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M05-13

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M05-13  [#permalink]

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New post 16 Sep 2014, 00:25
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In xy plane, line K passes through the points A(6, -7) and B(4, 5). Does line K also pass through point C?


(1) Coordinates of Point C are ( 5, -1)

(2) Point C is equidistant from Point A and Point B.

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Re M05-13  [#permalink]

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New post 16 Sep 2014, 00:25
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A
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Re: M05-13  [#permalink]

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New post 16 Dec 2017, 14:00
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Bunuel
Can you please tell me if we always assume that "the line" is straight?
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Re: M05-13  [#permalink]

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New post 16 Dec 2017, 14:10
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BobsterGMAT wrote:
Bunuel
Can you please tell me if we always assume that "the line" is straight?


Yes. In Euclidean geometry, a line is a straight curve. In coordinate geometry, lines in a Cartesian plane can be described algebraically by linear equations and linear functions.
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Re: M05-13  [#permalink]

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New post 17 Dec 2017, 08:44
Bunuel wrote:
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A


Bunuel

Can you expain how the highlited part can be the case?

Thanks
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Re: M05-13  [#permalink]

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New post 17 Dec 2017, 09:20
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Mo2men wrote:
Bunuel wrote:
Official Solution:


Notice that since we have two distinct points of line K, then we can find its equation.

(1) Coordinates of Point C are (5, -1). We know the equation of line K, hence we can find whether it passes through some particular point. Sufficient.

(2) Point C is equidistant from point A and point B. Point C may be the midpoint of the line segment AB, so on line K. But point C can also be anywhere on the line which is perpendicular to line K and passes through that midpoint. Not sufficient.


Answer: A



Bunuel

Can you expain how the highlited part can be the case?

Thanks


Check the image below:

Image

Point (5, -1), red point, is the midpoint of (6, -7) and (4, 5). Red line is passing through that midpoint and is perpendicular to the blue line. Any point on the red line will be equidistant from (6, -7) and (4, 5).

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.


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Re: M05-13  [#permalink]

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New post 14 Jan 2018, 01:15
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Hi Bunnel,
I might have a perpendicular line passing through C but C is on the line K right which in question we need to say whether k pass through C. In the above graph which you posted the point C is on line K. pls explain.
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Re: M05-13  [#permalink]

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New post 14 Jan 2018, 01:34
tejyr wrote:
Hi Bunnel,
I might have a perpendicular line passing through C but C is on the line K right which in question we need to say whether k pass through C. In the above graph which you posted the point C is on line K. pls explain.


Any point on red line is equidistant from point A and point B, so C could be anywhere on that red line. If it's on the intersection (red dot) then it'll be on blue line to but if it's anywhere else on red line then it won't be on blue line.
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Re: M05-13  [#permalink]

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New post 14 Jan 2018, 07:18
I was unable to understand how any point on the red line is equidistant from A and B.
For instance I took a point on the red perpendicular line (0,-11/6).
when i calculate the distance between (0,-11/6) and A
;(0,-11/6) and B are not equal.
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Re: M05-13  [#permalink]

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New post 14 Jan 2018, 07:49
tejyr wrote:
I was unable to understand how any point on the red line is equidistant from A and B.
For instance I took a point on the red perpendicular line (0,-11/6).
when i calculate the distance between (0,-11/6) and A
;(0,-11/6) and B are not equal.


The distance MUST be the same. The distance between (0, -11/6) and (4, 5) as well as the distance between (0, -11/6) and (6, -7) is \(\frac{\sqrt{2257}}{6}\).

I'll try to explain in another way. Consider points A = (0, 1) and B = (0, -1). Perpendicular bisector of AB is x-axis. Obviously any point on x-axis is equidistant from A and B.

The same way, any point on perpendicular bisector, of segment AB in our question, will be equidistant from A and B.
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Collection of Questions:
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Re: M05-13  [#permalink]

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New post 15 Jan 2018, 21:43
My bad any point in red line is equidistant from A,b. Tq for explanation.
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Re: M05-13 &nbs [#permalink] 15 Jan 2018, 21:43
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