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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 12 May 2006, 04:22
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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 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.
[Reveal] Spoiler: OA
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 [#permalink] New post 12 May 2006, 10:38
Well....its twice the horror for me man....below is the best I could come up with. If anyone can explain the solution clearly would be great. If this comes in the real thing, I'll most likely guess instead of trying to solve it.
------------------

Slope of line = (y2-y1)/(x2-x1)
We can assume (x2, y2) as (4,3)

From (1) either both x-intercepts are +ve or -ve so either (x1, 0) and (x2, 0)
If +ve, slope of L = (3 - 0)/(4 - x1) ... depends on value of x1
If -ve, slope of L = (3 - 0)/(4 - x1) ... which is positive.

Same goes for Line K.
Hence insufficient.

From (2) either one y-intercept is +ve and another is -ve so (0, y1) and (0, -y2)

Slope of L = (3 - y1)/(4 - 0) ... depends on value of y1
Slope of K = (3 - (-y2))/(4 - 0) ... which is positive

Hence insufficient

Taking both together, we still cannot say due to values of x1 and y1 unknown hence answer should be E.
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 [#permalink] New post 12 May 2006, 10:41
Here is a similar question...maybe this one will help somewhat (im going to take a look now)

http://www.gmatclub.com/phpbb/viewtopic.php?t=29364
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 [#permalink] New post 12 May 2006, 12:25
E should be the answer.

Deduced with help of rough sketch. Tried to make one with mspaint but too much hard work :roll:
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 [#permalink] New post 12 May 2006, 14:53
I'm going on answer (C)

For line L : 3 = 4*a(l) + b(l)
For line K : 3 = 4*a(k) + b(k)

1) Says us:

For L
xo(l)*a(l) + b(l) = 0
so,
xo(l)= -b(l)/ a(l)

Similarly, for k : xo(k)= -b(k)/ a(k)

xo(l) * xo(k) > 0
so, ( b(l)*b(n) ) / ( a(l)*a(n) ) > 0

Unsifficient

2) is : b(k)*b(l) < 0

Unsifficient

Combining (1) and (2), as b(k)*b(l) < 0 thus a(k)*a(l) < 0
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 [#permalink] New post 12 May 2006, 19:01
Just thought of taking diagramatic approach for this & getting "C". Looks like following is the only possible solution.
Attachments

File comment: x-y Plane
xy.JPG
xy.JPG [ 5.68 KiB | Viewed 3405 times ]

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 [#permalink] New post 15 May 2006, 02:30
Looking at vivek's plot - we don't care whether blue line intersects X on positive or negative side - so I'll for B
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 [#permalink] New post 15 May 2006, 11:41
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The question becomes much simpler if you see whether the lines passing through point (4, 3) has any bearing on the sign of the slopes -- NO.

So, it comes down to the simple slope equation:
slope = y-intercept / x-intercept.

We need slope(l)*slope(k)

slope(l)*slope(k) = ((y-intercept of l) / x-intercept of l) * ((y-intercept of k / x-intercept of k)
= (y-intercept of l * y-intercept of k) / (x-intercept of l * y-intercept of x)

Using 1) and 2),
= (-ve) (+ve)
= -ve
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Re: [#permalink] New post 07 Sep 2010, 06:13
ramk10 wrote:
The question becomes much simpler if you see whether the lines passing through point (4, 3) has any bearing on the sign of the slopes -- NO.

So, it comes down to the simple slope equation:
slope = y-intercept / x-intercept.

We need slope(l)*slope(k)

slope(l)*slope(k) = ((y-intercept of l) / x-intercept of l) * ((y-intercept of k / x-intercept of k)
= (y-intercept of l * y-intercept of k) / (x-intercept of l * y-intercept of x)

Using 1) and 2),
= (-ve) (+ve)
= -ve


oh bro you are cool, thanks kudos to you.

But in which situation we can say m=y/x

y=mx+b so why you think that b is 0?
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Re: GMATPrep DS: xy-coordinate plane [#permalink] New post 07 Sep 2010, 06:22
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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.


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.
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Re: GMATPrep DS: xy-coordinate plane [#permalink] New post 07 Sep 2010, 07:29
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: GMATPrep DS: xy-coordinate plane [#permalink] New post 07 Sep 2010, 07:52
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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: GMATPrep DS: xy-coordinate plane [#permalink] New post 07 Sep 2010, 19:23
C.

Sweet question. A kudos for you for that. If you draw the lines using conditions you will find that lines can intercept in quadrant 1 only when both the X intercepts are positive (both negative is not possible given the condition 2). When thats the case the angles each line make with the X-axis are acute and obtuse. Tan of such angles are opposite in sign. Hence the answer.

