Question

Let a, b, c be the distinct complex numbers such that $|a|=|b|=|c|and|b+c-a|=|a|$ then considering A, B, C as their geometrical images -

• A. $a+c=0$
• B. $b+c=0$
• C. (AB) is perpendicular to (AC)
• D. (BC) is perpendicular to (AB)

Answer 1

Answer: (B), (C)

$b+c=0$
(AB) is perpendicular to (AC)

If $a,b,c$ are complex numbers such that $|b+c-a|=|a|$ ....( i)
$|a|=|b|=|c|=R$
$\Rightarrow 2(R^2+b.c-a.c-a.b)=0$
$\Rightarrow R^2+b.c-a.c-a.b=0$
$\Rightarrow a.a+b.c-a.c-a.b=0$
$\Rightarrow (a-b).(a-c)=0$
hence AB is perpendicular to AC therefore BC is the diameter of circumcircle of $△ABC$ which has centre at 0
so $b+c=0$