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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.

Last edited by mrsmarthi on 05 Mar 2009, 20:30, edited 2 times in total.

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.

(1) Pick a point in quadrant I. (4,3) is here just to distract you. Start drawing the two lines that passes through that point. See if you can draw two lines that has same sign of slope and satisfies condition (1). The trivial answer would be L, K being the same line. Try to draw another situation where the slopes of L,K are off different signs. This is definitely feasible. Hence, not sufficient.

(2) Use the same drawing technique and see if you can find 2 different scenarios while satisfying condition (2). And yes you can.

When combined, you still can try to draw the two lines. But you'll realized the case when the two x-intercepts are negative is not possible any more. The only possible way to draw the lines is when their slopes are of different signs.

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.

Line L: y=m1x+c1 Line K: y=m2x+c2

question : is m1m2 is negative? (1) prod. x intercepts = c1c2>0 not sufficient

(2) prod. of y intercepts = -c1/m1 * -c2/m2 = c1c2/m1m2<0 not sufficient

combined clearly m1m2<0

sufficient

I guess u have made a mistake by taking c1 & c2 as the x-intercept. to find out the x-intercept either put y=0 in the equation of line or remember the formula as, if y = mx + c. then x-interrcept is -c/m.

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.

Line L: y=m1x+c1 Line K: y=m2x+c2

question : is m1m2 is negative? (1) prod. x intercepts = c1c2>0 not sufficient

(2) prod. of y intercepts = -c1/m1 * -c2/m2 = c1c2/m1m2<0 not sufficient

combined clearly m1m2<0

sufficient

I guess u have made a mistake by taking c1 & c2 as the x-intercept. to find out the x-intercept either put y=0 in the equation of line or remember the formula as, if y = mx + c. then x-interrcept is -c/m.

Similarly for the y-intercept.

you are right!! its flipped.

Thanks for pointing out.
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Re: In the xy-coordinate plane, line L and line K intersect at [#permalink]

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14 Nov 2016, 23:39

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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