Coordinate Geo

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Sat Feb 26, 2011 3:43 am

Lines r and s lie in the xy-plane. Is the y-intercept of line r less
than the y-intercept of line s ?

(1) At the intersection point of r and s, the x-coordinate
and y-coordinate are both negative.

(2) The slope of line r is greater than the slope of line s.

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Source: Beat The GMAT — Data Sufficiency | Sat Feb 26, 2011 3:43 am

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sat Feb 26, 2011 5:02 am

yellowho wrote:Lines r and s lie in the xy-plane. Is the y-intercept of line r less
than the y-intercept of line s ?

(1) At the intersection point of r and s, the x-coordinate
and y-coordinate are both negative.

(2) The slope of line r is greater than the slope of line s.

This should be solved by drawing some lines in your scratch pad.

Statement 1:
At the intersection point of r and s, the x-coordinate and y-coordinate are both negative.

=> The two lines pass through the 4th quadrant.
However, since we don't know the relationship between the slopes of the two lines, we can't compare the y-intercepts - Insufficient

Statement 2:
The slope of line r is greater than the slope of line s.

We know the relationship between the slopes however don't know where the two lines lie on the co-ordinate plane. So if you choose all the different quadrants and draw the two lines, you can see that the slope can't be compared. - Insufficient

Combined 1 and 2:

The quadrant is 4th and we know m(r) > m(s)
We can draw the lines and can conclude the relationshipC between the y-intercepts. It should be i(r) < i(s) - Sufficient

Make sure to check with the combinations of +ve and -ve slopes for the two lines while drawing (I didn't draw all of them while answering however feel it shouldn't matter).

C

Thanks
Anshu

(Every mistake is a lesson learned )

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Sat Feb 26, 2011 6:18 am

@anshu , I think the answer to this question should be none sufficient and E, as the function of a slope of a line is derived from x and y, and by knowing one coordinate (the cross point) lying on the lines r and s we may assume different directions of the lines - hence relationship of slopes. Both cases are present when i(r) > i(s) and i(r) < i(s).

anshumishra wrote:
yellowho wrote:Lines r and s lie in the xy-plane. Is the y-intercept of line r less
than the y-intercept of line s ?

(1) At the intersection point of r and s, the x-coordinate
and y-coordinate are both negative.

(2) The slope of line r is greater than the slope of line s.
This should be solved by drawing some lines in your scratch pad.

Statement 1:
At the intersection point of r and s, the x-coordinate and y-coordinate are both negative.

=> The two lines pass through the 4th quadrant.
However, since we don't know the relationship between the slopes of the two lines, we can't compare the y-intercepts - Insufficient

Statement 2:
The slope of line r is greater than the slope of line s.

We know the relationship between the slopes however don't know where the two lines lie on the co-ordinate plane. So if you choose all the different quadrants and draw the two lines, you can see that the slope can't be compared. - Insufficient

Combined 1 and 2:

The quadrant is 4th and we know m(r) > m(s)
We can draw the lines and can conclude the relationshipC between the y-intercepts. It should be i(r) < i(s) - Sufficient

Make sure to check with the combinations of +ve and -ve slopes for the two lines while drawing (I didn't draw all of them while answering however feel it shouldn't matter).

C

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sat Feb 26, 2011 6:41 am

Night reader wrote:@anshu , I think the answer to this question should be none sufficient and E, as the function of a slope of a line is derived from x and y, and by knowing one coordinate (the cross point) lying on the lines r and s we may assume different directions of the lines - hence relationship of slopes. Both cases are present when i(r) > i(s) and i(r) < i(s).
anshumishra wrote:
yellowho wrote:Lines r and s lie in the xy-plane. Is the y-intercept of line r less
than the y-intercept of line s ?

(1) At the intersection point of r and s, the x-coordinate
and y-coordinate are both negative.

(2) The slope of line r is greater than the slope of line s.
This should be solved by drawing some lines in your scratch pad.

