In the xy coordinate plane, line L and line K intersect at : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 08:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the xy coordinate plane, line L and line K intersect at

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 770 [0], given: 33

In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

06 May 2010, 11:31
17
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

52% (03:36) correct 48% (01:31) wrong based on 411 sessions

### HideShow timer Statistics

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

(1) The product of the x-intersects of lines L and K is positive.
(2) The product of the y-intersects of lines L and K is negative.
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 May 2012, 11:49, edited 1 time in total.
Edited the question and added the OA
Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7079

Kudos [?]: 93177 [6] , given: 10553

In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

06 May 2010, 12:24
6
KUDOS
Expert's post
6
This post was
BOOKMARKED
LM wrote:

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: $$y_l=m_1x+b_1$$ and $$y_k=m_2x+b_2$$. The question: is $$m_1*m_2<0$$?

Lines intersect at the point (4,3) --> $$3=4m_1+b_1$$ and $$3=4m_2+b_2$$

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of $$x$$ for $$y=0$$ and equals to $$x=-\frac{b}{m}$$ --> so $$(-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0$$ --> $$\frac{b_1b_2}{m_1m_2}>0$$.

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of $$y$$ for $$x=0$$ and equals to $$y=b$$ --> $$b_1*b_2<0$$.

(1)+(2) $$\frac{b_1b_2}{m_1m_2}>0$$ and $$b_1*b_2<0$$. As numerator in $$\frac{b_1b_2}{m_1m_2}>0$$ is negative, then denominator $$m_1m_2$$ must also be negative. So $$m_1m_2<0$$. Sufficient.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.
_________________
Director
Joined: 03 Sep 2006
Posts: 879
Followers: 6

Kudos [?]: 770 [1] , given: 33

Re: GMAT PREP (DS) [#permalink]

### Show Tags

06 May 2010, 12:28
1
KUDOS
Bunuel, thanks a lot.
Senior Manager
Joined: 25 Feb 2010
Posts: 481
Followers: 4

Kudos [?]: 84 [8] , given: 10

Re: Interesting Line Eqn DS problem [#permalink]

### Show Tags

19 Jul 2010, 07:29
8
KUDOS
1
This post was
BOOKMARKED
line L: y1 = m1x1 + b1; x-intercept = -b1/m1; y-intercept = b1
line K: y2 = m2x2 + b2; x-intercept = -b2/m2; y-intercept = b2

(4,3) lies on both.
3 = 4m1 + b1 = 4m2 + b2

Is m1m2 < 0?

1. (-b1/m1)*(-b2/m2) > 0
b1b2/m1m2 > 0
If b1b2 > 0, m1m2 > 0
If b1b2 < 0, m1m2 < 0
NOT SUFFICIENT.

2. b1b2 < 0
NOT SUFFICIENT.

Together, b1b2 < 0 and m1m2 < 0.
SUFFICIENT. Answer is C.
_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 957 [5] , given: 25

Re: GMAT PREP (DS) [#permalink]

### Show Tags

25 Sep 2010, 11:27
5
KUDOS
2
This post was
BOOKMARKED
The equation of a line can be written as :

$$\frac{x}{a} + \frac{y}{b} = 1$$

Here a is the X-intercept and b is the Y-intercept

The slope of this line is -b/a

Let the lines be :

$$\frac{x}{a_1} + \frac{y}{b_1} = 1$$
$$\frac{x}{a_2} + \frac{y}{b_2} = 1$$

We know both lines pass through (4,3)
So we know that
$$\frac{4}{a_1} + \frac{3}{b_1} = \frac{4}{a_2} + \frac{3}{b_2} = 1$$
The information about (4,3) doesn't tell us much about the signs of a1,a2,b1,b2

Question is if Slope1*Slope2 =(b1 b2) / (a1 a2) < 0

(1) a1 * a2 > 0
Depends on the sign of b1*b2

(2) b1 * b2 < 0
Depends on the sign of a1*a2

(1) + (2) : Clearly sufficient to say yes. The slope product is always < 0.
_________________
Director
Joined: 21 Dec 2009
Posts: 591
Concentration: Entrepreneurship, Finance
Followers: 18

Kudos [?]: 662 [5] , given: 20

Re: GMAT PREP (DS) [#permalink]

### Show Tags

27 Sep 2010, 07:39
5
KUDOS
under timed schedule, i considered using a quick graphical sketch and that helped.
Let M1 and M2 represent the gradients of lines K and L

(1)
from sketch (i), M1*M2 < 0
from sketch (ii), M1*M2 > 0
INSUFFICIENT

(2)
from sketch (i), M1*M2 < 0
from sketch (iii), M1*M2 > 0
INSUFFICIENT

Combining (1) and (2):
ONLY sketch i) satisfies --> SUFFICIENCY
option C is therefore correct.
Attachments

intersecting lines K and L.docx [11.19 KiB]

_________________

KUDOS me if you feel my contribution has helped you.

