Coordinate 2

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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

by MBA.Aspirant » Fri Jul 08, 2011 4:08 am

The lines 9x-3y+30=0 and 3x-y+6=0 represent opposite sides of a square. What is the perimeter of the square?

A.128/5
B.64/25
C.64/10
D.16/10

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Source: Beat The GMAT — Problem Solving | Fri Jul 08, 2011 4:08 am

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Fri Jul 08, 2011 4:24 am

Hi,
I am not sure whether this is a GMAT question.
Normal distance between two parallel lines ax+by+c1 =0 and ax+by+c2 = 0 is |c1-c2|/sqrt(a^+b^2)
Line-1 is 3x-y+10 =0
Line-2 is 3x-y+6 =0
So distance between them is |10-6|/sqrt(3^2 + (-1)^2) = 4/sqrt10
This the edge of square
So, Area is (4/sqrt10)^2 = 16/10

Hence, D

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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

by MBA.Aspirant » Fri Jul 08, 2011 4:47 am

Frankly Frank I don't know whether it's a GMAT or no. I got it from another forum called Urch.

Thanks for the formula. I tired to apply it to this other problem but the result is abstruse. Can you please clarify? https://www.beatthegmat.com/coordinate-t86705.html

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Fri Jul 08, 2011 5:11 am

Hi,
I posted on that thread. But, I don't think we need these formulas for GMAT. Just my opinion.

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Things are not what they appear to be... nor are they otherwise

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mehta003: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Apr 30, 2011 4:52 pm

by mehta003 » Fri Jul 08, 2011 1:30 pm

Thank you Frankenstein. However, I think you might have misread the question. They were asking us for the perimeter and you gave us the area.

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Fri Jul 08, 2011 8:41 pm

mehta003 wrote:Thank you Frankenstein. However, I think you might have misread the question. They were asking us for the perimeter and you gave us the area.

Hey..Thanks. I really misread it. Perimeter will be 4*4/sqrt 10 =16/sqrt10 and this option is absent in the given answer choices.

Cheers!

Things are not what they appear to be... nor are they otherwise

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amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members; Location: pune; Thanked: 36 times

by amit2k9 » Fri Jul 08, 2011 9:18 pm

distance between parallel lines y=mx+b and y=mx+c is |b-c|/(m^2+1)^1/2

thus, |10-6|/(9+1)^1/2 = 4/10^(1/2)
perimeter = 16/10^(1/2).

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