Right triangle RST can be constructed in the xy-plane such

Question

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

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

Bunuel · Accepted Answer

−3 ≤ x ≤ 4 and −7 ≤ y ≤ 3 gives a rectangle with 8*11 dimensions (8 horizontal and 11 vertical dots). We are given that RS is parallel to x-axis, RT is parallel to y-axis and the right angle is at R. Choose the (x,y) coordinates for vertex R: 7C1*10C1 (we are told that coordinates of R, S, and T must be nonzero integers, so we are choosing from 7 and 10 instead of 8 and 11 because we should exclude 0); Choose the x coordinate for vertex S (as y coordinate is fixed by R): 6C1, (7-1=6 as 1 horizontal dot is already occupied by R); Choose the y coordinate for vertex T (as x coordinate is fixed by R): 9C1, (10-1=9 as 1 vertical dot is already occupied by B). 7C1*10C1*6C1*9C1=3,780. Answer: A. Similar questions to practice: triangle-abc-will-be-constructed-in-a-xy-plane-according-to-132573.html right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-71597.html right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-88380.html right-triangle-abc-is-to-be-drawn-in-the-xy-plane-so-that-88958.html a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-94644.html a-right-triangle-abc-has-to-be-constructed-in-the-xy-plane-100675.html Also, please read and follow: 11-rules-for-posting-133935.html (rule #3)

mikemcgarry · Answer

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 https://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 :-)

GSBae · Answer

Tricky Question!

Imagine we have a grid, ranging from -3 to 7 on the x-axis, and from -7 to 3 on the y-axis. Every point on this grid could potentially be point R, the right angle of the triangle (EXCEPT ANY POINT ON THE AXES - the question says *non-zero* integers! They're not talking about the length of the side, they mean the coordinates of the points. This is very important!). Thus, we're left with 7*10 = 70 choices for our point R.

Now we know that the sides of the triangle are parallel with the axes, so the other points must lie on the same horizontal or vertical as our point R. Therefore, we have 7-1 = 6 choices for the one on the x axis, and 10-1= 9 choices for the one on the y-axis (we subtract one from each because we can't have the same point).

Therefore, in total, we have 7*10*6*9 = 54*70 = 3500 + 280 = 3780 possible choices.

Answer: A

Notice if they had NOT stated that the coordinates had to be non zero, we would have had 8*11 choices for R, and 7 and 10 choices for the other points, respectively. The number of possibilities here is 8*11*7*10 = 88*70 = 5600+560 = 6160, answer D. Very tricky!

rishi02 · Answer

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

felippemed · Answer

How many triangles you see here?

EMPOWERgmatRichC · Answer

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:  A

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

Fdambro294 · Answer

To create any right triangle in the coordinate plane, the unique coordinates required will be:

2 unique X coordinates

And

2 unique Y coordinates

Allowable X coordinates: -3 to 4, inclusive and excluding 0

7 X integer coordinates to choose from and we need to select 2.

Then once the 2 are selected whether we assign the one X coordinate to points R and T and the other X coordinate to point S matters - in other words, order matters.

(7 c 2) * 2! = 42

Allowable Y coordinates: -7 to +3, inclusive and excluding 0

That means we have 10 allowable Y integer coordinates to choose from and we need 2 unique ones to assign to our right triangle.

Again, it matters which points get the integer - in other words order matters

One Y coordinate goes to Point R and S (because side RS is perpendicular to the Y axis)

And the other Y coordinate goes to Point T

(10 c 2) * 2! = 90

Answer:

(42) * (90) = 3,780

Posted from my mobile device

rxb266 · Answer

official explanation -

The question asks for how many different triangles can be constructed, and the choices clearly indicate that there are too many to count by hand, so recognize this as a permutation problem. To do the permutation, multiply the number of possible points for each vertex of the triangle.

Start with point R. Since the coordinates of each point are to be nonzero integers that satisfy the inequalities −3 ≤ x ≤ 4 and −7 ≤ y ≤ 3, point R has 7 potential x-coordinates and 10 potential y-coordinates. Therefore point R could be 7 × 10 = 70 different points.

Next, evaluate point S. Since RS is perpendicular to the y-axis, the y-coordinate of S is the same as that of R. Therefore, only the x-coordinate needs to be calculated. The x-value of S cannot be the same as that of R. So, for each point R, point S could be 6 different points.

Next, evaluate point T. Since RT is perpendicular to RS, the x-coordinate of T is the same as that of R. Therefore, only the y-coordinate coordinate needs to be calculated. The y-value of T cannot be the same as that of R. So, for each point R, point T could be 9 different points.

Finally, do the permutation of all of the vertices by multiplying R × S × T = 70 × 6 × 9 = 3780. The correct answer is choice A.

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Right triangle RST can be constructed in the xy-plane such

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