Intern
Joined: 30 Mar 2010
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 4

Re: GMAT PREP (DS) [#permalink]

### Show Tags

28 Sep 2010, 13:47
I didn´t waste any time calculating this question, just pictured how the lines could be drawn in the coordinate plane after considering both statements!!! I got letter C!!!
Manager
Joined: 07 Aug 2010
Posts: 83
Followers: 1

Kudos [?]: 18 [0], given: 9

Re: GMAT PREP (DS) [#permalink]

### Show Tags

14 Oct 2010, 18:29
C

1) gives 1 point of lines l and k on the axes

2) gives 1 more set of points for lines l and k on the axes

pick some numbers and plot the intercepts on the xy co-ord system. in order for the lines to intersect in q1, they have to have opposite slopes (may not be equal)... SUFF
_________________

Click that thing - Give kudos if u like this

Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 406
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross) - Class of 2014
GMAT 1: 730 Q49 V39
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)
Followers: 7

Kudos [?]: 45 [0], given: 46

Re: GMAT PREP (DS) [#permalink]

### Show Tags

16 Dec 2010, 08:26
Bunuel wrote:
LM wrote:

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: $$y_l=m_1x+b_1$$ and $$y_k=m_2x+b_2$$. The question: is $$m_1*m_2<0$$?

Lines intersect at the point (4,3) --> $$3=4m_1+b_1$$ and $$3=4m_2+b_2$$

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of $$x$$ for $$y=0$$ and equals to $$x=-\frac{b}{m}$$ --> so $$(-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0$$ --> $$\frac{b_1b_2}{m_1m_2}>0$$.

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of $$y$$ for $$x=0$$ and equals to $$x=b$$ --> $$b_1*b_2<0$$.

(1)+(2) $$\frac{b_1b_2}{m_1m_2}>0$$ and $$b_1*b_2<0$$. As numerator in $$\frac{b_1b_2}{m_1m_2}>0$$ is negative, then denominator $$m_1m_2$$ must also be negative. So $$m_1m_2<0$$. Sufficient.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.

Wow, this explanation made something CLICK in my head.

This all makes sense to me now.

Thanks a lot Bunuel
_________________
Manager
Joined: 02 Oct 2010
Posts: 158
Followers: 2

Kudos [?]: 39 [0], given: 29

Re: GMAT PREP (DS) [#permalink]

### Show Tags

16 Dec 2010, 21:34
Bunuel wrote:
LM wrote:

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: $$y_l=m_1x+b_1$$ and $$y_k=m_2x+b_2$$. The question: is $$m_1*m_2<0$$?

Lines intersect at the point (4,3) --> $$3=4m_1+b_1$$ and $$3=4m_2+b_2$$

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of $$x$$ for $$y=0$$ and equals to $$x=-\frac{b}{m}$$ --> so $$(-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0$$ --> $$\frac{b_1b_2}{m_1m_2}>0$$.

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of $$y$$ for $$x=0$$ and equals to $$x=b$$ --> $$b_1*b_2<0$$.

(1)+(2) $$\frac{b_1b_2}{m_1m_2}>0$$ and $$b_1*b_2<0$$. As numerator in $$\frac{b_1b_2}{m_1m_2}>0$$ is negative, then denominator $$m_1m_2$$ must also be negative. So $$m_1m_2<0$$. Sufficient.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.

