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# Intersecting lines

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Manager
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Intersecting lines [#permalink]  13 Nov 2005, 16:42
I could swear I've posted this one already but can't find it anywhere so here goes:

In the XY coordinate plane, Line L and Line K intersect at point (4,3). Is the products of their slopes negative?

1. The product of the x-intercepts of lines L and K is positive

2. The product of the y-intercepts of lines L and K is negative.

Please explain the rules too! thx
SVP
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[#permalink]  13 Nov 2005, 18:20
Intern
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Re: Intersecting lines [#permalink]  15 Nov 2005, 21:32
Jennif102 wrote:
I could swear I've posted this one already but can't find it anywhere so here goes:

In the XY coordinate plane, Line L and Line K intersect at point (4,3). Is the products of their slopes negative?

1. The product of the x-intercepts of lines L and K is positive

2. The product of the y-intercepts of lines L and K is negative.

Please explain the rules too! thx

whoah! where did you get this problem Jen?

Well it took me some time but my answer is B. this is why:

we are give 1 point (4,3)

stmt. 1)
talks about x intercept so we have another point for L and K
L(X1, 0) and K(X2, 0)

slope L = m1 = -3/(x1- 4) --> using slope, m = (y2 - y1)/(x2 - x1)
slope K = m2 = -3/(x2-4)

m1.m2 = -9/(x1-4)(x2-4) ....can be -ve if both x1 and x2>4 or both <4. since the stmt says that x1.x2>0 it doesnt help
insufficient!

stmt 2)
here we are given points L(0,y1) and K(0,y2)

slope L = m1 = (y1-3)/4
slope K = m2 = (y2-3)/4

m1.m2 = (y1-3)(y2-3)/16. this can be negative is either y1 or y2 <3

since we are given that y1.y2<0, we know that one of them is negative and one positve. therefore m1.m2 <0

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Re: Intersecting lines [#permalink]  15 Nov 2005, 21:43
tank wrote:
Jennif102 wrote:
I could swear I've posted this one already but can't find it anywhere so here goes:

In the XY coordinate plane, Line L and Line K intersect at point (4,3). Is the products of their slopes negative?

1. The product of the x-intercepts of lines L and K is positive

2. The product of the y-intercepts of lines L and K is negative.

Please explain the rules too! thx

whoah! where did you get this problem Jen?

Well it took me some time but my answer is B. this is why:

we are give 1 point (4,3)

stmt. 1)
talks about x intercept so we have another point for L and K
L(X1, 0) and K(X2, 0)

slope L = m1 = -3/(x1- 4) --> using slope, m = (y2 - y1)/(x2 - x1)
slope K = m2 = -3/(x2-4)

m1.m2 = -9/(x1-4)(x2-4) ....can be -ve if both x1 and x2>4 or both <4. since the stmt says that x1.x2>0 it doesnt help
insufficient!

stmt 2)
here we are given points L(0,y1) and K(0,y2)

slope L = m1 = (y1-3)/4
slope K = m2 = (y2-3)/4

m1.m2 = (y1-3)(y2-3)/16. this can be negative is either y1 or y2 <3

since we are given that y1.y2<0, we know that one of them is negative and one positve. therefore m1.m2 <0

Take y1=-1<0 , y2=2> 0 , we have y1.y2=-1*2<0
but m1.m2= -4*-1/16= 1/4 >0
Take y1=-1, y2=4 , we have m1.m2= -4* 1= -4 <0
----> (2) insuff
Intern
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[#permalink]  15 Nov 2005, 22:40
ahh! you're right. y1 or y2 being +ve isn't good enough it needs to >3. SO, b is in fact insufficient.
Answer would be E then.
Manager
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[#permalink]  16 Nov 2005, 09:51
I think I got this by doing a GMAT PREP practice test. Pretty high level huh? Surprisingly high for someone who managed to score only 57th percentile on the real thing's math sectio

Anyway, I can go back and check the OA. But it seems like E ought to be right. Unless....
Is anyone aware of any rules concerning intersecting lines and X and Y intercepts? All I know is that the slopes will be negative recipricals.

thanks,
Jen
Manager
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[#permalink]  17 Nov 2005, 05:09
I wouldn't spend more than 2-3 min, in the real test. So, I would've answered E for this reason:

If product of x-intercepts is +ve, we can draw lines that intersect at (4,3) that have both +ve slopes and -ve slopes ( easy to tell- does it fall from left to right?) - insuff

If the product of y intercepts is -ve, we can draw lines that intersect at (4,3) that have both +ve slopes and -ve slopes ( easy to tell- does it fall from left to right?) - insuff

SVP
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[#permalink]  17 Nov 2005, 05:18
then how about the case in which two statements are combined?!
I'm stuck to C as i pointed out in my attached illustration.
VP
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[#permalink]  20 Mar 2006, 20:29
Let the two lines be y1 =m1x1 +c1 and y2 = m2x2+ c2.

1. The x-intercepts are -c1/m1 and -c2/m2
=> product is c1c2/m1m2 >0..INSUFF

2. the y-intercepts are c1 and c2 => product c1c2 < 0.....INSUFF

I+II => c1c2 < 0 AND c1c2/m1m2 > 0 => m1m2 < 0...SUFF.
Senior Manager
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[#permalink]  20 Mar 2006, 23:11
lhotseface wrote:
Let the two lines be y1 =m1x1 +c1 and y2 = m2x2+ c2.

1. The x-intercepts are -c1/m1 and -c2/m2
=> product is c1c2/m1m2 >0..INSUFF

2. the y-intercepts are c1 and c2 => product c1c2 < 0.....INSUFF

I+II => c1c2 < 0 AND c1c2/m1m2 > 0 => m1m2 < 0...SUFF.

Great way of solving, I was always thinking should I go memorize these intercept equation by heart...but now I have no doubt...
[#permalink] 20 Mar 2006, 23:11
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# Intersecting lines

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