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In the xy-coordinate plane, line L and line k intersect at the point

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In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   18 Feb 2008, 19:01
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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.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   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.
SUFFICIENT. Answer is C.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   06 May 2010, 13:24
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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?

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.

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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In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   23 Jun 2008, 14:17
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alimad wrote:
In the xy-coordinate plane, line L and line K intersect at point (4,3), Is the product of their slopes negative?

(1). The product of the x-intercept of line L and K is positive
(2). The product of the y-intercept of line L and K is negative.

Please provide some explaination. Thanks


Answer is C.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   19 Feb 2008, 23:15
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C

To make using this editor easy, I am naming lines as 1 and 2
and thier x and y intercepts as x1, x2, y1 and y2

slope of line 1 = -y1/x1
slope of line 2 = -y2/x2

product of slopes = (y1*y2)/(x1*x2)
since numerator is -ve and denominator is +ve, product of slopes is -ve

What is OA?
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   Updated on: 05 Mar 2009, 21:30
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It is given that lines l and k passes thru the point (4,3). Let m1 and b1 be the slope and y intercept of line l and m2 and b2 be the slope and y intercept of k. So we know, line l will be 3 = 4m1 + b1 and line k will be 3 = 4m2 + b2(slope intercept form equation replacing (x,y) with (4,3)). The question being asked is "Is product of the slopes negative ? ".

From stmt 1 : - Product of x intercepts is positive ==> we know that x intercept in general is -b/m. So it is given that
(-b1/m1) (-b2/m2) > 0 ==> b1b2/m1m2 > 0. Here we can understand that product of slopes and product of y intercetps should be of the same sign. But we can't say whether the product is positive or negative. So insufficient.

From stmt 2 : - product of y intercepts is negative ==> b1 * b2 < 0. But nothing is mentioned abt the slopes. So Insufficient.

Now combine both the clues -

From stmt1, product of x intercepts and product of slopes have the same sign.
From stmt2, product of y intercepts is negative.

Now we can conclude that product of slopes is negative. Because from stmt 1, we know that both the products are having the same sign.

IMO C.

Originally posted by mrsmarthi on 05 Mar 2009, 21:05.
Last edited by mrsmarthi on 05 Mar 2009, 21:30, edited 2 times in total.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   25 Jul 2009, 13:44
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Statment (1) is not sufficient ....plz refer image attached......possibility 2 will lead us to -ve product where as other two are +ve products of slopes....hence insufficient

Statment (2) Sufficient ...see image attached

Hence B....lemme know if its right


If u think this is right....Kudos plz...trying to access tests :wink:
Attachments

Stat 2.GIF
Stat 2.GIF [ 5.32 KiB | Viewed 25382 times ]

Stat 1.GIF
Stat 1.GIF [ 8.88 KiB | Viewed 25393 times ]

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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   25 Jul 2009, 22:38
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Thanks for your post. You have explained really well, but unfortunately OA is C. Anyway kudos for such a great explanation :) , after going through your post I was able to solve the problem.

Just posting my explanation for your reference:
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   07 Sep 2010, 08:29
Bunuel wrote:
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.


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.



Bunuel i just dont understand "But from this statement we can deduce the following: x-intersect is value of for " this part rest of them are ok. Can you tell me how did you decide y=0?
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   07 Sep 2010, 08:52
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fatihaysu wrote:
Bunuel i just dont understand "But from this statement we can deduce the following: x-intersect is value of for " this part rest of them are ok. Can you tell me how did you decide y=0?


X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point \((x,0)\) - the value of \(x\) when \(y=0\): \(y=mx+b\) --> \(0=mx+b\) --> \(x=-\frac{b}{m}\). So X-intercept of a line \(y=mx+b\) is \(x=-\frac{b}{m}\);

Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point \((0,y)\) - the value of \(y\) when \(x=0\): \(y=mx+b\) --> \(y=m*0+b\) --> \(y=b\). So Y-intercept of a line \(y=mx+b\) is \(y=b\).

Check Coordinate Geometry chapter for more (link in my signature).

Hope it's clear.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   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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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   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.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   27 Apr 2011, 11:54
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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 37703 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 37697 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 37625 times ]


Ans: "C"
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   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.

Answer - C
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   07 Jun 2011, 11:14
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bsjames2 wrote:
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-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.