Hey Bunnel.. Good Explanation..
Manager
Status: Preparing for GMAT - March 2011
Joined: 21 May 2010
Posts: 149
Location: London
Schools: INSEAD, RSM, HEC, St. Gallen, IF, IESE
WE 1: Finance 6 years
Followers: 5

Kudos [?]: 18 [0], given: 0

Re: GMAT PREP (DS) [#permalink]

### Show Tags

17 Dec 2010, 03:31
Brunel Thanks a lot for your explanation
_________________

In The World Full Of Duplicates,, I Am The Only Masterpiece..
Girl Power
http://gmatclub.com/forum/beat-the-beast-with-non-native-speaker-108349.html

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 161

Kudos [?]: 1705 [4] , given: 376

Re: GMAT PREP (DS) [#permalink]

### Show Tags

27 Apr 2011, 10:54
4
KUDOS
1
This post was
BOOKMARKED
In the xy-coordinate plant, line t and k intersect at (4,3). Is the product of their slopes negative?
1. Product of the x intercepts of the line t & k is positive
2. Product of the y intercepts of the line t & k is negative

Sol:
To draw a line with +ve slope through a point(x1, y1) on a 2D plane; draw two lines passing through the point perpendicular to each other with one line parallel to x-axis and the other parallel to y-axis(depicted by red lines in the images). These red-lines make 4-quadrants with origin(x1,y1). Any line that passes through this origin and lies on 1st and 3rd quadrant will have +ve slope. Any line passing through the point and lying on the 2nd and 4th quadrants will have negative slope.

Attached images show two possible scenarios for St1 and St2 making the statements insufficient individually. Last image shows the only possible scenario for the lines, making both the statements sufficient together.

St1:
Insufficient.
Attachment:

product_slopes_of_two_lines_3_n_4_St1.PNG [ 14.96 KiB | Viewed 8319 times ]

St2:
Insufficient.
Attachment:

product_slopes_of_two_lines_3_n_4_St2.PNG [ 15.7 KiB | Viewed 8316 times ]

Combined:
Sufficient.
Attachment:

product_slopes_of_two_lines_3_n_4_Combined.PNG [ 12.94 KiB | Viewed 8296 times ]

Ans: "C"
_________________
SVP
Joined: 16 Nov 2010
Posts: 1672
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 33

Kudos [?]: 514 [0], given: 36

Re: GMAT PREP (DS) [#permalink]

### Show Tags

27 Apr 2011, 19:13
(1)

So they both have +ve x-intercept or -ve x=intercept

This is not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have -ve and 1 +ve slope

(2)

So they both intersect y-axis on different sides of X-Axis

This is also not sufficient, as can be seen from diagram, both can have +ve slope or 1 can have -ve and 1 +ve slope

(1) and (2)

They are sufficient, as only the option with two lines on right of y-axis is possible.

Hence products of slopes is negative.

Attachments

Lines.JPG [ 16.69 KiB | Viewed 7515 times ]

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 30 Nov 2010
Posts: 47
Followers: 0

Kudos [?]: 10 [0], given: 27

Re: GMAT PREP (DS) [#permalink]

### Show Tags

01 May 2011, 10:41
In this problem description, why was the point (4,3) given?
Since, all we need from y1 = m1x1 + c1 and y2=m2x2 + c2 - were the x & y intercepts.
Senior Manager
Joined: 28 Dec 2010
Posts: 334
Location: India
Followers: 1

Kudos [?]: 201 [0], given: 33

Re: In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

25 May 2012, 11:10
picked C. Logic was that only x intercept or y intercept cannot define slope (since they are essentially just one point and infinite lines can be drawn through a point) you need tow point s to define a line. so x when x and y both intercepts are given, then a line can be defined. Since we now know the direction of the rise/fall of the line. we get the answer.
Manager
Joined: 06 Apr 2012
Posts: 51
Followers: 1

Kudos [?]: 10 [0], given: 25

Re: GMAT PREP (DS) [#permalink]

### Show Tags

26 May 2012, 00:11
Bunuel wrote:
LM wrote:

In the xy coordinate plane, line L and line K intersect at the point (4,3). Is the product of their slopes negative?

We have two lines: $$y_l=m_1x+b_1$$ and $$y_k=m_2x+b_2$$. The question: is $$m_1*m_2<0$$?

Lines intersect at the point (4,3) --> $$3=4m_1+b_1$$ and $$3=4m_2+b_2$$

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of $$x$$ for $$y=0$$ and equals to $$x=-\frac{b}{m}$$ --> so $$(-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0$$ --> $$\frac{b_1b_2}{m_1m_2}>0$$.

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of $$y$$ for $$x=0$$ and equals to $$x=b$$ --> $$b_1*b_2<0$$.

(1)+(2) $$\frac{b_1b_2}{m_1m_2}>0$$ and $$b_1*b_2<0$$. As numerator in $$\frac{b_1b_2}{m_1m_2}>0$$ is negative, then denominator $$m_1m_2$$ must also be negative. So $$m_1m_2<0$$. Sufficient.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.

