Question #23 (correct)
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?
(A) 72
(B) 576
(C) 4032 your answer correct
(D) 4608
(E) 6336
Explanation
We are asked to find out how many possible triangles could exist within an area on the xy-plane horizontally bounded by x = -3 and x = 4 (8 possible integer values of x, including 0) and by y = 3 and y = 11 vertically (9 possible integer values). Point L lies on one of the points within this rectangular area, which means that L can lie in any one of 72 possible points (8 x 9).
Because we know that LM is parallel to the x-axis (a horizontal line), we know that point M always has the same y-value as L. We do not know the orientation of the triangle, so point M could be either to the left or right of point L on the plane, meaning that it could lie on any point on the same horizontal line as L, within the boundaries of the problem, except the point on which L already lies. This means that there are 7 possible values of point M for every point that L lies on. We can thus multiply the 72 possible values of L with the 7 possible values of M to get all the possible placements of L and M: 504.
********** POSSIBLE ERROR
********** So shouldn't this be 8 possible values for point M? So 72 * 8 instead of 72* 7?
********** POSSIBLE ERROR
Every one of these possible triangles has a right angle at L, which means that point N has the same x-coordinate as L. However, it could be above or below L, and can thus be in any of 8 different points within the boundaries (the 9 possible y-values minus the one on which point L lies). This means that for every combination of L and M, there are 8 different triangles that can be made. We multiply 504 by 8 to get 4032.
Answer choice C is correct.
Right triangle LMN is to be constructed in the xy-plane so that the right angle is at point L and LM is parallel to the x-axis. The x- and y- coordinates of L, M, and N are to be integers that satisfy the inequalities -3 < x < 4 and 3 < y < 11. How many different triangles with these properties could be constructed?
(A) 72
(B) 576
(C) 4032 your answer correct
(D) 4608
(E) 6336
Explanation
We are asked to find out how many possible triangles could exist within an area on the xy-plane horizontally bounded by x = -3 and x = 4 (8 possible integer values of x, including 0) and by y = 3 and y = 11 vertically (9 possible integer values). Point L lies on one of the points within this rectangular area, which means that L can lie in any one of 72 possible points (8 x 9).
Because we know that LM is parallel to the x-axis (a horizontal line), we know that point M always has the same y-value as L. We do not know the orientation of the triangle, so point M could be either to the left or right of point L on the plane, meaning that it could lie on any point on the same horizontal line as L, within the boundaries of the problem, except the point on which L already lies. This means that there are 7 possible values of point M for every point that L lies on. We can thus multiply the 72 possible values of L with the 7 possible values of M to get all the possible placements of L and M: 504.
********** POSSIBLE ERROR
********** So shouldn't this be 8 possible values for point M? So 72 * 8 instead of 72* 7?
********** POSSIBLE ERROR
Every one of these possible triangles has a right angle at L, which means that point N has the same x-coordinate as L. However, it could be above or below L, and can thus be in any of 8 different points within the boundaries (the 9 possible y-values minus the one on which point L lies). This means that for every combination of L and M, there are 8 different triangles that can be made. We multiply 504 by 8 to get 4032.
Answer choice C is correct.
