Question

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

• 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 but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

Let $z=x+iy\Rightarrow z=(x,y)$
$\Rightarrow iz=ix-y\Rightarrow iz=(-y,x)$
so, $z+iz=(x-y,x+y)$
$\therefore$ Area $=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ x-y& x+y& 1\end{array}\right|$
$=\frac{1}{2}\left|\left[x\left(-y\right)-y\left(-x\right)-yx-y^2-x^2+xy\right]\right|$
$=\frac{1}{2}\left(x^2+y^2\right)=\frac{1}{2}{\left|z\right|}^2$
$\therefore$ Assertion (A) is true.
Again Area $=\frac{i}{4}\left|\begin{array}{ccc}{z}_1& {\bar{z}}_1& 1\\ z_2& {\bar{z}}_2& 1\\ z_2& {\bar{z}}_3& 1\end{array}\right|$
Replacing $z_1=z,z_2=iz,z_3=z+iz$, we get
$\frac{i}{4}\left|\begin{array}{ccc}z& \bar{z}& 1\\ iz& -i\bar{z}& 1\\ z+iz& \bar{z}-i\bar{z}& 1\end{array}\right|$
Using $R_2\to R_2-R_1,R_3\to R_3-R_1$, we get
$=\frac{i}{4}\left|\begin{array}{ccc}z& \bar{z}& 1\\ z\left(i-1\right)& -\bar{z}\left(1+i\right)& 0\\ iz& -i\bar{z}& 0\end{array}\right|$
$=\frac{i}{4}z\bar{z}\left\{-i\left(i-1\right)+i\left(1+i\right)\right\}$
$=\frac{i}{4}{\left|z\right|}^2\left\{2i\right\}=\frac{1}{2}{\left|z\right|}^2$,as area is always positive.
$\therefore$ Assertion (A) can be proved from Reason (R).