The conjecture that would be incorrect is the third conjecture "The rectangle has a length of 12." The perimeter is the sum of all the sides, and if 2 of the sides are 12, that means that half of the perimeter is already 24, which means that the other 2 sides would have to be 0, which is impossible.
![\vec{BE}=[-2-(-6);\ 4-10]=[4;\ -6]](https://tex.z-dn.net/?f=%5Cvec%7BBE%7D%3D%5B-2-%28-6%29%3B%5C%204-10%5D%3D%5B4%3B%5C%20-6%5D)
![D(x;\ y)\\\\\vec{ED}=[x-(-2);\ y-4]=[x+2;\ y-4]\\\\\vec{ED}=\vec{BE}\ therefore\\\\ \ [x+2;\ y-4]=[4;\ -6]\to x+2=4\ and\ y-4=-6\\\\x=2\ and\ y=-2\\\\Answer:\ \boxed{A.\ (2;\ -2)}](https://tex.z-dn.net/?f=D%28x%3B%5C%20y%29%5C%5C%5C%5C%5Cvec%7BED%7D%3D%5Bx-%28-2%29%3B%5C%20y-4%5D%3D%5Bx%2B2%3B%5C%20y-4%5D%5C%5C%5C%5C%5Cvec%7BED%7D%3D%5Cvec%7BBE%7D%5C%20therefore%5C%5C%5C%5C%20%5C%20%5Bx%2B2%3B%5C%20y-4%5D%3D%5B4%3B%5C%20-6%5D%5Cto%20x%2B2%3D4%5C%20and%5C%20y-4%3D-6%5C%5C%5C%5Cx%3D2%5C%20and%5C%20y%3D-2%5C%5C%5C%5CAnswer%3A%5C%20%5Cboxed%7BA.%5C%20%282%3B%5C%20-2%29%7D)
