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

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In the xy coordinate plane, line L and line K intersect at [#permalink] New post 06 May 2010, 11:31
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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
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Re: GMAT PREP (DS) [#permalink] New post 06 May 2010, 12:24
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LM wrote:
Please explain...


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.

Answer: C.

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.
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Re: GMAT PREP (DS) [#permalink] New post 06 May 2010, 12:28
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Re: Interesting Line Eqn DS problem [#permalink] New post 19 Jul 2010, 07:29
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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.
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Re: GMAT PREP (DS) [#permalink] New post 25 Sep 2010, 11:27
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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.
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Re: GMAT PREP (DS) [#permalink] New post 27 Sep 2010, 07:39
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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.
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Re: GMAT PREP (DS) [#permalink] New post 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!!!
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Re: GMAT PREP (DS) [#permalink] New post 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
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Re: GMAT PREP (DS) [#permalink] New post 16 Dec 2010, 08:26
Bunuel wrote:
LM wrote:
Please explain...


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.

Answer: C.

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
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Re: GMAT PREP (DS) [#permalink] New post 16 Dec 2010, 21:34
Bunuel wrote:
LM wrote:
Please explain...


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.

Answer: C.

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..
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Re: GMAT PREP (DS) [#permalink] New post 17 Dec 2010, 03:31
Brunel Thanks a lot for your explanation
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Re: GMAT PREP (DS) [#permalink] New post 27 Apr 2011, 10:54
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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
product_slopes_of_two_lines_3_n_4_St1.PNG [ 14.96 KiB | Viewed 4031 times ]



St2:
Insufficient.
Attachment:
product_slopes_of_two_lines_3_n_4_St2.PNG
product_slopes_of_two_lines_3_n_4_St2.PNG [ 15.7 KiB | Viewed 4034 times ]


Combined:
Sufficient.
Attachment:
product_slopes_of_two_lines_3_n_4_Combined.PNG
product_slopes_of_two_lines_3_n_4_Combined.PNG [ 12.94 KiB | Viewed 4012 times ]


Ans: "C"
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Re: GMAT PREP (DS) [#permalink] New post 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.

Answer - C
Attachments

Lines.JPG
Lines.JPG [ 16.69 KiB | Viewed 3236 times ]


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Re: GMAT PREP (DS) [#permalink] New post 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.
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Re: In the xy coordinate plane, line L and line K intersect at [#permalink] New post 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.
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Re: GMAT PREP (DS) [#permalink] New post 26 May 2012, 00:11
Bunuel wrote:
LM wrote:
Please explain...


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.

Answer: C.

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.
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In the xy coordinate plane, line L and line K intersect at the p [#permalink] New post 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. :wink:
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

Answer C.
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Re: In the xy coordinate plane, line L and line K intersect at [#permalink] New post 27 Jun 2013, 21:28
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Re: In the xy coordinate plane, line L and line K intersect at   [#permalink] 27 Jun 2013, 21:28
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