5
Question
Find the circumcentre of the triangle whose vertices are given by the complex numbers z1,z2,z3.
Find the circumcentre of the triangle whose vertices are given by the complex numbers z1,z2,z3.
Open in App
Solution
The correct option is A z=∑z1¯z1(z2−z3)∑¯z1(z2−z3)
Let C be the circum-center. Distance of this point from the vertices are equal.
Thus, |z1−c|=|z2−c|=|z3−c|
Hence, (z1−c)(¯z1−¯c)=(z2−c)(¯z2−¯c)=(z3−c)(¯z3−¯c)
From the first 2 terms, we get:
−c¯z1−z1¯c+z1¯z1=−c¯z2−z2¯c+z2¯z2
=> (z1−z2)¯c+(¯z1−¯z2)c=z1¯z1−z2¯z2
=> (z1−z2)¯c+(¯z1−¯z2)c=z1¯z1−z2¯z2
Similarly, from the last 2 terms:
(z2−z3)¯c+(¯z2−¯z3)c=z2¯z2−z3¯z3
Solving for C:
c=∑z1¯z1(z2−z3)∑¯z1(z2−z3)
Hence, (A) is correct.
Let C be the circum-center. Distance of this point from the vertices are equal.
Thus, |z1−c|=|z2−c|=|z3−c|
Hence, (z1−c)(¯z1−¯c)=(z2−c)(¯z2−¯c)=(z3−c)(¯z3−¯c)
From the first 2 terms, we get:
−c¯z1−z1¯c+z1¯z1=−c¯z2−z2¯c+z2¯z2
=> (z1−z2)¯c+(¯z1−¯z2)c=z1¯z1−z2¯z2
=> (z1−z2)¯c+(¯z1−¯z2)c=z1¯z1−z2¯z2
Similarly, from the last 2 terms:
(z2−z3)¯c+(¯z2−¯z3)c=z2¯z2−z3¯z3
Solving for C:
c=∑z1¯z1(z2−z3)∑¯z1(z2−z3)
Hence, (A) is correct.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