Statement 1:
At the intersection point of r and s, the x-coordinate and y-coordinate are both negative.

=> The two lines pass through the 4th quadrant.
However, since we don't know the relationship between the slopes of the two lines, we can't compare the y-intercepts - Insufficient

Statement 2:
The slope of line r is greater than the slope of line s.

We know the relationship between the slopes however don't know where the two lines lie on the co-ordinate plane. So if you choose all the different quadrants and draw the two lines, you can see that the slope can't be compared. - Insufficient

Combined 1 and 2:

The quadrant is 4th and we know m(r) > m(s)
We can draw the lines and can conclude the relationshipC between the y-intercepts. It should be i(r) < i(s) - Sufficient

Make sure to check with the combinations of +ve and -ve slopes for the two lines while drawing (I didn't draw all of them while answering however feel it shouldn't matter).

C

Night reader,

Here are the 3 scenarios I considered while solving. Please point out if I made mistakes with them or missed anything.

[/img]

Thanks
Anshu

(Every mistake is a lesson learned )

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Sat Feb 26, 2011 7:36 am

Am I reading correctly - it says at the intersection of lines r and s, x and y coordinates are negative? Why in three examples only y coordinate is negative?

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sat Feb 26, 2011 8:00 am

Night reader wrote:Am I reading correctly - it says at the intersection of lines r and s, x and y coordinates are negative? Why in three examples only y coordinate is negative?

you are right. Thanks for pointing that out!
Totally read it wrong, that why considered it to be in 4th quadrant while it should be in the 3rd quadrant.

And then the answer should be E.

Thanks
Anshu

(Every mistake is a lesson learned )

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yellowho: Master | Next Rank: 500 Posts; Posts: 233; Joined: Wed Aug 22, 2007 3:51 pmFollowed by: 2 members; Location: New York; Thanked: 7 times

by yellowho » Sat Feb 26, 2011 3:55 pm

Actually its still C. How did you organize this ash?

My approach:

1) 1 is not enough

2) Assume Slope R is positive. Test if there are different result done. not sufficient done. If only one result got to 2b

2b) Assume Slope R is negative.Test if there are different result. Not sufficient. done. If only one result. Compare to 2a. If both are the same then Sufficient. If not then insufficient.

3) Combine the statements. Repeat 2a and 2b.

I think what also makes this confusing is when the slopes are both negative. In this case the steeper slope is actually lower.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sun Feb 27, 2011 11:22 pm

yellowho wrote:Lines r and s lie in the xy-plane. Is the y-intercept of line r less
than the y-intercept of line s ?

(1) At the intersection point of r and s, the x-coordinate
and y-coordinate are both negative.

(2) The slope of line r is greater than the slope of line s.

Solution:
Let the equation of line r be y = m1x+c1, where m1 is the slope and c1 is the y intercept.
Let the equation of line s be y = m2x+c2, where m2 is the slope and c2 is the y intercept.
We need to know whether c1 < c2 or not.
First consider (1) alone.
At the intersection point x = (c2 - c1)/(m1-m2), y = (m1c2 - m2c1)/(m1-m2).
Now, x < 0 and y < 0.
Or (c2 - c1)/(m1-m2) < 0 and (m1c2 - m2c1)/(m1-m2) < 0.
But this is not enough to tell us whether c1 < c2 or not.
We need to know about m1 and m2 as well.
Or (1) alone is not sufficient.
Next consider (2) alone.
It says m1 > m2.
But this does not tell us whether c1 < c2 or not.
Or (2) alone is not sufficient.
Next, combine both the statements together and check.
Since m1 > m2, (m1 - m2) > 0.
Also, since (c2 - c1)/(m1-m2) < 0 from (1), c2 - c1 < 0.
Or c1 > c2.
So, the answer to the main question is NO.
Hence, both statements together are sufficient to answer the question.

The correct answer is (C).