Please explain your answer


This question has a great takeaway - something I am sure you know intuitively but you may not think of it while doing this question because it is seldom written out:
The slope of a line is -(y intercept)/(x intercept)

Above, sreehari uses this concept to solve the question very efficiently.

If you are wondering why it is so, think what 'intercept' represents...
The point of x intercept is (x, 0) (where y co-ordinate is 0)
The point of y intercept is (0, y) (where x co-ordinate is 0)

So slope = (y - 0)/(0 - x) = -y/x
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   18 May 2014, 07:06
VeritasPrepKarishma wrote:
bsjames2 wrote:
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-intercepts of lines l and k is positive.
2) The product of the y-intercepts of lines l and k is negative.


Please explain your answer


This question has a great takeaway - something I am sure you know intuitively but you may not think of it while doing this question because it is seldom written out:
The slope of a line is -(y intercept)/(x intercept)

Above, sreehari uses this concept to solve the question very efficiently.

If you are wondering why it is so, think what 'intercept' represents...
The point of x intercept is (x, 0) (where y co-ordinate is 0)
The point of y intercept is (0, y) (where x co-ordinate is 0)

So slope = (y - 0)/(0 - x) = -y/x



Hi karishma

Slope of line 1 is -y1/x1
slope of line 2 is -y2/x2

if we multiply both the slopes
wont the product be positive??

Kindly help
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   18 May 2014, 21:24
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adityagogia9899 wrote:
Hi karishma

Slope of line 1 is -y1/x1
slope of line 2 is -y2/x2

if we multiply both the slopes
wont the product be positive??

Kindly help


Remember that y1, x1, y2 and x2 are all variables. They could be positive or negative.

Product of slopes = (y1/x1)*(y2/x2) = (y1*y2)/(x1*x2)
The product depends on whether y1, x1, x2, y2 are positive or negative. The statements tell you that x1*x2 is positive and y1*y2 is negative. So product of slopes = negative/positive which is negative.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   02 Jun 2015, 03:13
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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.

.


Question: Is the product of slopes of line L and K negative?

Requirement: For the product of the slopes to be negative one of the lines must have positive slope (Sloping upward from left to right) and one of the lines must have Negative slope (Sloping Downward from left to right)

Statement 1: The product of the x-intercepts of lines l and k is positive.

With positive X intercept of both the lines, both the lines may have positive slope
OR
With positive X intercept of both the lines, One of the lines may have positive slope and other line may have negative slope therefore NOT SUFFICIENT


Statement 2: The product of the y-intercepts of lines l and k is negative.

With positive Y intercept of one of the lines and Negative Y intercept of other line, both the lines may have positive slope
OR
With positive Y intercept of one of the lines and Negative Y intercept of other line, One of the lines may have positive slope and other line may have negative slope therefore NOT SUFFICIENT

Combining the two Statements:

Case 1: Both Lines have positive X intercepts (For product of therefore X intecepts to be (+ve) )
One of the lines will have positive Y intercept and one Line will have Negative Y intercept

The line with (+ve) X-intecept and (+ve)Y intercept will bound to have (-ve)Slope
and
The line with (+ve) X-intecept and (-ve)Y intercept will bound to have (+ve)Slope
Hence Product of slopes will be Negative


Case 2: Both Lines have Negative X intercepts (For product of therefore X intecepts to be (+ve) )
One of the lines will have positive Y intercept and one Line will have Negative Y intercept

The line with (-ve) X-intecept and (+ve) Y intercept will bound to have (+ve)Slope
and
The line with (-ve) X-intecept and (-ve) Y intercept will bound to have (-ve)Slope
Hence Product of slopes will be Negative

Consistent answer
therefore, SUFFICIENT

Answer: Option
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C
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   09 Jun 2016, 23:26
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If we modify the original condition and the question, the equation for a line that crosses (2,1) and (4,3) is y=x-1. Then, if B(m,n), n=m-1. There are, 2 variables (m and n) and 1 equation (n=m-1). In order to match the number of variables to the number of equations, we need 1 equation. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that D is the correct answer.
Since 1)=2) (m,n)=(3,2), the answer is yes and the conditions are sufficient. The correct answer is D.
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink] New post   28 Aug 2017, 08:24
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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.


Always draw the figure to solve such questions

Answer option C
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Re: In the xy-coordinate plane, line L and line k intersect at the point [#permalink]
28 Aug 2017, 08:24
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