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GMATBLACKBELT
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line L and k intersect at point 4,3. Is the product of their 08 Sep 2007, 22:20
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line L and k intersect at point 4,3. Is the product of their slopes negative?


1) The product of x-intercepts of the lines is positive

2) The product of the y-intercepts of the lines is negative.




I'm stumped why the correct answer is not E



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line L and k intersect at point 4,3. Is the product of their 08 Sep 2007, 23:48
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?

1) The product of x-intercepts of the lines is positive
2) The product of the y-intercepts of the lines is negative.
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E.
1. x of lines could be both -ves or both +ves. similarly the slopes could be both -ves and +ves. nsf.
2. if y intercept of one line is -ve, the the same of the other line is +ve and vice versa. again the slops could be both -ves, and +ves. so nsf.
togather also not sufficient.
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 11:16
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?


1) The product of x-intercepts of the lines is positive

2) The product of the y-intercepts of the lines is negative.




I'm stumped why the correct answer is not E
Show more


Yes it is C. Look inside!!!
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Solution.doc [21 KiB]
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 12:11
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?

1) The product of x-intercepts of the lines is positive
2) The product of the y-intercepts of the lines is negative.

E.
1. x of lines could be both -ves or both +ves. similarly the slopes could be both -ves and +ves. nsf.
2. if y intercept of one line is -ve, the the same of the other line is +ve and vice versa. again the slops could be both -ves, and +ves. so nsf.
togather also not sufficient.
Show more


if the coordinates of Lines L and K are (-2, 1) and (-1, -15) respectively, their slopes are:
L = (1-3)/(-2-4) = -2/-6 = 1/3
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is +ve.

again, if the coordinates of Lines L and K are (-2, 15) and (-1, -15) respectively, their slopes are:
L = (15-3)/(-2-4) = 12/-6 = -2
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is -ve.


again, if the coordinates of Lines L and K are (2, 1) and (1, -15) respectively, their slopes are:
L = (1-3)/(2-4) = -2/-2 = 1
K = (-15-3)/(1-4) = -18/-3 = 6
their product is +ve.

again, if the coordinates of Lines L and K are (2, 15) and (1, -15) respectively, their slopes are:
L = (15-3)/(2-4) = 12/-2 = -6
K = (-15-3)/(1-4) = -18/-3 = 6
their product is -ve.

you have -ve and +ve as the product of the slopes. So how it is C?
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 12:18
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?


1) The product of x-intercepts of the lines is positive

2) The product of the y-intercepts of the lines is negative.




I'm stumped why the correct answer is not E
Show more


C

Say we have:
Y1 = M1*X1 + B1
Y2 = M2*X2 + B2

(1) We have
(-B1/M1) * (-B2/M2) > 0
We don't know the values of B, so INSUFFICIENT

(2) B1*B2<0
Don't know anything about M, so INSUFFICIENT

Together, we know that B1*B2 < 0
It must be the case that either B1 or B2 are negative.

Now, using what we have from (1):
(-B1/M1) * (-B2/M2) > 0
Take out negative sign:
(B1*B2)/(M1*M2) > 0
Knowing that B1*B2 is negative
This means that M1*M2 must also be negative.
SUFFICIENT.
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 12:39
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?


1) The product of x-intercepts of the lines is positive

2) The product of the y-intercepts of the lines is negative.




I'm stumped why the correct answer is not E

C

Say we have:
Y1 = M1*X1 + B1
Y2 = M2*X2 + B2

(1) We have
(-B1/M1) * (-B2/M2) > 0
We don't know the values of B, so INSUFFICIENT

(2) B1*B2<0
Don't know anything about M, so INSUFFICIENT

Together, we know that B1*B2 <0> 0
Take out negative sign:
(B1*B2)/(M1*M2) > 0
Knowing that B1*B2 is negative
This means that M1*M2 must also be negative.
SUFFICIENT.
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Perfect :)
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 19:15
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GMATBLACKBELT
line L and k intersect at point 4,3. Is the product of their slopes negative?

1) The product of x-intercepts of the lines is positive
2) The product of the y-intercepts of the lines is negative.

