AskiitiansExpert Mohit-IITD
Last Activity: 14 Years ago
Dear Naidu,
The point D is nothing but the point of intersection of AZ and BC.
Please check the options/question, the coordinates of P will be z1+z2+z3, coordinates of Z(e) will be (-z2z3/z1) and that of D will be 1/2(z1+z2+z3-z2z3/z1).
For Z:
Eqn of AZ: (z-z1)/(z'-z1') + (z2-z3)/(z2'-z3')=0
So, (e-z1)/(e'-z1') + (z2-z3)/(z2'-z3')=0
|z1|=|z2|=|z3|=|e|=1 (they lie on circle),
so on simplifying using z1'=1/z1 etc => Z(e)=-z2z3/z1
For D:
BD is perpendicular to AD, so (z-z1)/(z'-z1') + (z-z2)/(z'-z2')=0.
Solve with (z-z1)/(z'-z1') + (z2-z3)/(z2'-z3')=0 to get D as (1/2(z1+z2+z3-z2z3/z1)
For P:
use D is mid-pt of PZ(e) => P is z1+z2+z3
Hope this helps.
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Now you can 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.