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In the xy-plane, lines k and l intersect at the point (1,1). Is the y-

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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 07 Aug 2017, 19:37
gary391 wrote:
chetan2u wrote:
nalinnair wrote:
In the \(xy\)-plane, lines \(k\) and \(l\) intersect at the point \((1,1)\). Is the \(y\)-intercept of \(k\) greater than the \(y\)-intercept of \(l\)?

(1) The slope of \(k\) is less than the slope of \(l\).
(2) The slope of \(l\) is positive.



Hi,

An easy question if you are aware of the slopes....

slope is the Increase in Y/ Increase in X.......... OR Vertical increase / Horizontal increase..

If BOTH increase with increase in each other, the SLOPE is Positive... But if Y decreases with increase in X, slope is -ive....
so for a point (1,1), a line intercepting above Y as 1, it is -ive and with increase in Y, the slope will have even lesser value...
At y=1, slope is 0 and at y<1, the slopw will keep increasing as the value of Y increases...

So the relation of y-intercept is dependent on slope.. lesser the slope higher is the intercept..


(1) The slope of \(k\) is less than the slope of \(l\).
As seen above y intercept of line l will be more than line k...
Suff

(2) The slope of \(l\) is positive.
we require to know slope of line k also
Insuff

A


Thanks, chetan2u great response. I am trying to understand the relationship of the slope and y intercept. I think irrespective of the line is increasing or decreasing the slope defines lines steepness w.r.t the y -axis. Thus, as we increase the slope our y -intercept should increase. [considering both the line don't have a common intersecting point on the y-axis]. Please let me know if my understanding is correct?


Hi...

If you just take the absolute value, yes with change in x as constant, change in y will be as per your understanding..

The slope depends on change in value of both x and y.
Numerator has change in y, so more the change more the slope
Denominator has change in x, so less the change more the slope

But the slope can be of two types..
From left bottom to right top...... Both change in x and y will be POSITIVE, so slope is POSITIVE
From left top to bottom right......Change in y is NEGATIVE and change in x will be POSITIVE, so slope is -/+= NEGATIVE even if the absolute value or steepness is MORE
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 31 Aug 2017, 23:22
please see the attachment,
let say Ml= -4
Mk= -6
then the slope of k is less than slope of L
y intercept K is greater than L

so I think we need both statement to confirm.
isn't it?

Bunuel or anyone, please correct me
Attachments

line.PNG
line.PNG [ 5.52 KiB | Viewed 739 times ]

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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 31 Aug 2017, 23:40
pclawong wrote:
In the \(xy\)-plane, lines \(k\) and \(l\) intersect at the point \((1,1)\). Is the \(y\)-intercept of \(k\) greater than the \(y\)-intercept of \(l\)?

(1) The slope of \(k\) is less than the slope of \(l\).
(2) The slope of \(l\) is positive.


please see the attachment,
let say Ml= -4
Mk= -6
then the slope of k is less than slope of L
y intercept K is greater than L

so I think we need both statement to confirm.
isn't it?

Bunuel or anyone, please correct me
Image


If the slope of line L is -4 and the slope of line K is -6, the lines won't look like the way you've drawn. The will look like as below:

Image

[Reveal] Spoiler:
Attachment:
Untitled.png
Untitled.png [ 10.79 KiB | Viewed 780 times ]

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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 01 Sep 2017, 00:05
Bunuel wrote:
pclawong wrote:
please see the attachment,
let say Ml= -4
Mk= -6
then the slope of k is less than slope of L
y intercept K is greater than L

so I think we need both statement to confirm.
isn't it?

Bunuel or anyone, please correct me


If the slope of line L is -4 and the slope of line K is -6, the lines won't look like the way you've drawn. The will look like as below:

Image

[Reveal] Spoiler:
Attachment:
Untitled.png


Oh yea, I mean the red line is K
and the blue line is L
so then, even the slope L is larger than slope K,
y intercept of K is still larger than L
am I correct?
that means, statement 1 is not suff.
we also need to know if it is positive slope or negative
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Re: In the xy-plane, lines k and l intersect at the point (1,1) [#permalink]

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New post 01 Sep 2017, 00:24
1
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Expert's post
pclawong wrote:
Bunuel wrote:
pclawong wrote:
please see the attachment,
let say Ml= -4
Mk= -6
then the slope of k is less than slope of L
y intercept K is greater than L

so I think we need both statement to confirm.
isn't it?

Bunuel or anyone, please correct me


If the slope of line L is -4 and the slope of line K is -6, the lines won't look like the way you've drawn. The will look like as below:

Image

[Reveal] Spoiler:
Attachment:
Untitled.png


Oh yea, I mean the red line is K
and the blue line is L
so then, even the slope L is larger than slope K,
y intercept of K is still larger than L
am I correct?
that means, statement 1 is not suff.
we also need to know if it is positive slope or negative


(1) says that the slope of \(k\) is less than the slope of \(l\). So, in my image BLUE line is K (slope -6) and the RED line is L (slope -4).

As you can see the y-intercept of K is greater than the y-intercept of L. For (1) the y-intercept of K will always be greater than the y-intercept of L.
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 12 Sep 2017, 07:05
(k): y=a1*x+b1
(l): y=a2*x+b2
Line k and l intersect at(1,1):
a1+b1=a2+b2
(1): a1<a2 => b1> b2 sufficient
(2) a2>0 not sufficient

A

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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 19 Sep 2017, 09:00
chetan2u wrote:
nalinnair wrote:
In the \(xy\)-plane, lines \(k\) and \(l\) intersect at the point \((1,1)\). Is the \(y\)-intercept of \(k\) greater than the \(y\)-intercept of \(l\)?

