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# M24#34

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Senior Manager
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M24#34 [#permalink]

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29 Nov 2008, 00:19
Point $$(1, 0)$$ is closest to which of the following lines?

(A) $$y = x$$
(B) $$y = 1$$
(C) $$y + x = 3$$
(D) $$x = 2$$
(E) $$x + y = -1$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

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Manager
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Re: M24#34 [#permalink]

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29 Nov 2008, 05:48
A

We can get this answer by drawing the lines of equations given in answer choices and visually comparing the distances.

But can anyone find the algebraic way to solve this kind of questions? Of course, if it won't take 18 min, to calculate

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Re: M24#34 [#permalink]

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29 Nov 2008, 12:40
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ventivish wrote:
Point $$(1, 0)$$ is closest to which of the following lines?

(C) 2008 GMAT Club - m24#34

* $$y = x$$
* $$y = 1$$
* $$y + x = 3$$
* $$x = 2$$
* $$x + y = -1$$

Agree with A.
The shortest distance between point (1, 0) and the equation y = x is 1/sqrt(2).

Any of the equation other than y = x has distance more than 1/sqrt(2). A drawing of graph would have much clear visualization.
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Re: M24#34 [#permalink]

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05 Sep 2009, 22:37
My brain is fried.

Stupid question: How to draw the graphs of these lines?

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Manager
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Re: M24#34 [#permalink]

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06 May 2010, 06:11
sid3699 you will have to substitue values for X and Y and point those values on the X and Y graph.

example
for A) i.e y = x

X Y
0 0
1 1
2 2
-1 -1
-2 -2

For B, Y =1 the line is parallel to X axis at y = 1

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Intern
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Re: M24#34 [#permalink]

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06 May 2010, 15:42
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Here is the formula

The shortest distance from (X1,Y1) to the line ax+by+c=0 is

mod(aX1+bY1+c)/\sqrt{(a$$2+b2$$)}

Mod(aX1+bY1+c)/sqrt(a^2+b^2).

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Re: M24#34 [#permalink]

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06 May 2010, 20:13
Ans:A draw line for each option is fastest way to solve.
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Re: M24#34 [#permalink]

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07 May 2010, 01:30
if u want to do this question by elimination see x=1 and y =1 give the same answers , so both gets eliminated ..ofcourse A is the ans ..
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Senior Manager
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Re: M24#34 [#permalink]

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20 May 2010, 07:28
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Answer is A.

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

Image21-summary2.jpg [ 5.39 KiB | Viewed 7920 times ]

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Re: M24#34 [#permalink]

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14 May 2012, 06:13
ventivish wrote:
Point $$(1, 0)$$ is closest to which of the following lines?

(A) $$y = x$$
(B) $$y = 1$$
(C) $$y + x = 3$$
(D) $$x = 2$$
(E) $$x + y = -1$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

i got the ANSWER by drawing the lines in a sheet and then finding out...is there any other easier way plz... i want to lessen the number of calculations i do...help required from the SMEs here
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Re: M24#34 [#permalink]

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15 May 2013, 08:41
ventivish wrote:
Point $$(1, 0)$$ is closest to which of the following lines?

(A) $$y = x$$
(B) $$y = 1$$
(C) $$y + x = 3$$
(D) $$x = 2$$
(E) $$x + y = -1$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Point (1,0) is closest to which of the following lines?

A. y=x
B. y=1
C. y+x=3
D. x=2
E. x+y=−1

Look at the diagram below:
Attachment:

Untitled.png [ 15.99 KiB | Viewed 2327 times ]
As you can see point (1,0) is closest to line y=x.

Answer: A.
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Re: M24#34 [#permalink]

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15 May 2013, 12:33
Just try to eliminate the answers ...

for x = 1 , the distance is 1 , so with y =1 so both of these are not answers and rest two line equations except y=x are far from the point (1,0). so left with option y=x.

may need to calculate the distance in other examples.

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Re: M24#34 [#permalink]

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30 Apr 2014, 02:26
HI Bunnel

when the equation is y=1 then why is your line on the graph passing through y=2... also when we have x=2... the distance between y=x line and dst between x=2 line are same.
i am confused badly.
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Re: M24#34 [#permalink]

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30 Apr 2014, 03:27
nandinigaur wrote:
HI Bunnel

when the equation is y=1 then why is your line on the graph passing through y=2... also when we have x=2... the distance between y=x line and dst between x=2 line are same.
i am confused badly.

Hi,

The question asks out of all the lines in the option A to E, Which is closest to Point (1,0).

To your query:

Yes, the line should pass through Y=1. The distance between the point (1,0) and line Y=X and between Point (1,0) and line X=2 can be same so this tells you surely these 2 options are not the answers. You can eliminate these answer choices

Also Distance of Point (x1,y1) from a line $$ax+by+c=0$$ can be found using the formulae

|$$(ax1+by1+c)/$$$$\sqrt{a^2+b^2}$$ |
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Re: M24#34 [#permalink]

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30 Apr 2014, 03:33
but the answer is A (y=x)... I am again confused.
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Re: M24#34 [#permalink]

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30 Apr 2014, 04:54
nandinigaur wrote:
but the answer is A (y=x)... I am again confused.

Refer to the figure below. Pt(1,0) is closest to Blue line which is y=x and hence ans is A. The distance from line x=2 is more as can be seen in the graph.
Attachments

m 1.png [ 11.39 KiB | Viewed 2884 times ]

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Re: M24#34 [#permalink]

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30 Apr 2014, 06:34
Thanks.... i got it...I guess I am so stressed that i am not even getting the easy explanations
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Re: M24#34 [#permalink]

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01 May 2014, 09:27
nandinigaur wrote:
HI Bunnel

when the equation is y=1 then why is your line on the graph passing through y=2... also when we have x=2... the distance between y=x line and dst between x=2 line are same.
i am confused badly.

Edited the graph of y=1.
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Re: M24#34 [#permalink]

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01 May 2014, 11:07
Thanks Bunnel. I understood!.
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Re: M24#34 [#permalink]

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01 May 2014, 22:09
Here is how i did this Question without using the Distance formula

After drawing all the lines
We can see tht there is a close call between
(A) $$y = x$$
(B) $$y = 1$$
(D) $$x = 2$$

Rest 2 could be eliminated just by seeing that their distance would be greater than the above 3 lines.

As for
(B) $$y = 1$$
(D) $$x = 2$$

Both have Perpendicular Distance =1 from the point $$(1, 0)$$
Now we can guess that since we have to find the shortest distance & no 2 options can have the minimum value & be the answer
Thus $$y = x$$ will be the answer

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Re: M24#34   [#permalink] 01 May 2014, 22:09
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# M24#34

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