Line \(Q\) has the equation \(5y – 3x = 45.\) If Line \(S\) is perpendicular to \(Q,\) has an integer for its \(y\)-inte

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Wed May 20, 2020 5:54 am

Line \(Q\) has the equation \(5y – 3x = 45.\) If Line \(S\) is perpendicular to \(Q,\) has an integer for its \(y\)-intercept, and intersects \(Q\) in the second quadrant, then how many possible Line \(S\)’s exist? (Note: Intersections on one of the axes do not count.)

(A) 25
(B) 33
(C) 36
(D) 41
(E) 58

[spoiler]OA=B[/spoiler]

Solution:

Expressing line Q in slope-intercept form, we have y = (3/5)x + 9. We see that line S must have a slope of -5/3 since lines S and Q are perpendicular. Furthermore, we see that line Q has a y-intercept of 9 and an x-intercept of -15. Since S intersects Q in the second quadrant, its y-intercept must be less than 9 (if it’s 9, then S and Q will intersect on the y-axis, and if it’s greater than 9, then S and Q will intersect in the first quadrant).

Now, let b be the y-intercept of line S such that S and Q will intersect on the x-axis. We can see that if the y-intercept of line S is actually greater than b (and less than 9), then lines S and Q will intersect in the second quadrant. We need to determine the value of b, and we see that the slope-intercept form of line S will be y = (-5/3)x + b. Since S and Q intersect on the x-axis and line Q has an x-intercept of -15, S also has an x-intercept of 15. Therefore, we have:

y = (-5/3)x + b

0 = (-5/3)(-15) + b

0 = 25 + b

-25 = b

Therefore, if the y-intercept of line S is greater than -25 and less than 9, then S and Q will intersect in the second quadrant. Since the y-intercept of line S has to be an integer, the number of possible possible lines for S is 8 - (-24) + 1 = 33.

Answer: B