E.
1. x of lines could be both -ves or both +ves. similarly the slopes could be both -ves and +ves. nsf.
2. if y intercept of one line is -ve, the the same of the other line is +ve and vice versa. again the slops could be both -ves, and +ves. so nsf.
togather also not sufficient.

if the coordinates of Lines L and K are (-2, 1) and (-1, -15) respectively, their slopes are:
L = (1-3)/(-2-4) = -2/-6 = 1/3
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is +ve.

again, if the coordinates of Lines L and K are (-2, 15) and (-1, -15) respectively, their slopes are:
L = (15-3)/(-2-4) = 12/-6 = -2
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is -ve.


again, if the coordinates of Lines L and K are (2, 1) and (1, -15) respectively, their slopes are:
L = (1-3)/(2-4) = -2/-2 = 1
K = (-15-3)/(1-4) = -18/-3 = 6
their product is +ve.

again, if the coordinates of Lines L and K are (2, 15) and (1, -15) respectively, their slopes are:
L = (15-3)/(2-4) = 12/-2 = -6
K = (-15-3)/(1-4) = -18/-3 = 6
their product is -ve.

you have -ve and +ve as the product of the slopes. So how it is C?
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Bud, if I understand you correctly, I think you are misinterpreting the stem here. (1) says that the product of the x-intercepts are positive. (2) says that the product of the y-intercepts are negative.
x-intercepts & y-intercepts aren't the slope.

Correct me if I am wrong. Otherwise, please show the multiplication of each x-y intercepts so I can understand this better.
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line L and k intersect at point 4,3. Is the product of their 09 Sep 2007, 19:31
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Bud, if I understand you correctly, I think you are misinterpreting the stem here. (1) says that the product of the x-intercepts are positive. (2) says that the product of the y-intercepts are negative.
x-intercepts & y-intercepts aren't the slope.

Correct me if I am wrong. Otherwise, please show the multiplication of each x-y intercepts so I can understand this better.
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I believe i have not misunderstood the terms and used them as under:

Since (1) the product of x-intercepts of the lines is positive and (2) the product of the y-intercepts of the lines is negative, x intercepts of lines L and K are either both -ves, or both +ves. Similarly, y intercepts of lines L and K are opposite to each other. if so, then we have 4 scenarios as under:

1. if the coordinates of Lines L and K are (-2, 1) and (-1, -15) respectively, their slopes are:
L = (1-3)/(-2-4) = -2/-6 = 1/3
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is +ve.

2. if the coordinates of Lines L and K are (-2, 15) and (-1, -15) respectively, their slopes are:
L = (15-3)/(-2-4) = 12/-6 = -2
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is -ve.

3. if the coordinates of Lines L and K are (2, 1) and (1, -15) respectively, their slopes are:
L = (1-3)/(2-4) = -2/-2 = 1
K = (-15-3)/(1-4) = -18/-3 = 6
their product is +ve.

4. if the coordinates of Lines L and K are (2, 15) and (1, -15) respectively, their slopes are:
L = (15-3)/(2-4) = 12/-2 = -6
K = (-15-3)/(1-4) = -18/-3 = 6
their product is -ve.

where i am missing??????????????????
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Okay, let's take a look at this more carefully. To satisfy (1) and (2)
x-intercept must be both negative or both positive <=basically the same sign
y-intercept must be one positive, one negative <=basically different sign

1. if the coordinates of Lines L and K are (-2, 1) and (-1, -15) respectively, their slopes are:
L = (1-3)/(-2-4) = -2/-6 = 1/3
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is +ve.

In this case,
L intersect (4,3) and (-2,1) then
x-intercept is negative, y-intercept is positive
K intersect (4,3) and (-1,-15), then
x-intercept is positive, y-intercept is negative
Since x-intercepts have different sign, this doesn't satisfy the stems.

2. if the coordinates of Lines L and K are (-2, 15) and (-1, -15) respectively, their slopes are:
L = (15-3)/(-2-4) = 12/-6 = -2
K = (-15-3)/(-1-4) = -18/-5 = 18/5
so the product of their slopes is -ve.

In this case,
L intersect (4,3) and (-2,15) then
x-intercept is positive, y-intercept is positive
K intersect (4,3) and (-1,-15), then
x-intercept is positive, y-intercept is negative
This satisfy the stem.


3. if the coordinates of Lines L and K are (2, 1) and (1, -15) respectively, their slopes are:
L = (1-3)/(2-4) = -2/-2 = 1
K = (-15-3)/(1-4) = -18/-3 = 6
their product is +ve.

In this case,
L intersect (4,3) and (2,1) then
x-intercept is positive, y-intercept is negative
K intersect (4,3) and (1,-15), then
x-intercept is positive, y-intercept is negative
Since y-intercepts have the same sign, this doesn't satisfy the stems.


4. if the coordinates of Lines L and K are (2, 15) and (1, -15) respectively, their slopes are:
L = (15-3)/(2-4) = 12/-2 = -6
K = (-15-3)/(1-4) = -18/-3 = 6
their product is -ve.

In this case,
L intersect (4,3) and (2, 15) then
x-intercept is positive, y-intercept is positive
K intersect (4,3) and (1,-15), then
x-intercept is positive, y-intercept is negative
This satisfy the stems.


The only valid comparison that you made are case 2 & 4 and they both produces negative multiplication of the slopes.
Clear?



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