Question

Assertion :If vertices of a triangle are represented by complex numbers $z,iz,z+iz$, then area of triangle is $\frac{1}{2}|z{|}^2$. Reason: Area of triangle whose vertices are $z_1,z_2,z_3$ is given by $\frac{i}{4}\left|\begin{array}{ccc}{z}_1& {\overline{z}}_1& 1\\ z_2& {\overline{z}}_2& 1\\ z_3& {\overline{z}}_3& 1\end{array}\right|$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true (R) is false.
• D. (A) is false (R) is true.

Answer 1

The correct option is A Both (A) & (R) are individually true & (R) is correct explanation of (A).

Let $z=x+iy$
$\therefore z=(x,y)$
$iz=ix-y$
$\therefore iz=(-y,x)$ & $z+iz=(x-y,x+y)$
$\therefore Area=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ x-y& x+y& 1\end{array}\right|$
$=\frac{1}{2}|\left[x\right(-y)-y(-x)+-yx-y^2-x^2+xy]|$
$\frac{1}{2}(x^2+y^2)=\frac{1}{2}|z{|}^2$
$\therefore$ Assertion (A) is true.
Again Area $=\frac{i}{4}\left|\begin{array}{ccc}{z}_1& \overline{z_1}& 1\\ z_2& \overline{z_2}& 1\\ z_3& \overline{z_3}& 1\end{array}\right|$
replacing $z_1=z,z_2=iz,z_3=z+iz$
$=\frac{i}{4}\left|\begin{array}{ccc}z& \overline{z}& 1\\ iz& -iz& 1\\ z+iz& \overline{z}-i\overline{z}& 1\end{array}\right|$
using $R_2\to R_2-R_1,R_3\to R_3,-R_1$
$\frac{i}{4}\left|\begin{array}{ccc}z& \overline{z}& 1\\ z(i-1)& -\overline{z}(1-i)& 0\\ iz& -i\overline{z}& 0\end{array}\right|$
$\frac{i}{4}z\overline{z}-i(i-1)+i(1+i)$
$=\frac{i}{4}|z{|}^{2}2i=\frac{1}{2}|z{|}^2$ as area is always positive
$\therefore$ Assertion (A) can be proved from Reason (R).