Question

If vertices of a triangle are represented by complex numbers $z,iz,z+iz$, then area of triangle is _____

• A. ${\left|z\right|}^2$
• B. $\frac{1}{2}{\left|z\right|}^2$
• C. $2{\left|z\right|}^3$
• D. $3{\left|z\right|}^2$

Answer 1

Answer: (A)

${\left|z\right|}^2$

Let $A=z,B-iz$ and $c=z+iz$

Let $z=x+iy,$ then if you take out the value of $|A-B|,|B-C|$ and $|C-A|$ you will see that it is an isocles triangle with $AC=BC.$

It is a property of an isocles triangle that if we join C and mid point of $A$ and $B$ ( let mid point be P ) it will be perpendicular to $AB.$

$\Rightarrow$ Area $={\frac{1}{2}}^{\ast }A{B}^{\ast }PC$

$\Rightarrow P=$ mid point of $A$ and $B=A+\frac{B}{2}=\frac{(z+iz)}{2}$

Now $PC=|z+iz-\frac{(z+iz)}{2}|=\left|\frac{(z+iz)}{2}\right|$

$\Rightarrow AB=|z-iz|$

$\Rightarrow$ area $={\frac{1}{2}}^{\ast }|z+iz|{\frac{1}{2}}^{\ast }|z-iz|=z^{\wedge }2+z^{\wedge }\frac{2}{4}={\left|z\right|}^{\wedge }\frac{2}{2}$

Hence, the answer is $z^2.$