Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R. The x and y-coordinates of R, S, and T are to be nonzero integers that satisfy the inequalities −3 ≤ x ≤ 4 and −7 ≤ y ≤ 3. Given these restrictions, how many different triangles can be constructed?

A. 3780
B. 4200
C. 4900
D. 6160
E. 7744

Dear kanusha,

I'm happy to help.

This question requires a little visualization and some facility with the x-y plane. The question tells us "Right triangle RST can be constructed in the xy-plane such that RS is perpendicular to the y-axis and the right angle is at R." This tells us
(a) there is a right angle at R, so RS is perpendicular to RT
(b) RS is parallel to the x-axis, perpendicular to the y-axis --- i.e. horizontal
(c) RT must be parallel to the y-axis, perpendicular to the x-axis --- i.e. vertical

Then, we have to recognize --- all points on the same horizontal line have the same height, so they all have the same y-coordinate. Therefore, we absolutely know that R & S, endpoints of the horizontal line, must have the same x-coordinate.

Similarly ---- all points on any vertical line share the same distance from the y-axis, the same length to the right or the left of the y-axis, so all points on the same vertical line must have the same x-coordinate. Thus, R & T, the endpoints of a vertical line, have the same y-coordinate.

Next, inclusive counting. See
http://magoosh.com/gmat/2012/inclusive- ... -the-gmat/
From x = -4 to x = +5, there are (+5) - (-4) + 1 = 10 points, and, similarly, 11 possible y-points.

Does this clear up your question? On precisely which part do you have questions?
Mike
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No of ways of choosing R(x,y) = 10*7
No of ways of choosing S =1*6
no of ways of choosing T= 9*1

Therefore total number of ways = 10* 7 *9 *6
=3780

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Hi All,

This question is about following graphing rules and keeping track of individual possibilities. Here's how you can solve it:

We're told a number of facts:
1) The X and Y co-ordinates must be NON-ZERO INTEGERS.
2) RS is perpendicular to the Y-axis; this means that points R and S will have the SAME Y-coordinate.
3) Angle R is a right angle; this means that points R and T will have the SAME X-coordinate.
4) The two co-ordinates are in the following ranges: -3 ≤ x ≤ 4 and -7 ≤ y ≤ 3

Let's start with point R. It can have any of 7 different X-coordinates (-3,-2,-1, 1, 2, 3, 4) and any of the 10 different Y-coordinates (-7,-6,-5,-4,-3,-2,-1,1,2,3).

That is 7x10 different possibilities = 70 different spots for point R

Once we FIX R in one spot, S MUST have the SAME Y-coordinate, BUT a DIFFERENT X-coordinate. We already used one of the X-coordinates when we placed point R, which means….

There are 7-1 = 6 possible spots for point S

With R still in its spot, we know that point T must have the SAME X-coordinate, but a DIFFERENT Y-coordinate. We already used one of the Y-coordinates when we place point R, which means….

There are 10-1 = 9 possible spots for Point T

We multiply the options: (70)(6)(9) = 3780 different triangles.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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