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# M13 Q12

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Intern
Joined: 20 Aug 2010
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Schools: Duke,Darden,Chicago University
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M13 Q12 [#permalink]  04 Dec 2011, 18:56
1
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Difficulty:

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Question Stats:

40% (01:03) correct 60% (01:26) wrong based on 5 sessions
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = \frac{3}{4}x - 3 ?

1.4
\sqrt{2}
1.7
\sqrt{3}
2.0

Can anyone Please explain this Question??
[Reveal] Spoiler: OA
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 4687
Location: Pune, India
Followers: 1079

Kudos [?]: 4841 [1] , given: 163

Re: M13 Q12 [#permalink]  04 Dec 2011, 23:45
1
KUDOS
Expert's post
CracktheGmat2010 wrote:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = \frac{3}{4}x - 3 ?

1.4
\sqrt{2}
1.7
\sqrt{3}
2.0

Can anyone Please explain this Question??

The question has been discussed before. This is my take on it.

Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? (shown by the bold line)

Attachment:

File.jpg [ 17.76 KiB | Viewed 983 times ]

Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer.

Then, let's try to find the length of the bold line, x.
Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line
x = 2.4
So minimum distance is 2.4 - 1 = 1.4
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Karishma
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4687 Location: Pune, India Followers: 1079 Kudos [?]: 4841 [1] , given: 163 Re: M13 Q12 [#permalink] 13 Dec 2011, 02:28 1 This post received KUDOS Expert's post sandeeepsharma wrote: what is the difficulty level of this question? 700+, actually closer to 750. It takes a degree of imagination to figure out the solution quickly. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Intern
Joined: 07 Nov 2011
Posts: 31
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Kudos [?]: 0 [0], given: 9

Re: M13 Q12 [#permalink]  07 Dec 2011, 07:15
VeritasPrepKarishma wrote:
CracktheGmat2010 wrote:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = \frac{3}{4}x - 3 ?

1.4
\sqrt{2}
1.7
\sqrt{3}
2.0

Can anyone Please explain this Question??

The question has been discussed before. This is my take on it.

Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? (shown by the bold line)

Attachment:
File.jpg

Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer.

Then, let's try to find the length of the bold line, x.
Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line
x = 2.4
So minimum distance is 2.4 - 1 = 1.4

i have one serious doubt
if we find the perpendicular distance (d) given by the formula
|ax+by+c|/(a^2+b^2)^1/2

the value comes to be 4 against your value which comes out to be 2.4
tell me where i am going wrong.
Intern
Joined: 07 Nov 2011
Posts: 31
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Kudos [?]: 0 [0], given: 9

Re: M13 Q12 [#permalink]  07 Dec 2011, 07:17
sorry i did some calculation mistake ... i am also getting 2.4
Manager
Joined: 26 Apr 2011
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Re: M13 Q12 [#permalink]  12 Dec 2011, 23:15
what is the difficulty level of this question?
Manager
Joined: 26 Apr 2011
Posts: 226
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Kudos [?]: 31 [0], given: 12

Re: M13 Q12 [#permalink]  13 Dec 2011, 03:37
thanks karishma..
Manager
Joined: 22 Jan 2011
Posts: 59
Location: London, UK
Followers: 1

Kudos [?]: 6 [0], given: 17

Re: M13 Q12 [#permalink]  13 Dec 2011, 10:00
That's an awesome question.
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Re: M13 Q12   [#permalink] 13 Dec 2011, 10:00
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