Right triangle ABC is to be drawn in the xy-plane so that

Question

Right triangle ABC is to be drawn in the xy-plane so that the right angle is at A and AB is parallel to the y-axis. If the x- and y-coordinates of A, B, and C are to be integers that are consistent with the inequalities -6 ≤ x ≤ 2 and 4 ≤ y ≤ 9 , then how many different triangles can be drawn that will meet these conditions?

A.54

B.432

C.2160

D.2916

E.148,824

i ve got it right.but this problem is very time consuming.can anyone suggest shorter method

A. 54
B. 432
C. 2,160
D. 2,916
E. 148,824

Bunuel · Accepted Answer

We have the rectangle with dimensions 9*6 (9 horizontal dots and 6 vertical). AB is parallel to y-axis and AC is parallel to x-axis.

Choose the (x,y) coordinates for vertex A: 9C1*6C1;
Choose the x coordinate for vertex C (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 B (as x coordinate is fixed by A): 5C1, (6-1=5 as 1 vertical dot is already occupied by A).

9C1*6C*8C1*5C1=2160.

Answer: C.

caioguima · Answer

Slightly different way of thinking:

On the 9x6 grid of possibilities, I can imagine a bunch of rectangles (with sides parallel to x and y axes). Each of these rectangles contains 4 triangles that fit the description of the question stem.

therefore:

Answer = ( # of Rectangles I can make on the grid) x 4

To create the rectangle, I need to pick 2 points on the x direction, and 2 points on the y direction. Therefore:

Answer = C(9,2) * C(6,2) * 4 = 36 * 15 * 4 = 2160 (OPTION C)

hrish88 · Answer

OA is C.very nice explanation.you rock man as always.

BlitzHN · Answer

so what about the triangles that look like the mirror images of the ones above? - think, switching the co-ords of A and C along x axis and switching A and B along y axis....

Bunuel · Answer

Above solution counts all position: AC and CA; A B and B A; For example point C with 8C1 can be placed to the right as well to the left of A and point B with 5C1 can be placed below as well as above of A. So all cases are covered. More here: arithmetic-og-question-88380.html?hilit=dimensions Hope it helps.

arundas · Answer

C.

I am not sure if this approach is correct. I used Elimination. There can be only 5 possible values of C if we fix A. So the number of triangles possible has to be multiple of 5. The only answer that satisfies the criterion is C.

mbaiseasy · Answer

First, get the integer points available for x-axis: 2 - (-6) + 1 = 9
Second, get the interger points available for y-axis: 9-4+1 = 6

How many ways to select the location of line AB in the x-axis? 9
How many ways to select the location of point C in the x-axis? 8 (Note: we cannot select the location of line AB)
How many ways to select the location of the base? 2 (Is it BC or AB?)
How many ways to position line AB parallel to y axis? 6!/2!4! = 15

Multiple all that: 9*8*2*15 = 2,160

Answer: C

venkat18290 · Answer

Hi Bunuel ,

That was a fantastic solution , but i have a small doubt . How do we ensure that by selecting points in this way the properties of a triangle are satisfied always . Could there be some points through which we cannot even form a triangle leave alone right angled triangle. I hope i am clear in my question .

Bunuel · Answer

ANY 3 non-collinear points on a plane form a triangle.

cjliu49 · Answer

Bunuel... you're a freaking genius. Get a job with NASA already.

GMATinsight · Answer

Please check solution as attached.

Answer:Option C

ganeshvenugopal · Answer

Hi @bunnel ,
I have a doubt.
If AB is parallel to Y-axis, how can we count X=0 as a possibility for vertex A?
If X=0, when vertex A lies of the Y-axis and therefore, AB can't be parallel to Y-axis..
With this in mind, I got 8*6 possibilities for vertex A.

please help.

GMATinsight · Answer

Here comes some cotradiction.

One definition says that parallel lines are lines that never intersect but they lack mentioning the fact that the lines should lie in one plane for it to happen.

Here the line is parallel to Y axis and Y-Axis is a direction while origin is just a point of reference from where the Y direction may be referenced so I believe that X=0 may be taken for the line which is parallel to Y-Axis.

Bambi2021 · Answer

If I pick a point for the right angle (A,A), I can pick this point in 9*6 ways. The other two points that forms the triangle must have one point at (A,y) and one point at (x,A). The total number of ways to pick the points becomes 9*6*8*5 = 2160 triangles.

Thelegend2631 · Answer

Once you know it forms a 6*9 Rectangle - IT becomes easy.

Since AB - Parallel to Y Axis -

Select two points ( AB ) from 6 vertical points : 6C2 
- Multiply this by 2 Since there can be right angle above or below 
- Multiply this by 9 Since AB Can be anywhere on 9 vertical bars
- Multiple this with 8 since we have 8 choices for Point C

6c2*2*8*9 = 2160
.

yashsharma1 · Answer

This is how I went about solving this.
In the triangle, there will be 3 points A, B & C. Each point will have (x,y) coordinates. Now let's fix point B. There are total 54 (x,y) combinations possible for point B (9 values of x * 6 values of y as given in question)

Let B be situated at (-6,4). Now A is vertically in same line as B (because AB is parallel to y-axis) and hence its x coordinate is fixed as -6. Now for y coordinate, A can have any value except 4 (because otherwise point B will lie on point A and no triangle will be formed) from the possible values allowed as per question. That leaves 5 possible y coordinates for point A.

Similarly, point C is horizontal to A and hence y coordinate for C is fixed at 4. For x coordinate, B can have any value other than -6 and that allows for 8 values of x coordinate for point C.

So, for each combination of (x,y) for point B, there are 40 (5 possibilities of y coordinate for point A and 8 possibilities of x coordinate for point C) different combinations of positioning point B & C. And because point B itself has 54 combinations, final answer is 54*40 = 2160

This is what Bunuel has explained crisply with combination formulas

Right triangle ABC is to be drawn in the xy-plane so that

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