Let z1, z2, z3 represent vertices of triangle A B C and |z1|=|z2|=|z3|. If arg ((z3-z1/z2-z1))=(π/6), then

Question

Let z1, z2, z3 represent vertices of triangle A B C and |z1|=|z2|=|z3|. If arg ((z3-z1/z2-z1))=(π/6), then (A) centroid of triangle ABC is at origin. (B)  arg ((z1-z2/z1-z3))=(-π/6) (C)  arg (( z 2/ Z 2))=(-π/3) (D)  arg ((z1/z2))=(π/12)

Tardigrade Admin · Accepted Answer

From given information, z 1, z 2, z 3 lie on circumference of circle whose. circum-centre is at origin. If its centroid is also at origin then its orthocentre <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/bbd058d721aaf39a490d10d8a6dcb47b-.png" /> will coincide with its centroid. Hence in this case triangle will be an equilateral. ∵ arg (( z 1/ z 2))=- arg (( z 2/ z 1)) Hence option (B) is justified Again apply rotation theorem on triangle B O C, we get arg (( z 3/ z 2))=(π/3) ⇒ arg (( z 2/ z 3))=-(π/3)

Q. Let $z_{1}, z_{2}, z_{3}$ represent vertices of triangle $A BC$ and $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣$ . If $ar g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}}) = \frac{π}{6}$ , then

centroid of triangle $A BC$ is at origin.

$ar g (\frac{z _{1} - z _{2}}{z _{1} - z _{3}}) = \frac{- π}{6}$

$ar g (\frac{z _{2}}{Z _{2}}) = \frac{- π}{3}$

$ar g (\frac{z _{1}}{z _{2}}) = \frac{π}{12}$

Q. Let z1​,z2​,z3​ represent vertices of triangle ABC and ∣z1​∣=∣z2​∣=∣z3​∣. If arg(z2​−z1​z3​−z1​​)=6π​, then

centroid of triangle ABC is at origin.

arg(z1​−z3​z1​−z2​​)=6−π​

arg(Z2​z2​​)=3−π​

arg(z2​z1​​)=12π​

Q. Let $z_{1}, z_{2}, z_{3}$ represent vertices of triangle $A BC$ and $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣$ . If $ar g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}}) = \frac{π}{6}$ , then

centroid of triangle $A BC$ is at origin.

$ar g (\frac{z _{1} - z _{2}}{z _{1} - z _{3}}) = \frac{- π}{6}$

$ar g (\frac{z _{2}}{Z _{2}}) = \frac{- π}{3}$

$ar g (\frac{z _{1}}{z _{2}}) = \frac{π}{12}$