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In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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12 May 2006, 05:22
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In the xycoordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative? (1) The product of the xintercepts of lines l and k is positive. (2) The product of the yintercepts of lines l and k is negative.
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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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12 May 2006, 11:38
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 = (y2y1)/(x2x1)
We can assume (x2, y2) as (4,3)
From (1) either both xintercepts 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 yintercept 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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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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12 May 2006, 15:53
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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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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12 May 2006, 20:01
Just thought of taking diagramatic approach for this & getting "C". Looks like following is the only possible solution.
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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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15 May 2006, 12:41
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 = yintercept / xintercept.
We need slope(l)*slope(k)
slope(l)*slope(k) = ((yintercept of l) / xintercept of l) * ((yintercept of k / xintercept of k)
= (yintercept of l * yintercept of k) / (xintercept of l * yintercept of x)
Using 1) and 2),
= (ve) (+ve)
= ve



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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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07 Sep 2010, 07:22
M8 wrote: In the xycoordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative? 1) The product of the xintercepts of lines l and k is positive. 2) The product of the yintercepts 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. 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 xintersects of lines L and K is positive. Now, one of the lines can intersect xaxis 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 xaxis 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: xintersect 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 yintersects of lines L and K is negative. Now, one of the lines can intersect yaxis 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 yaxis 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: yintercept 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: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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07 Sep 2010, 20:23
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 Xaxis are acute and obtuse. Tan of such angles are opposite in sign. Hence the answer.
Thank you. Hemanth



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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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07 Oct 2010, 17:30
satishreddy wrote: hi guys, below are few DS questions i had trouble answering,,,,,,,,,,need help.....thanks in advance 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. xintercept of line l, that is when y=0=> x1=c1/m1 and xintercept 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. yintercept of line l, that is when x=0=> y1=c1 and yintercept 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: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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08 Oct 2010, 08:05
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 xycoordinate plane, line l and line k intersect at the point [#permalink]
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22 Dec 2011, 07:13
Line L and Line K intersect at the point (4,3). Is the product of their slopes negative?
(1) The product of the xintercepts 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 yintercepts 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. yintercept (where b1*b2<0, Log: Combination A2 and B2 Log: From A2 and B2, select only pos. xintercept (b1*b2)/(m1*m2) > 0 Log: A2
This restriction is made more cogent by noting that a line with a negative yintercept and destination (4,3) must cross the xaxis in positive territory and have a positive slope.
Since the other line must have a positive yintercept and is restricted to a positive xintercept, 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 xycoordinate plane, line l and line k intersect at the point [#permalink]
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10 Jun 2015, 02:49
very hard question on gmatprep. this question can be done by drawing the lines and see that C is correct.
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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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23 Jun 2015, 07:02
thangvietnam wrote: very hard question on gmatprep.
this question can be done by drawing the lines and see that C is correct. Yes, solving by drawing is much easier in this question



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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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13 Nov 2015, 01:46
Bunuel wrote: M8 wrote: In the xycoordinate plane, line l and line k intersect at the point (4,3). Is the product of their slope negative? 1) The product of the xintercepts of lines l and k is positive. 2) The product of the yintercepts 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. 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 xintersects of lines L and K is positive. Now, one of the lines can intersect xaxis 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 xaxis 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: xintersect 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 yintersects of lines L and K is negative. Now, one of the lines can intersect yaxis 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 yaxis 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: yintercept 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. In the above explanation 'But from this statement we can deduce the following: yintercept is value of \(y\) for \(x=0\) and equals to \(x=b\) ' I don't understand how x=b. I think it would be y = b. Am I right?
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Re: In the xycoordinate plane, line l and line k intersect at the point [#permalink]
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16 Oct 2017, 02:44
so, we do not need the information that the two lines intersect at (4,3)
is that right?
the question is not good. right.
redundant information?




Re: In the xycoordinate plane, line l and line k intersect at the point
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