In the XY-coordinate plane, line L and line K intersect

[Harish Dorai]

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 line L and K is positive.

2) The product of the y-intercepts of line L and k is negative.

[unique]

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 line L and K is positive.

2) The product of the y-intercepts of line L and k is negative.

I think solution is C.

Yl*Yk < 0

So one line has Y intercept -ve and the other has Y intercept +ve

Xl*Xk > 0

both lines shud intercept X axis with x> 0 for the other condition to be valid and to pass thru point (4,3)

Please someone post how to solve such problems quickly - it took me some time.

[givemeanid]

line L: y1 = m1x1 + b1; x-intercept = -b1/m1; y-intercept = b1
line K: y2 = m2x2 + b2; x-intercept = -b2/m2; y-intercept = b2

(4,3) lies on both.
3 = 4m1 + b1 = 4m2 + b2

Is m1m2 < 0?

1. (-b1/m1)*(-b2/m2) > 0
b1b2/m1m2 > 0
If b1b2 > 0, m1m2 > 0
If b1b2 < 0, m1m2 < 0
NOT SUFFICIENT.

2. b1b2 < 0
(3-4m1)(3-4m2) < 0
NOT SUFFICIENT.

Together, b1b2 < 0 and m1m2 < 0.
SUFFICIENT. Answer is C.


This is the same answer I would get even if the point (4,3) wasn't provided. I am wondering what did I miss?

[Harish Dorai]

C is the right answer. Great explanations!

[unique]

Lines L and K cut at (4,3) is the KEY piece of data.

1. X1*X2 > 0

means both lines could have -ve x-intercepts or +ve x-intercepts

cond 1:
if both lines have -ve x-intercepts they have +ve slopes as they pass thru point (4,3) so m1m2>0

cond 2:
if both lines have +ve x-intercepts.

the line(s) could go from left to right passing thru (4,3) which means +ve slope
the line(s) could go from right to left passing thru (4,3) which means -ve slope

We cannot be sure how Lines L and K intercept X axis - INSUFFICIENT

2. Y1*Y2 < 0

means Y1 < 0 OR Y2 < 0

if Y1< 0 line has pass thru (4,3) = +ve slope
other line can be pass from down up or bottom down thru (4,3) so slope is unknown INSUFFICIENT

TOGETHER one line passes down up and other bottom down for both conditions to be true and hence SUFFICIENT.

Man - this took a long time to figure out!

[Guest]

A quick question-Is it ok to assume that the y coordinate in statement 1 is 0, since the term x-intercept is used? If not, why?

Thanks....

[RonPurewal]

A quick question-Is it ok to assume that the y coordinate in statement 1 is 0, since the term x-intercept is used? If not, why?

Thanks....

yes, "x-intercept" always refers to the point at which y = 0.

make sure you take a look at the intuitive approach (the one appearing in the post above yours), which uses an understanding of positive/negative slopes and pictorial interpretation rather than straight algebra. if you can understand, and employ, an approach like that one, you will save yourself a lot of time and effort on these types of problems.

[trying_hard]

Lines L and K cut at (4,3) is the KEY piece of data.

1. X1*X2 > 0

means both lines could have -ve x-intercepts or +ve x-intercepts

cond 1:
if both lines have -ve x-intercepts they have +ve slopes as they pass thru point (4,3) so m1m2>0

cond 2:
if both lines have +ve x-intercepts.

the line(s) could go from left to right passing thru (4,3) which means +ve slope
the line(s) could go from right to left passing thru (4,3) which means -ve slope

We cannot be sure how Lines L and K intercept X axis - INSUFFICIENT

2. Y1*Y2 < 0

means Y1 < 0 OR Y2 < 0

if Y1< 0 line has pass thru (4,3) = +ve slope
other line can be pass from down up or bottom down thru (4,3) so slope is unknown INSUFFICIENT

TOGETHER one line passes down up and other bottom down for both conditions to be true and hence SUFFICIENT.

Man - this took a long time to figure out!

Ron,

I wasnt able to make the jump to "C" from this statement, can you help?

thanks.

[RonPurewal]

Lines L and K cut at (4,3) is the KEY piece of data.

1. X1*X2 > 0

means both lines could have -ve x-intercepts or +ve x-intercepts

cond 1:
if both lines have -ve x-intercepts they have +ve slopes as they pass thru point (4,3) so m1m2>0

cond 2:
if both lines have +ve x-intercepts.

the line(s) could go from left to right passing thru (4,3) which means +ve slope
the line(s) could go from right to left passing thru (4,3) which means -ve slope

We cannot be sure how Lines L and K intercept X axis - INSUFFICIENT

2. Y1*Y2 < 0

means Y1 < 0 OR Y2 < 0

if Y1< 0 line has pass thru (4,3) = +ve slope
other line can be pass from down up or bottom down thru (4,3) so slope is unknown INSUFFICIENT

TOGETHER one line passes down up and other bottom down for both conditions to be true and hence SUFFICIENT.

Man - this took a long time to figure out!

