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Co-ordinates on a line segment - PS Set 42 Question 13

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by gmatrant » Mon Oct 15, 2007 5:47 pm
PS Set 42 Question 13

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?
(A) 5
(B) 8
(C) 10
(D) 11
(E) 20

I did not know how to attache an image? Hence P co-ordinate is (0,30)
while Q cordinate is (50,0).So POQ is a right angled triangle where O is the origin.

a)What is the approach for such problems?
b)What if the above question(3) had otherwise asked the number of positive integer co-ordinates (both x and y must be positive) below the line segment PQ.

Answer : D

My Apporach (But i got the wrong answer)
I found the slope which is 3/5, i then took the equation of the line to be
y=mx+c where m is the slope . So x must be a multiple of 5 and this leads to x taking in 10 values from 0 -50. Guess this approach is wrong.
Please let me know how to solve this.

Thanks
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by pahwa » Mon Oct 15, 2007 8:17 pm
2 points P and Q are given. So, use 2-point fomula to know the equation of line. (What you did is correct).

y-y1 = [(y2-y1)/(x2-x1)]*(x-x1)

once you know the line equation, calculate x in terms of y. Lets assume that equation you get is 3x-2y=9. Bring this in the form that x=(9+2y)/3

Put various values of y (integer) and see what all results of x are integers
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by gmatrant » Tue Oct 16, 2007 6:47 am
pahwa wrote:2 points P and Q are given. So, use 2-point fomula to know the equation of line. (What you did is correct).

y-y1 = [(y2-y1)/(x2-x1)]*(x-x1)

once you know the line equation, calculate x in terms of y. Lets assume that equation you get is 3x-2y=9. Bring this in the form that x=(9+2y)/3

Put various values of y (integer) and see what all results of x are integers
I tried solving the way you had suggested, but couldn't get to the answer

x-50 =(30/-50)(y-0)
x-50-(-3/5)(y)
5x-250=-3y
x=(250-3y)/5
x= 50 - 3y/5
y can take values between 0 and 30 in multiples of 5
y = 0,5,10,15,20,25,30
So there can be only seven values.

Can anyone please confirm if this is right.
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by gmatrant » Tue Oct 16, 2007 8:35 pm
Anonymous wrote:why can't y be negative?
y Can't be negative because y is between 0 and 30. This is as given in the graph.
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by samirpandeyit62 » Tue Oct 16, 2007 10:26 pm
lenght of pq = sqrt(50^2 + 30^2)

=10sqrt(34)

now if we take any point on pq & join it with x & y axis we will get a triangle similar to poq

hence ratio of sides will be same

i.e y/10sqrt(34) = x/50 =z/40

here x & z will be integers only if y = sqrt(34) i.e y cancels out the decimal part of 10sqrt(34)

hence if we divide pq = 10sqrt(34) into 10 parts each of sqrt(34) then the corresponding x, y coordinates will be integers (as each part will be a multiple of sqrt(34))

since PQ is divided into 10 parts hence nos of points on it will be 11

so ans should be D 11
Regards
Samir
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by ri2007 » Thu Oct 18, 2007 1:42 pm
Hi
can any one tell me where i can find the figure for this question?
thanks
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by gmatrant » Sat Oct 20, 2007 5:19 am
It took some time for me to understand the solution.. but thats a great approach.

Is there a way to find out the number of positive integer co-ordinates (x,y) below the line PQ?
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by agni_mba » Sat Oct 20, 2007 11:39 am
gmatrant,

your approach is absolutely okay. just that you forgot to count (0,30) as a possible positive integral solution to the problem y = -(3/5)x+ 30

cheers
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by gmatrant » Sat Oct 20, 2007 9:07 pm
Are you referring to this approach
My Apporach (But i got the wrong answer)
I found the slope which is 3/5, i then took the equation of the line to be
y=mx+c where m is the slope . So x must be a multiple of 5 and this leads to x taking in 10 values from 0 -50.
I recalculated it, yes it give me 11 co-ordinates and the slope is negative...,thanks.. but at hindsight I seem to be making a lot of silly mistakes....
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