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Re: GMAT PREP (DS) [#permalink]
06 May 2010, 12:24

5

This post received KUDOS

Expert's post

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. _________________

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. _________________

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!!!

Re: GMAT PREP (DS) [#permalink]
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 _________________

Re: GMAT PREP (DS) [#permalink]
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.

Re: GMAT PREP (DS) [#permalink]
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.

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 4031 times ]

St2: Insufficient.

Attachment:

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 [ 12.94 KiB | Viewed 4012 times ]

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

Re: GMAT PREP (DS) [#permalink]
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.