(1) The slope of \(k\) is less than the slope of \(l\).
(2) The slope of \(l\) is positive.



Hi,

An easy question if you are aware of the slopes....

slope is the Increase in Y/ Increase in X.......... OR Vertical increase / Horizontal increase..

If BOTH increase with increase in each other, the SLOPE is Positive... But if Y decreases with increase in X, slope is -ive....
so for a point (1,1), a line intercepting above Y as 1, it is -ive and with increase in Y, the slope will have even lesser value...
At y=1, slope is 0 and at y<1, the slopw will keep increasing as the value of Y increases...

So the relation of y-intercept is dependent on slope.. lesser the slope higher is the intercept..


(1) The slope of \(k\) is less than the slope of \(l\).
As seen above y intercept of line l will be more than line k...
Suff

(2) The slope of \(l\) is positive.
we require to know slope of line k also
Insuff

A



Hi,

from (1) Shouldn't the value of y-intercept of K always be greater than the y-intercept of L?
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 15 Oct 2017, 10:03
gary391 wrote:
We can write the equation of the line in slope-intercept form as follow: [Where M is the slope and C is the y-intercept]

For Line K : Yk = Mk.Xk +Ck
For Line L : Yl = Ml.Xl + Cl

Now, since these lines intercept at (1,1), it must satisfy the equation

For Line K : 1 = Mk.1 +Ck
For Line L : 1 = Ml.1 + Cl

=> Mk + Ck = Ml + Cl ------ (equation I)

Now, From Statement 1 - Mk > Ml. Thus, for equation I to hold true. Y intercept of line k must be less than Y intercept of line l (Ck < Cl). Therefore Statement 1 is Sufficient.

Statement 2: Is not sufficient as it doesn't provide information regarding the slope of line L.


You can further refine equation: 1 = Mk + ck => Mk = 1-Ck. Similarly, Ml = 1-Cl. Therefore, from statement 1, Mk < Ml => 1-Ck < 1 - Cl => Ck > Cl
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 20 Feb 2018, 04:59
MarkusKarl Asked this on the first page and I have the same question.

I played devils advocate and got E. What if the slope of L is 1 and the slope of K is zero or undefined? Then K would never intercept the axis, correct?

Thanks in advance for the explanation.
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 03 Mar 2018, 08:31
I agree with msurls, if for line K we have equation X= costant, how do you define the slope and therefore the intercept with Y (which doesn't exist) ? Do you consider the slope of a line parallel to Y axis as +infinite or (-) infinite...? In this case how can you say that 1) is suff, since you don't have any value of y-intercept for line k ?
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 05 Mar 2018, 01:13
teone83 wrote:
I agree with msurls, if for line K we have equation X= costant, how do you define the slope and therefore the intercept with Y (which doesn't exist) ? Do you consider the slope of a line parallel to Y axis as +infinite or (-) infinite...? In this case how can you say that 1) is suff, since you don't have any value of y-intercept for line k ?


A vertical line has no slope. Or put another way, for a vertical line the slope is undefined. This, by the way, does NOT mean that its slope is 0, horizontal lines has slope equal to 0.

Statement (1) says that "The slope of k is less than the slope of l". If any of the lines were vertical, their slopes would be undefined and it would not make sense to compare an undefined slope to anything. Thus, (1) implies that neither of the lines is vertical.
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Collection of Questions:
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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 05 Mar 2018, 01:17
Bunuel wrote:
teone83 wrote:
I agree with msurls, if for line K we have equation X= costant, how do you define the slope and therefore the intercept with Y (which doesn't exist) ? Do you consider the slope of a line parallel to Y axis as +infinite or (-) infinite...? In this case how can you say that 1) is suff, since you don't have any value of y-intercept for line k ?


A vertical line has no slope. Or put another way, for a vertical line the slope is undefined. This, by the way, does NOT mean that its slope is 0, horizontal lines has slope equal to 0.

Statement (1) says that "The slope of k is less than the slope of l". If any of the lines were vertical, their slopes would be undefined and it would not make sense to compare an undefined slope to anything. Thus, (1) implies that neither of the lines is vertical.


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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y- [#permalink]

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New post 11 May 2018, 14:31
gary391 wrote:
We can write the equation of the line in slope-intercept form as follow: [Where M is the slope and C is the y-intercept]

For Line K : Yk = Mk.Xk +Ck
For Line L : Yl = Ml.Xl + Cl

Now, since these lines intercept at (1,1), it must satisfy the equation

For Line K : 1 = Mk.1 +Ck
For Line L : 1 = Ml.1 + Cl

=> Mk + Ck = Ml + Cl ------ (equation I)

Now, From Statement 1 - Mk > Ml. Thus, for equation I to hold true. Y intercept of line k must be less than Y intercept of line l (Ck < Cl). Therefore Statement 1 is Sufficient.

Statement 2: Is not sufficient as it doesn't provide information regarding the slope of line L.


No offense, but I think that more than 1 person here slightly misread statement 1. In other words, Mk < Ml according to the statement. So, the following is what I think:

From statement 1 ---> Mk < Ml. Thus for equation I to hold true the Y intercept of line k must be > the Y intercept of line l. (Ck > Cl). Therefore, statement 1 is sufficient.
Re: In the xy-plane, lines k and l intersect at the point (1,1). Is the y-   [#permalink] 11 May 2018, 14:31

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