On the coordinate plane (6, 2) and (0, 6)

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abhi332: Master | Next Rank: 500 Posts; Posts: 117; Joined: Wed Dec 30, 2009 1:57 pm; Location: India; Thanked: 4 times; Followed by:1 members

On the coordinate plane (6, 2) and (0, 6)

by abhi332 » Wed Feb 24, 2010 12:01 pm

On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)
[spoiler]
OA : C[/spoiler]

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Source: Beat The GMAT — Problem Solving | Wed Feb 24, 2010 12:01 pm

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pm; Location: Arabian Sea; Thanked: 125 times; Followed by:2 members

by ajith » Wed Feb 24, 2010 9:22 pm

abhi332 wrote:On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)
[spoiler]
OA : C[/spoiler]

Equation of the diagonal x-6/(6) = y-2/(-4)

-4x+24 = 6y -12
2x+3y -18 =0

The other diagonal will be perpendicular to the first and bisects the first at (3,4)

the equation is 3x-2y +c =0
since it passes through (3,4) 3*3-2*4 +c =0
c = 1
2x-3y= -1

Half the length of the diagonal = sqrt(3^2+2^2) (distance from (3,4) to (0,6))

Now we have to find a point which is at sqrt(3^2+2^2) distance from (3,4) on the line 2x-3y= -1

(1,1) and (4,7) fits the bill of which, (1,1) is closer to the origin

Distance of (1,1) to the origin = sqrt(1^2+1^2) =[spoiler]sqrt(2)[/spoiler]

[spoiler]This is neither an elegant method nor an easy method![/spoiler]

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