Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q

Dear TheLordCommander,

I'm happy to respond. This is a hard question, so questions about it are not "stupid."

Part of the requirement of the question is that the y-intercept has to be an integer. If we mark off the boundaries on the y-intercept, then we simply can count integers along the y-axis.

You see, if the x-intercept is an integer, that doesn't guarantee that the y-intercept is an integer (unless the slope is +1 or -1). Certainly for any non-integer slope, the general rule is that for most x-intercepts that are integers, the y-intercept is not an integer, and vice versa. If we start looking at points on the x-axis, we know they have to be -15 and 5, but within that range, we have no idea what spacings of the values of the x-intercept would result in integer values on the y-intercept.

Therefore, it's much easier simply to stick to the y-intercept and count integers.

Does all this make sense?
Mike

5y - 3x = 45 --> 5y = 3x + 45
at x= 0, y = 9
at y=0, x = -15
Now slope = 3/5
perpendicular 's slope will be -5/3
at ( -15,0) perp line eqn = y-0 = -5/3(x-(-15))
thus y = -5/3 x - 25
thus at x=0, y = - 25
24 + 1 + 8 = 33

Hello Mike,

Well this was a difficult question for me.

I understood the solution but i have a silly doubt-intersection at the axes means the perpendicular line S that passes through the origin should be excluded??

Dear Ashishsteag,

I'm happy to respond. Yes, this is an extremely difficult question.

Your question is not silly. Let's look at the prompt:
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.)
In this problem, the intersection of concern, the only intersection being discussed, is the intersection of Line Q and Line S. That is the only intersection that matters in the problem. If these two lines intersect on the axis---for example, at the point (0, 9)---then we wouldn't count that as an intersection in QII, precisely because points on either axis are not points in QII. Line S will have its own intersections with the x- and y-axes, but those are irrelevant to the problem. It doesn't matter where Line S intersects the axes, even at the origin. The only thing that matters is the intersection point of Line Q and line S.

Does all this make sense?
Mike

Alternate Solution:

As \(S\) is a Line perpendicular to \(5y-3x=45\), hence it will have slope \(-\frac{5}{3}\)
Let the equation of \(S\) be \(y=-(\frac{5}{3})x+c\) ; where \(c\) is the y-intercept

Solving the 2 lines for point of intersection
\((5y-3x=45)*3\) ...eqn of Q
\((3y+5x=3c)*5 \)...eqn of S

Solving for x we get,
\(x=\frac{(15c-135)}{34}\) ..\(eqn(1)\)

As the intersection point of \(Q\) and \(S\) is in II Quadrant, the value of x will vary from \((-15,0)\). [\(-15\) and \(0\) are not included as point of intersection cannot be on axis as stated in question]

hence,

\(-15<x<0\)
replacing \(x\) from \(eqn(1)\)

\(-15<\frac{(15c-135)}{34}<0\)

\(-25<c<9\)

or \(c\) belongs to \([-24,8]\)

as \(c\) is given to be integral, number of integral points between \(-24\) and \(8\) will be \(8-(-24)+1=33\)

How do we get 24 + 1 + 8 = 33
Please explain?

Finally get the concept but a little too difficult for myself

Hi Mike
thanks - great explanation...+1

Actually,I got confused with this part of the question written at the very end:"(Note: Intersections on one of the axes do not count.)",and then the question also speaks about y-intercept in the part:"If Line S is perpendicular to Q, has an integer for its y-intercept",so I thought that since the perpendicular line S does not have any sort of y-intercept at the origin (0,0),so we need to exclude that part?
Even one of the solutions mentioned above with the required figure drawn is subtracting 1 at the very end of the solution.I understand your point that only intersections between line S and line Q need to be considered and they need not be counted on either of the axes.Thanx a lot for your help.

my question is - we calculated the intersections in the y axis between 9 and -25 to get the answer; why did we not calculate the intersections in the x axis between -15 and 5 instead? Please bear with me if this is a stupid question. Thank you.

Makes a lot of sense Mike. Thanks

First, for the line S to be perpendicular to the given line, the slope must be the Negative Reciprocal of Line Q’s slope

5y - 3x = 45

y = (3/5)x + 9

The X intercept will be at (-15 , 0)

The Y intercept will be at (0 , 9)

And the slope will be positive upward sloping = (3/5)

The negative reciprocal ——> (3/5) * m = -1

Where m = -(5/3)


So the Line that will be perpendicular through one of the points on Line Q must have the slope = m = -(5/3)

Line S will take the form of:

y = -(5/3)x + b

The question constraints are the following:

Line S must interest Line Q at a perpendicular angle in the 2nd Quadrant

Line S must NOT intersect any of the Axis (Y axis or X Axis)

The Y intercept (given by b) must be an Integer (can be positive or negative)

(1st) we can find the equation of the line that passes through the X Intercept of Line Q

y = -(3/5)x + b

Plug in (-15 , 0)

And you get: b = -25

So the equation of:

y = (-3/5)x - 25

Although the line would be perpendicular to line Q, it will be perpendicular at the intersection of the X Axis, which is not allowed.

However, lines such as:

y = -(3/5)x - 24

y = -(3/5)x - 23


All the way up the value of b = +9 must intersect Line Q in the 2nd Quadrant

For the line given by:

y = -(3/5)x + 9

The line will be perpendicular to Line Q at the Y intercept of (0 , 9) - which is NOT allowed


So in the end, it becomes a counting problem.

How many consecutive integers exist from -24 to +8, inclusive

(24 negative Integers) + (1 zero) + (8 positive integers) =

24 + 1 + 8 = 33

B

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