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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]

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06 May 2010, 12:31
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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, 12:49, edited 1 time in total.
Edited the question and added the OA

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

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06 May 2010, 13:24
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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.
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06 May 2010, 13:28
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Bunuel, thanks a lot.

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19 Jul 2010, 08:29
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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.
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25 Sep 2010, 12:27
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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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27 Sep 2010, 08:39
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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.
Attachments

intersecting lines K and L.docx [11.19 KiB]

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28 Sep 2010, 14:47
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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14 Oct 2010, 19:29
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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16 Dec 2010, 09:26
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.

This all makes sense to me now.

Thanks a lot Bunuel
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16 Dec 2010, 22:34
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..

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17 Dec 2010, 04:31
Brunel Thanks a lot for your explanation
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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 [ 14.96 KiB | Viewed 9712 times ]

St2:
Insufficient.
Attachment:

product_slopes_of_two_lines_3_n_4_St2.PNG [ 15.7 KiB | Viewed 9705 times ]

Combined:
Sufficient.
Attachment:

product_slopes_of_two_lines_3_n_4_Combined.PNG [ 12.94 KiB | Viewed 9682 times ]

Ans: "C"
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27 Apr 2011, 20:13
(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 8892 times ]

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01 May 2011, 11:41
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]

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25 May 2012, 12:10
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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26 May 2012, 01:11
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

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]

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10 May 2013, 12:58
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.

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

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

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27 Jun 2013, 22:28
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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

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25 Nov 2014, 08:01
shrouded1 wrote:
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.

Did this way using the intercept form of a line. Have made an observation that using intercept equations in gmat questions significantly reduces the time to solve a question.
Thanks.

A thumb rule I use in co-ordinate geometry for lines:
1. Use intercept form x/a + y/b = 1 - If intercepts are given, for area questions usually, for ladder against wall type questions, triangle questions
2. Use slope form y=mx+c - when slope is implicit or explicitly stated in the question
3. Use one point slope form y-y1=m(x-x1) - when we know slope and line passes through a point
4. Use 2 point form y-y1/y2-y1 = x-x1/x2-x1 - when line passes through two points
5. When all else fails we resort to pure algebra of a linear equation ax+by+c=0
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Re: In the xy coordinate plane, line L and line K intersect at [#permalink]

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13 Nov 2016, 17:29
Can the question be solved the following way? Or, am I just lucky with the solution particular to this question?

Slope of a line = $$\frac{rise}{run}$$ = $$\frac{y-intercept}{x-intercept}$$

With this in mind,

(1) The product of the x-intercepts of lines $$l$$ and $$k$$ is positive

Translation:

$$\frac{y}{x}$$ = $$+$$ or $$\frac{-y}{x}$$ = $$-$$

Insufficient

(2) The product of the y-intercepts of lines $$l$$ and $$k$$ is negative

Translation:

$$\frac{-y}{-x}$$ = $$+$$ or $$\frac{-y}{x}$$ = $$-$$

Insufficient

(1) & (2):

$$\frac{-y}{x}$$ = $$-$$ is the only answer in common and therefore, sufficient.

Thank you!

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Re: In the xy coordinate plane, line L and line K intersect at   [#permalink] 13 Nov 2016, 17:29

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