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Co-ordinate Geomentry

by priyasaibaba » Thu Apr 22, 2010 12:44 pm
Can anyone help solve this problem?

108. Does line L pass through the first quadrant?
(1) L is perpendicular to the line with equation 3y-4x+5=0
(2) L passes through the third quadrant.

A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Thanks,
Priya
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by tpr-becky » Thu Apr 22, 2010 2:38 pm
In order to know if a line passes through the first quadrant you need a point and slope of the line to determine if any of the points are (+,+) but you can also think about it more conceptually

1. if you manipulate the line equation you get y=4/3x- 5 which means that the line crosses the y axis at -5 and has a postitive slope. because our line is perpendicular you only know that it has a slope of -3/4 but you don't know where it intersects the line so you don't have a poitn on the line. Since we don't have a point on the line we can't determine whether it will eventually pass through the first quadrant.(here you can make a quick sketch and see that you can draw a line in the first quadrant as well as one that doesn't pass through the first quadran.

BCE

2. Just saying it passes through the third quadrant dosn't give us any information about the slope or any points - insufficient.

If we put the two facts together we now know that our line has a negative slope and passes through the third quadrant - which means that it crosses the y axis in the negative and thus cannot pass through the first quadrant

Th answer is C
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by chrsrook » Thu Apr 22, 2010 3:26 pm
I read somewhere that if a line has a negative slope, it will pass through quadrants 2 and 4 and if it has a positive slope, it will pass through quadrants 1 and 3. If this were true, then won't statement 1 alone be sufficient to answer the question? We know that line L will have a negative slope and will therefore not pass through quadrant 1?

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by tpr-becky » Thu Apr 22, 2010 4:01 pm
while it is true that a negative slope line will pass through 2 and 4 it is not true that it will ONLY pass through 2 and 4 - a negative slope line can pass through 1, 2 and 4 or 1, 3 and 4. The only time it passes through just 2 and 4 is when it crosses the y axis at the origin. same is true for a postive slope line with quadrants 1 and 3. Try drawing it out.
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by eaakbari » Thu Apr 22, 2010 10:42 pm
tpr-becky wrote:In order to know if a line passes through the first quadrant you need a point and slope of the line to determine if any of the points are (+,+) but you can also think about it more conceptually

1. if you manipulate the line equation you get y=4/3x- 5 which means that the line crosses the y axis at -5 and has a postitive slope. because our line is perpendicular you only know that it has a slope of -3/4 but you don't know where it intersects the line so you don't have a poitn on the line. Since we don't have a point on the line we can't determine whether it will eventually pass through the first quadrant.(here you can make a quick sketch and see that you can draw a line in the first quadrant as well as one that doesn't pass through the first quadran.

BCE

2. Just saying it passes through the third quadrant dosn't give us any information about the slope or any points - insufficient.

If we put the two facts together we now know that our line has a negative slope and passes through the third quadrant - which means that it crosses the y axis in the negative and thus cannot pass through the first quadrant

Th answer is C
Thats a great answer Becky. But in general I fail to visualize the slope of a line. Do help me here. If I say a line has slope -3 , what is the mental picture formed in your mind on the xy plane
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by tpr-becky » Fri Apr 23, 2010 6:24 am
Image


a negative slope always goes downhill from left to right so it always starts in 4 and ends in 2 - depends on where it crosses the y axis.

A positive slope goes uphil from left to right so it always starts in 3 and goes to 1 - where the line is depends on the y intercept

While specific problems may be affected by the value of the slope this problem is not.
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by eaakbari » Fri Apr 23, 2010 6:29 am
tpr-becky wrote:Image


a negative slope always goes downhill from left to right so it always starts in 4 and ends in 2 - depends on where it crosses the y axis.

A positive slope goes uphil from left to right so it always starts in 3 and goes to 1 - where the line is depends on the y intercept

While specific problems may be affected by the value of the slope this problem is not.
Thanks but I was actually asking about the value of a slop, how is a slope 1/6 different from 3. Is there an immediate picture formed in ones head or do I think about intercepts using -y/x.
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by akhpad » Fri Apr 23, 2010 6:56 am
Hello eaakbari,

If slope m = -3 = tanTHETA

THETA > 90 degree

which can be the left one of the above picture.

I can see above that Small question but big explanation. Good.
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by akhpad » Fri Apr 23, 2010 7:15 am
eaakbari wrote:
tpr-becky wrote:Image


a negative slope always goes downhill from left to right so it always starts in 4 and ends in 2 - depends on where it crosses the y axis.

A positive slope goes uphil from left to right so it always starts in 3 and goes to 1 - where the line is depends on the y intercept

While specific problems may be affected by the value of the slope this problem is not.
Thanks but I was actually asking about the value of a slop, how is a slope 1/6 different from 3. Is there an immediate picture formed in ones head or do I think about intercepts using -y/x.
tanA = 1/6
tanB = 3

0 < A < B < 90 degree
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by tpr-becky » Fri Apr 23, 2010 11:47 am
I actually never get into trig on the GMAT so I see slope as the value of rise/run or alternately the y2-y1/x2-x1 -

Therefore a slope of three would rise 3 and move to the right 1 and a slope of 1/6 would rise one and move to the right 6.

Negative slopes move to the left as they rise.

Hope this helps.
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by frank1 » Sat Apr 24, 2010 2:31 am
Great thanks
tpr-becky wrote: 1. if you manipulate the line equation you get y=4/3x- 5 which means that the line crosses the y axis at -5 and has a postitive slope.
By the way isn't it
y=(4x/3)-5/3?
i was trying to understand the whole thing ...line by line and when i encountered this line....i really felt at that point that i have been out from the school for long time now as i had to rethink over it...lol
any way doesnt seem to have big bearing in this question

any way thanks
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by govind_raj_76 » Sat Apr 24, 2010 6:40 am
I have the same issue as frank.

By the way isn't it
y=(4x/3)-5/3?

please explain how the slope comes to y=4/3x- 5 ?

Thanks,
Govind
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by akhpad » Sat Apr 24, 2010 6:57 am
Equation of line 3y-4x+5=0

You can find slope in various ways

1.
3y = 4x -5
y = 4x / 3 - 5 / 3
y = mx + c, where m is the slope of line and c is the y-intercept.

Slop = m = 4/3

2.
m = - (coefficient of x) / (coefficient of y) = - (-4) / 3 = 4/3

etc
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by govind_raj_76 » Sat Apr 24, 2010 9:20 am
As per Becky, she has said it is a negative slope
----------------------------------------------------------------
2. Just saying it passes through the third quadrant dosn't give us any information about the slope or any points - insufficient.

If we put the two facts together we now know that our line has a negative slope and passes through the third quadrant - which means that it crosses the y axis in the negative and thus cannot pass through the first quadrant .

But the slope is Slop = m = 4/3. Is this is a positive slope ??? My mind wavers on this. Please clarify this.
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by frank1 » Sat Apr 24, 2010 10:48 am
govind_raj_76 wrote:I have the same issue as frank.

By the way isn't it
y=(4x/3)-5/3?

please explain how the slope comes to y=4/3x- 5 ?

Thanks,
yup the slope still is 4/3 but constant c should be -5/3 not -5
Thank god that In this question,value of c didnt have any role to play....
but it created some confusion for me initially

thanks
Last edited by frank1 on Sat Apr 24, 2010 7:39 pm, edited 1 time in total.
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