Question 84108
OK AS DESIRED I AM GIVING THE FULL SOLUTION.
BUT WHOEVER GAVE THIS PROBLEM APPEARS TO BE MORE KEEN ON GETTING LABORIOUS WORK DONE WITH HORRID NUMBERS THAN CHECKING THE INTELLIGENCE IN METHOD OF APPLICATION
Working on a puzzle.

Assume an equilateral triangle. I'm given 2 points:						
Point A: -13, -116						
Point B: 7, -75						
LET C BE (H,K)						
SLOPE OF AC=(K+116)/(H+13)=M SAY						
LET D BE MID POINT OF AB. D  IS (7-13)/2,(-75-116)/2=(-3,-95.5)						
SLOPE OF CD =-1/SLOPE 0F AB						
SLOPE OF AB = (-75+116)/(7+13)=41/20						
SLOPE OF CD =-1*(7+13)/(-75+116)=-20/41=(K+95.5)/(H+3)						
-20H/41 -K = 95.5+60/41 				
-0.4878H-K=	96.96341463	……………………………………….1		
TAN(ANGLE CAB)=[M-(41/20)]/[1+41M/20]=TAN(60)=SQRT(3)				
M[1-41{SQRT(3)/20}]=SQRT(3)+(41/20)				
            M=	-1.482747734	=(K+116)/(H+13)		
-1.48275H-K=116+13*1.48275=			135.27575	………………………..2
EQN.1-EQN.2 GIVES				
0.99495	H  = 	-38.31233537		
            H = 	-38.50679468			
PUTTING IN EQN.1				
              K=	-78.17980019			
HENCE  C IS (-38.5068,-78.1798)				
NOW THERE WILL BE ANOTHER C' SATISFYING THE ABOVE REQUIREMENT				
YOU CAN GET C' BY TAKING  TAN (ANGLE CAB)=-SQRT(3) IN TO GET EQN.2				

CHECK THE RESULT				
         AB^2=	2081	        AB=	45.61797891			
         BC^2=	2080.979974	        BC=	45.61775942			
         CA^2=	2080.964374	        CA=	45.61758843