Question

If $z,iz$ & $z+iz$ are the vertices of a triangle whose area is $2$ units then the value of $|z{|}^2$ will be

• A. $4$
• B. $2$
• C. $\sqrt{2}$
• D. $0$

Answer 1

Answer: (A)

$4$

Let $z=x+iy$

$iz=-y+ix$

$z+iz=(x-y)+i(x+y)$

Hence, the vertices of the triangle are

$(x,y),(-y,x)(x-y,x+y)$

Hence, area of triangle by determinant method

$=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ (x-y)& (x+y)& 1\end{array}\right|$

Applying $R_2=R_2+R_1$ we get
$=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ (x-y)& (x+y)& 2\\ (x-y)& (x+y)& 1\end{array}\right|$

Now $R_3=R_3-R_2$
$=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ (x-y)& (x+y)& 2\\ 0& 0& -1\end{array}\right|$

Interchanging rows, we get
$\frac{1}{2}\left|\begin{array}{ccc}0& 0& -1\\ x& y& 1\\ (x-y)& (x+y)& 2\end{array}\right|$

$△=\frac{1}{2}(x^2+xy-(xy-y^2)$ ...(negative sign is inserted due to interchanging of columns).

$=\frac{1}{2}(x^2+y^2)$

Now Area is $2$ square units,

Hence,

$(x^2+y^2)=4$

$|z{|}^2=4$