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M12-20

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M12-20 [#permalink]

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New post 15 Sep 2014, 23:47
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Re M12-20 [#permalink]

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New post 15 Sep 2014, 23:47
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Re: M12-20 [#permalink]

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New post 24 Aug 2016, 05:54
Bunuel wrote:
Official Solution:

Which of the following points is closest to line \(y = x\)?

A. (2, -1)
B. (2, 4)
C. (-1, 1)
D. (2, 1)
E. (-2, 0)


Look at the diagram below:

Image

As you can see point (2, 1) is the closest to line \(y=x\).


Answer: D


Hi Banuel,

I understand the solution, but I was wondering if there's an algebraic approach to this problem.

Thanks.
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Re: M12-20 [#permalink]

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New post 24 Aug 2016, 06:17
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Hi toby001,

This can be solved algebraically, but it will consume a lot of your time.
I feel this approach is most appropriate for GMAT, since you would be given a scratch paper, which is having graph available. You just need to plot the co-ordinates... :)

Coming to your question.
Find the line perpendicular and that passes through the points in options.
Once you get the line again solve for distance between the line and point of interception of both lines.

The following is the formula to find the distance from a point to the line.

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

d = {​∣Am+Bn+C∣}/ {\sqrt{(A^2 + B^2}}

+1 Kudos. If found helpful

Cheers,
Harry
​​
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Re: M12-20 [#permalink]

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New post 24 Aug 2016, 08:04
HARRY113 wrote:
Hi toby001,

This can be solved algebraically, but it will consume a lot of your time.
I feel this approach is most appropriate for GMAT, since you would be given a scratch paper, which is having graph available. You just need to plot the co-ordinates... :)

Coming to your question.
Find the line perpendicular and that passes through the points in options.
Once you get the line again solve for distance between the line and point of interception of both lines.

The following is the formula to find the distance from a point to the line.

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

d = {​∣Am+Bn+C∣}/ {\sqrt{(A^2 + B^2}}

+1 Kudos. If found helpful

Cheers,
Harry
​​


I agree that this is the best solution. I was just curious because when I got the problem, for some silly reason, I didn't think of drawing it out, even though I had all the info I needed. I just decided to try (Y-X) for each option, and somehow arrived at the same answer. I was sure it was just a coincidence; just wanted to be sure that there's no reason it would work out that way.

Thanks!
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Re: M12-20 [#permalink]

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New post 24 Aug 2016, 20:47
I just looked at each pair and did a quick, what's the difference between x and y? The lowest is the answer. Way faster than drawing a graph.
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Re: M12-20 [#permalink]

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New post 25 Aug 2016, 06:33
wainsdaylion wrote:
I just looked at each pair and did a quick, what's the difference between x and y? The lowest is the answer. Way faster than drawing a graph.


I did the same, like I mentioned above. However, I'm doubtful that this will apply in every case.
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Re: M12-20 [#permalink]

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New post 07 Mar 2017, 02:29
I solved it in the following manner:

The X and Y coordinate of any given point on the given line would be equal. Therefore, X=Y=A (assume A as that point)

Then slope of the line = A-0/0-A = A/-A = -1

Now we just have to calculate the slope f each given point and find out which is closest to -1.

Options:

a) -1-0/0-2 = 0.5
b) 4-0/0-2 = -2
c) 1-0/0-(-1) = 1
d) 1-0/0-2 = -0.5
e) 0-0/0-(-2) = 0

Clearly the slope of point d) is closest to the slope of the given line.

Is this a correct way of solving this problem?
Re: M12-20   [#permalink] 07 Mar 2017, 02:29
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