Several points to address here:
First, we have the presence of the arrowhead, which makes A distinct from B and makes the orientation of the points matter.This basically doubles our possibilities, which we'll do at the end of each part of our work.
Second, we have to make sure that we use the "or equal" condition to allow 10 different values each for X and Y (zero through nine).
Third, we have to DISQUALIFY all arrows that don't stay in this range and, very importantly, don't have integer values for X and Y.
The easiest to consider are those line segments that are parallel to the X-axis or Y-axis. We can go from X=0 to X=5, X=1 to X=6, and so on until X=4 to X=9 for EACH possible Y. This is 5 possible horizontal segments for each possible Y, yielding 50 (5 times 10) possible horizontal segments. As we turn 90 degrees and consider the VERTICAL segments, we will arrive at the same total of 50 (5 vertical possibilities for each X), for a total of 100 so far. OK, now consider that we can "turn around" each of these arrows by reversing A and B, yielding 200 already. There's more to come, so don't answer C! (Look like a trap?)
Now... we must ALSO consider those 5-unit segments that have a slope between zero and one. These will basically be the hypotenuses of various 3-4-5 triangles. So, we have to see how many places we can fit five units. So if we start at the origin (0,0), we can draw a 5-segment line to (3,4) or (4,3), as shown by the pythagorean theorum or distance formula. In this way we can go "over 3, up 4" for a number of points [all up to (6, 5)] and "over 4, up 3" for a number of points [all up to (5,6)]. With 7 possibilities for one variable (0-6) and and 6 possibilities for the other variable (0-5), we find 42 of each shape, for a total of 84 different segments NOT parallel to an axis. Again, when we consider switching the positions of A and B, this will give us 168 arrows, for a TOTAL of 368, or answer choice D!
-Brian K., Instructor,
bkalar@veritasprep.com
