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I'm given the coordinates of two vertices of a square: $(5,1)$ and $(8,5)$ and I'm asked to find the other two.

I suppose there are two squares which meet these conditions. I can obtain a point $(x,y)$ by solving two equation derived from the distance formula for the aforementioned points. One equation implies that the side length of the square is $5$ and the other implies that the diagonal is $5 \sqrt{2}$

$$x^2-16x+y^2-10y+64=0$$

$$x^2-10x+y^2-2y-24=0$$

Solving these equations would give me two points each in a different square, and I can do the same for the other two points. But this approach is tiring and time-consuming. Is there a simpler approach?

ZaIROuS
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    I think there are 3 squares! – imranfat Oct 24 '15 at 21:11
  • Note: There are three. Your vertices might lie on a diagonal, not a side. If you want $PQ$ to be a side then note that side $QX$, say, is perpendicular to $PQ$ and $d(Q,X)=d(P,Q)$. – lulu Oct 24 '15 at 21:15

1 Answers1

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First translate the problem to be a little simpler.

Now there is one point at $(0,0)$ and the other is at $(3,4)$. So if that is one side of the square, you can put the other two to the upper left. You need one of them to be a distance 5 from the origin and give a perpendicular edge so it should be at $(-4,3)$. Now fill in the top corner and translate back.

If this is the diagonal of the square you have that the sides are distance $5/\sqrt{2}$ so the other two are given by the intersections of two circles. The first one being $x^2 + y^2 = 50$ and the second being $(x-3)^2 + (y-4)^2 = 50$. But this is the procedure you already put, but it should be simpler now that you've gotten rid of some of the extraneous numbers by translation.

AHusain
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