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madhan_dc: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Mon Jan 24, 2011 1:37 pm

by madhan_dc » Mon Feb 28, 2011 11:34 am

Since the intersection point is both negative, i picked a point in that quadrent and since r has greater slope it will be steeper than s. From that can i say the y intercept of r will be greater than s?

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Mon Feb 28, 2011 12:01 pm

Using the graphical approach (or better say drawing six lines to be exact) here is why it is "C", as in all the 3 possible cases i(r) > i(s) .

Thanks
Anshu

(Every mistake is a lesson learned )

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Feb 28, 2011 12:36 pm

@anshu, we are both missing one point made by Anurag in his post - he has put the coordinate geometry relationship between two lines and their intersection point into an algebraic form (function, equation). I will ask Anurag for additional reference points (sources) to back up this useful theory.

@madhan, I have rechecked conclusion made by Anurag the same way as you suggest in your post, picked one common point x (-ve) and y (-ve) for both lines and selected various intercept points on y below 0 and above 0 to test whether the slopes increase and intercepts decrease. I revealed that the increase in slopes will always result in increase in intercepts.

anshumishra wrote:Using the graphical approach (or better say drawing six lines to be exact) here is why it is "C", as in all the 3 possible cases i(r) > i(s) .

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Mon Feb 28, 2011 12:59 pm

hi all
i also tried to solve this problem with math, but my approach is slightly different from that of Anurag, it will be great somebody comment on this
r=y=k1*x+b1
s=y=k2*x+b2, we need to estimate whether b2>b1.
b1=y-k1*x, and
b2=y-k2*x,
here, if b2>b1, then y-k2*x>y-k1*x, from here i canceled y and left k1*x-k2x>0 i did not cancel x as i don`t know the sign
finally, does x(k1-k2)>0,
and start to solve it as pure math inequality
for x(k1-k2)>0, it is possible in two cases, x>0 k1>k2. or x<0 k1<k2
(1) says x<0, not suff,
(2) says k1>k2, not suff as we have no info about x
but both says that x(k1-k2)<0 (-)(+)=(-) so suff -to me. the answer is no
is it right or i got answer by chance?

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Mon Feb 28, 2011 1:00 pm

Night reader wrote:@anshu, we are both missing one point made by Anurag in his post - he has put the coordinate geometry relationship between two lines and their intersection point into an algebraic form (function, equation). I will ask Anurag for additional reference points (sources) to back up this useful theory.

Yes, Anurag's method is the most general approach to solve this problem, reducing it to an algebra problem.
He is using the famous "slope intercept" form for a straight line., i.e y=mx+c (where m = slope and c is y intercept).

My approach is to use the graph to solve it.

Thanks
Anshu

(Every mistake is a lesson learned )

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Mon Feb 28, 2011 1:10 pm

clock60 wrote:hi all
i also tried to solve this problem with math, but my approach is slightly different from that of Anurag, it will be great somebody comment on this
r=y=k1*x+b1
s=y=k2*x+b2, we need to estimate whether b2>b1.
b1=y-k1*x, and
b2=y-k2*x,
here, if b2>b1, then y-k2*x>y-k1*x, from here i canceled y and left k1*x-k2x>0 i did not cancel x as i don`t know the sign
finally, does x(k1-k2)>0,
and start to solve it as pure math inequality
for x(k1-k2)>0, it is possible in two cases, x>0 k1>k2. or x<0 k1<k2
(1) says x<0, not suff,
(2) says k1>k2, not suff as we have no info about x
but both says that x(k1-k2)<0 (-)(+)=(-) so suff -to me. the answer is no
is it right or i got answer by chance?

That is perfect as well !
It is similar to Anurag's solution, the steps are different. However in both cases we are using to equivalent equations :
y = m1x + c1 or k1x + b1 and
y = m2x + c2 or k2x + b2
and proving the same thing : c1> c2 or b1>b2

Thanks
Anshu

(Every mistake is a lesson learned )

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Mon Feb 28, 2011 1:36 pm

thank you anshumishra
you reply is encouraging

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