Thank you.
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Re: GMATPrep DS: xy-coordinate plane [#permalink] New post 09 Sep 2010, 06:00
whats the difficulty level of this question???? Are such type of questions asked in actual gmat??Mans are these above 50 points level qustions??
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math DS questions: need help solving [#permalink] New post 07 Oct 2010, 15:49
hi guys, below are few DS questions i had trouble answering,,,,,,,,,,need help.....thanks in advance

:lol:
159) In xy coordinate plane, line l and line k intersect at point (4, 3). Is product of the slopes negative? C
a. Product of x intercepts of lines l and k is positive
b. Product of y intercepts of lines l and k is negative

OA: C
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Re: math DS questions: need help solving [#permalink] New post 07 Oct 2010, 16:30
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hi guys, below are few DS questions i had trouble answering,,,,,,,,,,need help.....thanks in advance

:lol:
159) In xy coordinate plane, line l and line k intersect at point (4, 3). Is product of the slopes negative? C
a. Product of x intercepts of lines l and k is positive
b. Product of y intercepts of lines l and k is negative

OA: C


Question: Product of the slope of line l and k -ve.

Let line l be y1= m1x1+c1 and k be y2 = m2x2+c2. General equation of the line => y = mx+c.

Hence to determine whether m1*m2 = -ve

Now x intercept is when y=0 and y intercept is when x=0.

Statement A: Product of x intercepts of line l and k is +ve.

x-intercept of line l, that is when y=0=> x1=-c1/m1 and x-intercept of line k is x2=-c2/m2.

Given their product is +ve. Now -c1/m1 * -c2/m2 is +ve. Either c1*c2 and m1*m2 could be both +ve or both -ve. We cannot determine the whether m1*m2 is +ve or -ve since we do not have information about c1*c2.

Statement B: Product of y intercepts of line l and k is -ve.

y-intercept of line l, that is when x=0=> y1=c1 and y-intercept of line k is y2=c2.

Given their product is -ve that is c1*c2=-ve. However with this statement alone we cannot determine whether m1*m2=-ve or +ve.

Combining both the statement we know that since c1*c2 = -ve, m1*m2 has to be -ve to satisfy Statement A.

Hence answer is C.
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Re: math DS questions: need help solving [#permalink] New post 07 Oct 2010, 23:27
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Re: math DS questions: need help solving [#permalink] New post 08 Oct 2010, 07:05
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Equation for a line in intercept form, X/A+Y/B=1 and slope will be m=-(B/A)

Now take the equation of first line as X/A1+Y/B1=1 and second line as X/A2+Y/B2=1

Product of their slope will be (B1/A1)*(B2/A2)=(B1*B2)/(A1*A2)

with first (A1 *A2) positive
With 2nd (B1*B2) -ve

So with both we can say Product of their slope is -ve. that is C.

Consider KUDOS if you like my solution.
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Re: In the xy-coordinate plane, line l and line k intersect at [#permalink] New post 22 Dec 2011, 06:13
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 Line L and Line K is positive.

Combination A1 m b

(-2, 0) (4,3) y = 1/2x + 1
(-3, 0) (4,3) y = 3/7x + 9/7

Combination B1 m b

(1, 0) (4,3) y = x - 1
(6, 0) (4,3) y = -3/2x + 9

(2) The product of the y-intercepts of Line L and Line K is negative.

Combination A2 m b

(0, 6) (4,3) y = -3/4x + 6 (8,0)
(0, -2) (4,3) y = 5/4x - 2 (8/5, 0)

Combination B2 m b

(0,-2) (4,3) y = 5/4x - 2 (8/5,0)
(0, 2) (4,3) y = 1/4x + 2 (-8, 0)

(C) Log: From S(2) result, select only neg.
y-intercept (where b1*b2<0,
Log: Combination A2 and B2
Log: From A2 and B2, select only pos.
x-intercept (b1*b2)/(m1*m2) > 0
Log: A2

This restriction is made more cogent by noting that a line with a negative y-intercept and destination (4,3) must cross the x-axis in positive territory and have a positive slope.

Since the other line must have a positive y-intercept and is restricted to a positive x-intercept, its slope
must be negative.

Both statements combined are sufficient.

Algebraic Solution

Is m1*m2 < 0


(1) -(b1/m1) * -(b2/m2) > 0

(b1*b2)/(m1*m2) > 0

(2) b1*b2 < 0

(1)&(2)

b1*b2/m1*m2 > 0, and b1*b2 < 0

m1*m2 < 0
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Re: In the xy-coordinate plane, line l and line k intersect at [#permalink] New post 29 Nov 2013, 21:41
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Re: In the xy-coordinate plane, line l and line k intersect at   [#permalink] 29 Nov 2013, 21:41
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