Right triangle ABC is to be drawn

[BTGmoderatorDC]

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. 2,160
D. 2,916
E. 148,824

I'm confused how to set up the formulas here. Can any experts help?

OA C

[elias.latour.apex]

Well, first we must determine how many possible starting points there are for the A. Since there are 9 possibilities for x, and 6 possibilities for y, there are 54 starting points for the triangle.

But don't think that (A) is the answer. That number is far too low. Once we have selected A, we can select choice C. It will have the same x, but a different y. Since there are 5 remaining possibilities for y (we used one for A) we can multiply 54 * 5 to get 270.

Then we will have to choose a location for B. But why bother? We can see that whatever number we pick, the answer will end in a 0, because any integer multiplied by 270 will end in zero. Don't waste time on math. Only answer choice (C) has the right characteristics.

Don't do math. Just pick the right answer and move on.

[Brent@GMATPrepNow]

Here's a similar question to practice with: https://www.beatthegmat.com/a-right-tria ... 97187.html

Cheers,
Brent