Dear student,

LET A,B,C BE THE VERTICES REPRESENTED BY Z1,Z2,Z3...ABC IS EQUILATERAL HENCE

AB=BC=CA=R SAY............
AB=Z2-Z1...|Z2-Z1|=R
BC=Z3-Z2...|Z3-Z2|=R
CA=Z1-Z3...|Z1-Z3|=R

ANGLES..ABC=BCA=CAB=60 DEG.
ANGLE BETWEEN AB & BC IS ANGLE ABC..THAT IS BETWEEN Z2-Z1 AND Z3-Z2..ETC,..

HENCE IF WE TAKE
AB=Z2-Z1=RE^(iT)............I..........WHERE T IS SOME ANGLE....THEN
BC=Z3-Z2= RE^i(T+60)...........II
CA=Z1-Z3=RE^i(T-60).....................III
SQUARING EQNS.I,II,III AND ADDING
(Z2-Z1)^2+(Z3-Z2)^2+(Z1-Z3)^2=R^2[E^(2iT)+E^2i(T+60)+E^2i(T-60)]
2[Z1^2+Z2^2+Z3^2-Z1Z2-Z2Z3-Z3Z1]=(R^2)(E^2iT)[1+E^(120i)+E^(-120i)]

[1+COS(120)+iSIN(120)+COS(-120)+iSIN(-120)]=(R^2)(E^2iT)[1-0.5+iSIN(120)-0.5-iSIN(120)]=0
Z1^2+Z2^2+Z3^2-Z1Z2-Z2Z3-Z3Z1=0
Z1^2+Z2^2+Z3^2=Z1Z2+Z2Z3+Z3Z1

Ultimately it gives (Z1/Z2 + Z2/Z1)=1

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com