Question 37610
Find the two remaining vertices of square ABCD that has A=(-2,1), B=(6,-5) and a vertex in quadrant III.
AB=SQRT((6+2)^2+(5+1)^2)=10
EQN. OF AB IS 
(Y-1)=[(1+5)/(-6-2)](X+2)}
4y-4=-3x-6
3X+4Y=-2
ITS DISTANCE FROM ORIGIN IS 2/5
CD IS PARALLEL TO AB AND IS AT 10 DIUSTANCEB FROM AB OR 10+2/5 =52/5 DISTANCE FROM ORIGIN.
HENCE ITS EQN,IS
3X+4Y=-52................................I..OR...3H+4K=-52 AS C H,K LIES ON IT.
SLOPE OF AB =-3/4
SLOPE OF BC =4/3=
LET C BE H,K
SLOPE OF BC = (K+5)/(H-6)=4/3
3K+15=4H-24
4H-3K=39...........................II
SOLVING I AND II
H=0...K=-13.SO C IS (0,-13)
SIMILARLY FOR D WE CAN FIND 

SLOPE OF AD =4/3=
LET D BE H,K
SLOPE OF AD = (K-1)/(H+2)=4/3
3K-3=4H+8
4H-3K=-11...........................III
SOLVING I AND III
H=-8 AND K=-7