Absolute 700+ level question. Please use concepts of Coordinate Geometer and Permutation & Combination to solve the problem.

Give that A, B, and C are non-zero integers and satisfy the inequalities -4 ≤ x ≤ 5 and -2≤ y ≤ 5. Hence x can have 9 possible values and y can have 7 possible values (zero cannot be a possible value since A, B, C are non-zero integers).

Step 1: There are 9 possible values of x coordinate and 7 possible values of y coordinate. Hence, point B (the right angle vertex) can be 9x7 = 63 possible values of coordinate B.

Step 2: Since AB is parallel to the y axis, point A has the same x coordinate but different y coordinate as point B. Since point B has already has 80 different values, we can have 7-1 = 6 possible values of coordinate A, each corresponding to those 80 possible values of coordinate B.

Step 3: Since ABC is a right angle triangle, point c has the same y coordinate but different x coordinate as point B. Since point B has already has 80 different values, we can have 9-1 = 8 possible values of coordinate C, each corresponding to those 80 possible values of coordinate B.

Therefore, the total possible number of triangle is the same as the the total possible values of A, B, & C, i.e. 63x6x8 = 3024 ways

Hence, the Correct Answer is Option A. 3,024
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ATS:
Coordinates of A,B and C are non-zero integers
AB is parallel to Y-axis
BC is parallel to X-axis

As vertex B is common to both sides (AB and BC), lets start with B:
Options along Y-axis= (-2 to +5 excluding 0) 7
Options along X-axis= (-4 to +5 excluding 0) 9
Total= 63

Vertex A:
Out of 7 options along Y-axis, one is taken by B, so 6 left for A.

Vertex C:
Out of 9 options along X-axis, one is taken by B, so 8 left for A.

Total possible triangles within the coordinate grid: 63x6x8= 3024, Ans A

Shouldn't this be 63 possible values of B instead of 80?