Ron,

I wasnt able to make the jump to "C" from this statement, can you help?

thanks.

from your wording it seems that you understand why (1) and (2) individually are insufficient, so i'll skip to using both of them together. if you don't actually understand why they're insufficient on an individual basis, then post back and say so.

look at it this way: the actual coordinates (4, 3) don't really matter. the only thing that matters is that the line goes through the 1st quadrant.
the reason we know this is we're only concerned with positives and negatives - and those concepts depend only on quadrants and signs, not the magnitude of the actual coordinates. so, if it helps you to think about (1, 1) - or just the first quadrant in general - instead of (4, 3), then go ahead.

statement (1) means that the x-ints are either both positive or both negative; statement (2) means that the y-ints have opposite signs.

let's try the 2 possible cases for statement (1):
both negative: this is impossible, because one of the y-ints has to be negative - and, if you have a line with negative x-int and negative y-int, that line doesn't touch the first quadrant.

so, it MUST be true that...
both x-ints are positive: in this case, one line has a positive x-int and a negative y-int, so it must slope upward: positive slope. (if you don't see why this is true, just draw it. please, please don't use algebra to prove this to yourself; that's a tremendous waste of time.) the other line has a positive x-int and a positive y-int, so it must slope downward: negative slope. (again, draw it to see this for yourself.)
so the product is negative.
sufficient.

--

ironically, if you have the 2 statements together, you don't even have to have the point (4, 3) anymore. as a challenge to the readers of this forum, try proving this: if you have ONLY statements (1) and (2), even if you have no idea where the intersection point is (in fact, even if you aren't even told that the lines intersect in the first place), see if you can prove that
* the product of the slopes must be negative, and
* the lines must intersect.
(hint: the second result follows from the first one)

[victorgsiu]

Ron,

I am not able to make (2) insufficent. Could you post back the subcase that eliminates B?

Here's what I did:

(2) Rephrase: Y1 * Y2 < 0. So Neg x Pos or Pos x Neg. (i.e., either (Y1>0 and Y2<0) or (Y1<0 and Y2>0).)

Let's plug in some numbers for both.

First Scenario:
Use Y1 = (0,-5) and given point (4,3). M1 = (rise/run) = (-5-3/0-4) = -8/-4. M1= 2
Use Y2 = (0, 5) and given point (4,3). M2 = (rise/run) = (5-3/0-4) = 2/-4. M2=-1/2
Find the product of their slopes: M1 x M2 = Pos x Neg = Negative.

Second Scenario:
Use Y1 = (0, 5) and given point (4,3). M1 = (rise/run) = (5-3/0-4) = 2/-4. M1=-1/2
Use Y2 = (0,-5) and given point (4,3). M2 = (rise/run) = (-5-3/0-4) = -8/-4. M2= 2
Find the product of their slopes: M1 x M2 = Neg x Pos = Negative.

Each scenario shows that M1 x M2 = Negative. I wasn't able to create a subcase that makes this statement insufficient.

[Ben Ku]

Ron,

I am not able to make (2) insufficent. Could you post back the subcase that eliminates B?

In the question: If the product of slopes is negative, that means one line has a positive slope and the other has a negative slope.

In statement (2), if the product of y-intercepts is negative, then one is above the x-axis (let's say this is line L) and the other is below the x-axis (let's say this is line K).

It is obvious that line K must have a positive slope. We can take, for example, the y-intercept (0, -3) and (4,3); here our slope is 3/2.

Statement (2) is insufficient because line L can have either a positive or negative slope.

For line L to have a positive slope, then the y-intercept must be below 3. The slope if the y-intercept is 1, is 1/2.

However, if the y-intercept is above 3, then line L will have a negative slope. The slope if the y-intercept is 7 is -1.

Because L can either have a positive or negative slope, then the product of slopes of K and L can be either positive or negative, so statement (2) is insufficient. Hope that helps.

[rajiv.sridhar]

Hi,
This seems to be old thread, but I'll post my approach anyway :)

From the question stem, lets assume points on both lines:
L = (a,b) => slope of L = b/a
K = (c,d) => slope of K = d/c

We need to evaluate if b/a * d/c = -1 -----Deduction 1

Statement 1: (ac) > 0
=> (i) a, c both > 0 OR
(ii) a, c both < 0

No information is given on b, d therefore NOT SUFFICIENT

Statement 2: (bd) < 0
=> b > 0 & d < 0 OR
b < 0 & d > 0

No information is given on a, c therefore NOT SUFFICIENT

Combining Stmts 1 & 2 and inputting into Deduction 1,
the denominator (ac) will be positive regardless of the signs since either both are +ve or both are -ve

The numerator, (bd) will be negative since bd < 0.

Hence (bd) / (ac) is -ve. SUFFICIENT

[mschwrtz]

From the question stem, lets assume points on both lines:
L = (a,b) => slope of L = b/a
K = (c,d) => slope of K = d/c

Only if you assume that both lines pass through the origin (0,0). The slope is not y/x, it's change in y over change in x.

[lj6871849]

Hi - To confirm my understanding, B ALONE is not sufficient because

the line that crosses the y-axis above the line y=0, (+ve Y intercept or vise-versa) can be a horizontal line as well. In that case the slope of that line will be zero..... and the product will be zero. Also we can make the two line perpendicular , as show in the above post. Hence option B here gives us "MAY BE" so NS.

Is this understanding correct?

Cheers

[jnelson0612]

Hi - To confirm my understanding, B ALONE is not sufficient because

the line that crosses the y-axis above the line y=0, (+ve Y intercept or vise-versa) can be a horizontal line as well. In that case the slope of that line will be zero..... and the product will be zero. Also we can make the two line perpendicular , as show in the above post. Hence option B here gives us "MAY BE" so NS.

Is this understanding correct?

Cheers

Yes. Good thinking!