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

Question

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

For a bank of challenging coordinate geometry problems, as well as the OE to this one, see:
https://magoosh.com/gmat/2014/challengin ... questions/

Mike

arunspanda · Accepted Answer

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

The equation of the given line Q is  5y-3x=45  or  y = \frac{3}{5}x +9

Therefore, slope of the line Q =  \frac{3}{5}

Slope of line S, which is perpendicular to line Q =  \frac{-5}{3}

X-intercept of line Q =  -15

Highest possible value of Y-intercept of the line S that is perpendicular to line Q and that intersects in the second quadrant 
= Y-intercept of the line Q = 9

Lowest possible value of Y-intercept of the line that is perpendicular to line Q and that intersects in the second quadrant 
= - (slope of line Q) *(X-intercept of line Q or X-intercept of line S that intersects Q at X-axis=-15) =  -25

The number of possible perpendicular line S that intersects line Q in the second quadrant and that its Y-intercept is a integer quantity (excluding intersection at axes)= Integer values between 9 and -25= 9 – (-25)-1 =9+25-1 = 33

Answer:  (B)

tushain · Answer

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

RonnieRon · Answer

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

mikemcgarry · Answer

Dear RonnieRon , Please see a full explanation here: https://magoosh.com/gmat/2014/challengin ... questions/ It's #7 on that page. Mike :-)

ramalo · Answer

Finally get the concept but a little too difficult for myself

RonnieRon · Answer

Hi Mike
thanks - great explanation...+1

Ashishsteag · Answer

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??

:roll:

mikemcgarry · Answer

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 :-)

Ashishsteag · Answer

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.

OreoShake · Answer

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.

mikemcgarry · Answer

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 :-)

OreoShake · Answer

Makes a lot of sense Mike. Thanks

amoljain · Answer

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

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

Fdambro294 · Answer

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

Posted from my mobile device

Kinshook · Answer

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.)

Line S perpendicular to Line Q: -
5x + 3y = k

Intersection of Line Q & Line S: -
5y – 3x = 45 (1)
5x + 3y = k (2)

5*(1) + 3*(2)
25y + 9y = 225 + 3k
y = (225 + 3k)/34 >0 0 since y>0 in second quadrant
225 + 3k > 0
k > -225/3 = -75

5*(2) - 3*(1)
25x + 9x = 5k - 135
x = (5k - 135)/34 < 0 since x<0 in second quadrant
5k - 135 < 0
k < 135/5 = 27

-75 < k < 27

y-intercept of Line S: -
y = k/3 is an integer

-25 < y = k/3 < 9

Number of possible Line S = 9 - (-25) + 1 - 3 = 33
Since extreme and y = 0 are not allowed

IMO B

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards