Coordinate 2

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

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

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

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.

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.

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).