Question

If origin, $1-\sqrt{3}i$ represents two vertices of equilateral triangle, then its third vertex is

• A. $2$ or $-1-i\sqrt{3}$
• B. $2$ or $1+i\sqrt{3}$
• C. $2$ or $-2$
• D. $1+\sqrt{3}i$ or $3+i\sqrt{3}$

Answer 1

Answer: (A)

$2$ or $-1-i\sqrt{3}$

Let the third vertex is equal to (x,y)

apply distance formula,

$\sqrt{(}x-0{)}^2+(y-0{)}^2$=$\sqrt{(}x-1{)}^2+(y+\sqrt{3}{)}^2$

squaring both the sides,

$(x-0{)}^2+(y-0{)}^2$=$(x-1{)}^2+(y+\sqrt{3}{)}^2$

$x^2+y^2=x^2+y^2-2x+2\sqrt{3}y+4$

the locus is,

$2x-2\sqrt{3}y-4=0$

checking options,

optain (A) is correct