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Coordinate 2

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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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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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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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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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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.
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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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