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# In the xy-plane, line k passes through the point (1, 1) and line m

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In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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11 Jun 2010, 06:24
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In the xy-plane, line k passes through the point (1, 1) and line m passes through the point (1, -1). Are the lines k and m perpendicular to each other ?

(1) Lines k and m intersect at the point (1, -1)
(2) Line k intersects the x-axis at the point (1, 0)
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Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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11 Jun 2010, 07:01
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gautamsubrahmanyam wrote:
In the xy-plane ,line k passes through the point (1,1) and line m passes through the point (1,-1).Are the lines k and m perpendicular to each other

(1) Lines k and m intersect at the point (1,-1)
(2) Line k intersects the x axis at the point (1,0)

For one line to be perpendicular to another, their slopes must be negative reciprocals of each other (if slope of one line is $$m$$ than the slope of the line perpendicular to this line is $$-\frac{1}{m}$$). In other words, the two lines are perpendicular if and only the product of their slopes is $$-1$$.

So basically the question is can we somehow calculate the slopes of these lines.

From stem we have one point for each line.

(1) gives us the second point of line $$k$$, hence we can get the slope of this line, but we still know only one point of line $$m$$. Not sufficient.

(2) again gives the second point of line $$k$$, hence we can get the slope of this line, but we still know only one point of line $$m$$. Not sufficient.

(1)+(2) we can derive the slope of line $$k$$ but for line $$m$$ we still have only one point, hence we can not calculate its slope. Not sufficient.

For more on this issue please check Coordinate Geometry chapter of Math Book (link in my signature).

ALSO:
 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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16 May 2012, 07:05
I did not understand why you said you can find out only line K slope from statement 1 - why not m

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Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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16 May 2012, 08:17
venmic wrote:
I did not understand why you said you can find out only line K slope from statement 1 - why not m

Even when considering the statements together we still know only one point of line m: (1, -1). We cannot get the slope based on just one point of a line.

For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Hope it helps.
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Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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07 Jun 2012, 15:36
Could you tell me how to determine whether the lines are perpendicular just in general??
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In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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07 Jun 2012, 15:46
Val1986 wrote:
Could you tell me how to determine whether the lines are perpendicular just in general??

For one line to be perpendicular to another, their slopes must be negative reciprocals of each other (if slope of one line is $$m$$ than the slope of the line perpendicular to this line is $$-\frac{1}{m}$$). In other words, the two lines are perpendicular if and only the product of their slopes is $$-1$$.

For more check Coordinate Geometry chapter of Math Book: http://gmatclub.com/forum/math-coordina ... 87652.html

Hope it helps.
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Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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20 Dec 2013, 11:32
gautamsubrahmanyam wrote:
In the xy-plane, line k passes through the point (1,1) and line m passes through the point (1,-1). Are the lines k and m perpendicular to each other

(1) Lines k and m intersect at the point (1,-1)
(2) Line k intersects the x axis at the point (1,0)

They don't mention that they have to be different lines do they? Can it be the same vertical line as well?

Thanks
Cheers!
J
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Posts: 50670
Re: In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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21 Dec 2013, 04:47
jlgdr wrote:
gautamsubrahmanyam wrote:
In the xy-plane, line k passes through the point (1,1) and line m passes through the point (1,-1). Are the lines k and m perpendicular to each other

(1) Lines k and m intersect at the point (1,-1)
(2) Line k intersects the x axis at the point (1,0)

They don't mention that they have to be different lines do they? Can it be the same vertical line as well?

Thanks
Cheers!
J

Yes, nothing in the question prevents m and k to be the same line.
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Re: In xy plane, line k passes through (1,1) and m through (1, -1). Are  [#permalink]

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27 Oct 2015, 10:54
2
Timmimmit wrote:
First time I am posting in the forum. I could not find the solution to this question from GMATPrep. Please help me.

In xy plane, line k passes through (1,1) and m through (1, -1). Are lines k and m perpendicular?

a. k and m intersect at (1, -1)

b. k intersects x –axis at (1, 0)

From the question, all we know are points on each k and m. To determine whether the lines are perpindicular, we need to know the slopes of each line.
Stmt 1) tells us that k goes through (1,1) and (1, -1), therefore we can figure out its slope. But we still have no information to determine the slope of line m. NOT SUFFICIENT
Stmt 2) gives us no informatin that we cannot glean from Stmt 1. NOT SUFFICIENT

Using both Stmts, we still cannot determine the slope of line m. NOT SUFFICIENT

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In xy plane, line k passes through (1,1) and m through (1, -1). Are  [#permalink]

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Updated on: 16 Apr 2018, 12:01
3
Timmimmit wrote:
First time I am posting in the forum. I could not find the solution to this question from GMATPrep. Please help me.

In xy plane, line k passes through (1,1) and m through (1, -1). Are lines k and m perpendicular?

a. k and m intersect at (1, -1)

b. k intersects x –axis at (1, 0)

Target question: Are lines K and m perpendicular to each other?

IMPORTANT: Since line K passes through the point (1, 1), statements 1 and 2 both have the same effect of "locking" line K into exactly one position. In fact, statements 1 and 2 essentially provide the exact same information. As such, it's either the case that each statement is sufficient (D) or each statement is not sufficient (E).

