coordinate geometry

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coordinate geometry

by datonman » Thu Oct 22, 2015 11:19 am
Right triangle PQR is to be constructed in the xy-plane so that the right angle is a P and PR (line on top of both letters)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 equal or less than x which is less than or equal to 5" and "6 less than or equal to y which is less than or equal to 16". How many different triangles with these properties could be constructed?

A.110
B.1,100
C.9,900
D.10,000
E.12,100

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by Brent@GMATPrepNow » Thu Oct 22, 2015 11:22 am
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) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

C
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 cfisher1 » Mon Sep 12, 2016 7:43 pm
Can it also be solved using the method found in this post? https://www.beatthegmat.com/wonderful-p- ... 71001.html

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by Jeff@TargetTestPrep » Tue Sep 13, 2016 10:13 am
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) 1,100
(C) 9,900
(D) 10,000
(E) 12,100
We can start by drawing the triangle in the xy-plane. This will help us visualize how we are to solve the problem.

Image

As we can see, the right triangle has a right angle at point P and side PR is parallel to the x-axis; side PQ must be parallel to the y-axis.

To solve this question we need to determine how many ways we can construct points P, Q, and R, and then we will multiply those possibilities together. We are given that: -4 ≤ x ≤ 5 and 6 ≤ y ≤ 6. This means that there are 10 possible integer values for the x-coordinate and 11 possible integer values for the y-coordinates.

Let's start by determining how many ways we can construct point P.

Since P is the first point we are trying to determine, we have all the options for the available x- and y-coordinates. Since there are 10 possible x-coordinates and 11 possible y-coordinates we have (10)(11) = 110 possible options. Next, we determine how many options we have for point R.

In determining point R, we must recognize that side PR of triangle PQR must remain parallel to the x-axis. This means that the y-coordinate of point R, must match the y-coordinate of point P. Thus there is only 1 option for the y-coordinate of point R. However there are 9 total options for the x-coordinate of point R. Since point P has already exhausted 1 option out of the 10 total options, there are only 9 x-coordinates left for designating point R.

Finally, we determine how many ways in which we can construct point Q.

In a similar fashion to the method used for determining the number of ways to construct point R, we must remember that side PQ of triangle PQR must remain parallel to the y-axis. Thus, the x-coordinate of point Q must match the x-coordinate of point P. Therefore, there is only 1 option for the x-coordinate of point Q. However, there are 10 total options for the y-coordinate of point Q. Since point P has already exhausted 1 option out of the 11 total options, there are only 10 y-coordinates left for designating point Q.

In summary, we know the following:

There are 110 ways to select point P, 9 ways to select point R, and 10 ways to select point Q. Because we need to determine the total number of ways to create triangle PQR, we must use the fundamental counting principle. Thus, we multiply these three options together:
110 x 9 x 10 = 9,900

The answer is C.

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Head of GMAT Instruction
[email protected]

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by Matt@VeritasPrep » Thu Sep 15, 2016 7:46 pm
A quick approach:

We need to choose two (distinct) x coordinates for the two points on the base, and we need to choose two (distinct) y coordinates for the two points on the height.

We've got 10 choices (-4 through +5) for the x-coordinates, and 11 choices (6 through 16) for the y-coordinates. So when we choose x, we have (10 * 9) options (10 for the first, 9 remaining for the second), and when we choose y, we've got (11 * 10) options.

Multiplying these together, we get (10 * 9) * (11 * 10), or 9900.