[JEE Advanced 1993] abcd is a rhombus diagonals meet at m with bd 2ac if d 1

Answer: $A=3-\dfrac i2$ (with $C=1-\dfrac{3}2 i$).

$M$ is the midpoint of both diagonals; $\vec{MD}=-1+2i$ with $|MD|=\sqrt5=\frac{BD}2$, so $\frac{AC}2=\frac{\sqrt5}2$. Moving $\frac{\sqrt5}2$ from $M$ perpendicular to $MD$ gives $A=3-\frac i2$.

JEE Advanced 1993 · Complex Numbers — verified solution by the Vidaara Team.