Question 84103
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