int that satisfy the inequalities x>=-4 and x<=5 & y>=6 and y<=16. how many different triangles with these properties could be constructed?

A.110
B.1100
C. 9900
D. 10000
E. 12100

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by GMATGuruNY » Tue Sep 27, 2016 1:12 pm
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by Brent@GMATPrepNow » Tue Sep 27, 2016 2:26 pm
Arsenal wrote:Right triangle PQR is to be constructed in XY plane so that right angle is at P and PR is parallel to x-axis. X and Y cordinates of P,Q & R are to be int that satisfy the inequalities x>=-4 and x<=5 & y>=6 and y<=16. how many different triangles with these properties could be constructed?

A.110
B.1100
C. 9900
D. 10000
E. 12100
Take the task of building triangles and break it into stages.

Stage 1: Select any point where the right angle will be (point P).
The point can be selected from a 10x11 grid. So, there 110 points to choose from.
This means that stage 1 can be completed in 110 ways.

Stage 2: Select a point that is on the same horizontal line as the first point. This point will be point R.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the y-coordinate of point R.
In how many ways can we select the x-coordinate of point R?
Well, we can choose any of the 10 coordinates from -4 to 5 inclusive EXCEPT for the x-coordinate we chose for point P (in stage 1).
So, there are 9 coordinates to choose from.
This means that stage 2 can be completed in 9 ways.

Stage 3: Select a point that is on the same vertical line as the first point. This point will be point Q.
The 2 legs of the right triangle are parallel to the x- and y-axes.
The first point we select (in stage 1) dictates the x-coordinate of point Q.
In how many ways can we select the y-coordinate of point Q?
Well, we can choose any of the 11 coordinates from 6 to 16 inclusive EXCEPT for the y-coordinate we chose for point P (in stage 1).
So, there are 10 coordinates to choose from.
This means that stage 3 can be completed in 10 ways.

So, by the Fundamental Counting Principle (FCP), the total number of triangles = (110)(9)(10) = [spoiler]9900 = C[/spoiler]

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by ewekkin12 » Thu Dec 08, 2016 12:55 pm
Thanks for the detailed description for arriving at the solution. One point of clarification: should the question frame the x and y ranges as greater than or equal to? The way I'm interpreting the question is that the grid would be 8 x 9. If not, I'm lost as to how we get the 9 x 11 grid. Thanks!

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by Brent@GMATPrepNow » Thu Dec 08, 2016 1:12 pm
ewekkin12 wrote:Thanks for the detailed description for arriving at the solution. One point of clarification: should the question frame the x and y ranges as greater than or equal to? The way I'm interpreting the question is that the grid would be 8 x 9. If not, I'm lost as to how we get the 9 x 11 grid. Thanks!
Good question.
Unfortunately, the original poster used some unorthodox notation (x>=-4 and x<=5 & y>=6 and y<=16)
This was meant to represent the following conditions:
x ≥ -4 and x ≤ 5 and y ≥ 6 and y ≤ 16

Put even better, the question should read...
Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties can be constructed?
(A) 110
(B) 1100
(C) 9900
(D) 10000
(E) 12100
Does that help?
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by ewekkin12 » Thu Dec 08, 2016 1:58 pm
Perfect. Thanks for the quick response!