Drawing a Square

Let $A = ( 9, 6 )$ , $B = ( 12, 10 )$ and $C = (5, 9 )$ . Observe that length of the line segments are $\overline{AB} \rightarrow \sqrt{(12 - 9)^2 + (10 - 6)^2} = 5 \\ \overline{BC} \rightarrow \sqrt{(12 - 5)^2 + (10 - 9)^2} = 5\sqrt{2} \\ \overline{AC} \rightarrow \sqrt{(9 - 5)^2 + (6 - 9)^2} = 5$ After these calculations, we see that we want a point whose distance from $B$ and $C$ is 5. The only point listed whose distance from $B$ and $C$ is 5 is $(8, 13)$

Let A = (5,9), B = (9,6) and C = (12,10) are the three vertices of a square. Let coordinates of the last vertex is D(p,q).

We know that, the intersecting point (x,y) of two diagonals of a square is the middle point for each of them. We can get the intersecting point from the coordinates of the diagonal (AC).

So, x = (5+12)/2 = 8.5 and y = (9+10)/2 = 9.5

Now, for diagonal (BD)

8.5 = (p+9)/2 , p = 8
and 9.5 = (q+6)/2 , q = 13

   So, D(8,13).

The coordinates of the midpoint of the diagonal is

$x_{m} = \frac{5+12}{2} = \frac{17}{2}$

$y_{m} = \frac{9+10}{2} = \frac{19}{2}$

Using the values above, we can solve for the coordinate of the fourth vertex using the other diagonal.

$\frac{17}{2} = \frac{9+x}{2}$ ; $x = 8$

$\frac{19}{2} = \frac{6+y}{2}$ ; $y = 13$

The fourth vertex is at point $(8,13)$ .