Let `A,B,C` are `3` points on the complex plane represented by complex number `a,b,c` respectively such that `|a|= |b|= |c|=1,a+b+c = abc=1`, then
A
area of triangle `ABC` is `2` (square unit)
B
triangle `ABC` is an equilateral triangle
C
traingle `ABC` is right isosceles triangle.
D
orthocentre of traingle `ABC` lies outside the triangle
Text Solution
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The correct Answer is:
C
`bar(a)+bar(b)+bar(c )=1` `implies ab+bc+ca = 1` implies a,b,c are roots of cubic `z^(3)-z^(2)+z-1=0` `(z^(2)+1)(z-1)=0` `implies z=pm i,1`
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