Since neither statement locks line M into any certain position, line M is free to be in lots of different positions, as long as it passes through the point (1, -1)

Okay, let's jump right to . . .
Statements 1 and 2 combined:
Here are two possible scenarios that satisfy statements 1 and 2.
Scenario a:

In this instance, lines M and K are perpendicular.

Scenario b:

In this instance, lines M and K are not perpendicular.

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Aside: This concept of "locking in" shapes on Geometry DS questions is discussed in much greater detail in our free video: http://www.gmatprepnow.com/module/gmat- ... cy?id=1103

Cheers,
Brent
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Originally posted by GMATPrepNow on 27 Oct 2015, 11:20.
Last edited by GMATPrepNow on 16 Apr 2018, 12:01, edited 1 time in total.
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Re: In xy plane, line k passes through (1,1) and m through (1, -1). Are  [#permalink]

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27 Oct 2015, 17:21
Timmimmit wrote:
In xy plane, line k passes through (1,1) and m through (1, -1). Are lines k and m perpendicular?

(1) k and m intersect at (1, -1)
(2) k intersects x –axis at (1, 0)

This question can be efficiently solved by plotting (imagining) the graph of the lines k and m, rather than trying to find out slopes of the lines algebraically.
To know the slope of a line, we should be able to 'fix' the line on the X-Y plane. And to fix a line we need at least two points that the line passes through. With this in mind, lets see what's given to us.

Question stem:
k passes through $$(1,1)$$, and
m passes through $$(1,-1)$$
At this point none of the lines are fixed. Both of them can rotate $$360^{\circ}$$ about the points they are passing through.

S1)
This gives another point for line K.
So now we have two points that line k passes through- $$(1,1)$$ and $$(1,-1)$$. Now we can fix our line k. (at this point we realize that like k is parallel to Y-axis).
But with this, we still do not have line m fixed. We cannot comment about their angle. Hence insufficient.

S2) This given yet another point that the line k passes through. This is not required as our like k is already fixed. All we needed was another point on line m to fix it. Since we do not have any info on line m and it can still rotate about the point $$(1,-1)$$ and can take any angle w.r.t. line k, this info is insufficient.

S1+S2) As we do not have any new info regarding line m and we cant fix it yet, this is again insufficient.

Hope that helps.
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In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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14 Jul 2016, 22:12
gautamsubrahmanyam wrote:
In the xy-plane, line k passes through the point (1, 1) and line m passes through the point (1, -1). Are the lines k and m perpendicular to each other ?
(1) Lines k and m intersect at the point (1, -1)
(2) Line k intersects the x-axis at the point (1, 0)

Two lines are parallel only if their slopes are inverse reciprocal of each other.
(1) Lines k and m intersect at the point (1, -1)
Using the (x,y) pair of (1,-1) and other (x,y) pair of (1,1) of line k from the question stem we can see that the slope is undefined.
Slope of K = $$\frac{1-(-1)}{1-1} = \frac{-2}{0} ==> undefined$$
Meaning Line K is parallel to Y axis and pass through the x axis at x=1, y=0
INSUFFICIENT we have no info about the slope of line K

(2) Line k intersects the x-axis at the point (1, 0)
We already know from statement 1 that k passes from x=1.
INSUFFICIENT

Merging both statement
WE NEED SLOPE OF LINE M
But there are only one (x,y) pair given for line M and thus we cannot calculate the slope. and thus cannot compare it with slope of line k
Merging also doesn't gives any new information Hence insufficient

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In the xy-plane, line k passes through the point (1, 1) and line m  [#permalink]

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Updated on: 16 Apr 2018, 11:18
Top Contributor
1
gautamsubrahmanyam wrote:
In the xy-plane, line k passes through the point (1, 1) and line m passes through the point (1, -1). Are the lines k and m perpendicular to each other ?

(1) Lines k and m intersect at the point (1, -1)
(2) Line k intersects the x-axis at the point (1, 0)

Target question: Are lines K and m perpendicular to each other?

IMPORTANT: Since line K passes through the point (1, 1), statements 1 and 2 both have the same effect of "locking" line K into exactly one position. In fact, statements 1 and 2 essentially provide the exact same information. As such, it's either the case that each statement is sufficient (D) or each statement is not sufficient (E).

Since neither statement locks line M into any certain position, line M is free to be in lots of different positions, as long as it passes through the point (1, -1)

Okay, let's jump right to . . .
Statements 1 and 2 combined:
Here are two possible scenarios that satisfy statements 1 and 2.
Scenario a:

In this instance, lines M and K are perpendicular.

Scenario b:

In this instance, lines M and K are not perpendicular.

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Cheers,
Brent
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Originally posted by GMATPrepNow on 18 Nov 2017, 10:19.
Last edited by GMATPrepNow on 16 Apr 2018, 11:18, edited 1 time in total.
In the xy-plane, line k passes through the point (1, 1) and line m &nbs [#permalink] 18 Nov 2017, 10:19
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