P.S. This question can be easily solved by drawing the lines without any calculations.

great explanation....

i just plotted the lines on a coordinate figure....

st 1: product of x-intercepts are positive, so from point (4,3) in first quadrant we could have two possibilities:
a) 1st line - negative slope and a positive x-intercept and 2nd line - positive slope and positive x-intercept (y-intercepts positive - line 1 and negative = line 2)
b) 1st line - positive slope and a positive x-intercept and 2nd line - positive slope and positive x-intercept (y-intercepts negative)

hence 2 cases are possible... not sufficient

st 2: approach the statement with the same philosophy as in st 1 and arrived at two cases,
a) 1st line - negative slope and a positive y-intercept and 2nd line - positive slope and negative y-intercept (x-intercepts positive)
b) 1st line - positive slope and a negative y-intercept and 2nd line - positive slope and positive y-intercept (x-intercepts positive - line 1 and negative - line 2)

hence 2 cases are possible... not sufficient

from both the statements, we can arrive at a unique case case a) from st 1 and case a) from st 2

answer therefore is C

the cases look very much conceivable on a graph plot.
Intern
Joined: 20 Apr 2013
Posts: 24
Concentration: Finance, Finance
GMAT Date: 06-03-2013
GPA: 3.3
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 2 [0], given: 99

In the xy coordinate plane, line L and line K intersect at the p [#permalink]

### Show Tags

10 May 2013, 11:58
M8 wrote:
In the xy-coordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope 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.

I cracked this question right.
But these types of questions are real horror for me.

Guys please elaborate your solution for me. Thank you.

I'll post the OA later.

Let X1 and X2 are the X intercepts for the lines l and K.

Calculate slope for L [(4,3) and (X1, 0)] and m [ (4,3) and (X2, 0)] such that X1*X2=+VE

Similarly take Y1 and Y2. and calculate slope for L [(4,3) and (0, Y1)] and m [ (4,3) and (0, Y2)] such that Y1*Y2=-VE

Now combine these two statement and calculate slope for L [(0, Y1) and (X1, 0)] and m [ (0, Y2) and (X2, 0)]

Slope of L*M = Y1/(-X1)*Y2/(-X2) = Y1*Y2/ X1*X2.

Given that Y1*Y2= -VE and X1*X2=+VE.

Hence Slope of L*M = -VE

Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7079

Kudos [?]: 93177 [0], given: 10553

Re: In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

27 Jun 2013, 21:28
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Coordinate Geometry: math-coordinate-geometry-87652.html

All DS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=41
All PS Coordinate Geometry Problems to practice: search.php?search_id=tag&tag_id=62

_________________
Current Student
Joined: 19 Dec 2011
Posts: 12
Concentration: Healthcare, Operations
Schools: Olin '17 (M)
GPA: 3.24
WE: Supply Chain Management (Military & Defense)
Followers: 0

Kudos [?]: 7 [0], given: 54

Re: In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

08 Nov 2014, 20:49
Hi Bunel,

You wrote:

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of y for x=0 and equals to x=b --> b_1*b_2<0.

Don't you mean y=b? y=mx+b --> y=m(0)+b --> y=b

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7079

Kudos [?]: 93177 [0], given: 10553

Re: In the xy coordinate plane, line L and line K intersect at [#permalink]

### Show Tags

09 Nov 2014, 04:44
jrawls wrote:
Hi Bunel,

You wrote:

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of y for x=0 and equals to x=b --> b_1*b_2<0.

Don't you mean y=b? y=mx+b --> y=m(0)+b --> y=b

Thanks

Yes, there was a typo x instead of y. Edited. Thank you.
_________________
Re: In the xy coordinate plane, line L and line K intersect at   [#permalink] 09 Nov 2014, 04:44

Go to page    1   2    Next  [ 23 posts ]

Similar topics Replies Last post
Similar
Topics:
In the xy-coordinate plane, line L 1 and line L 2 intersect at the 1 05 Mar 2016, 11:59
35 Lines k and l intersect in the coordinate plane at point (3, 19 30 Nov 2012, 14:57
8 In the xy-plane, line l and line k intersect at the point 6 26 Sep 2010, 10:49
In the xy-plane, line l and line k intersect at the point 2 05 Jan 2010, 07:00
15 In the xy-coordinate plane, line l and line k intersect at 12 18 Feb 2008, 18:01
Display posts from previous: Sort by

# In the xy coordinate plane, line L and line K intersect at

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.