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 $\frac{i}{4}\begin{vmatrix}z_{1}&\bar{z}_{1}&1\\ z_{2}&\bar{z}_{2}&1\\ z_{3}&\bar{z}_{3}&1\end{vmatrix}$

• 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 $z = x + iy \Rightarrow z = (x, y )$
$\Rightarrow iz = i x - y \Rightarrow iz = (- y, x)$
so, $z + iz = (x - y, x + y)$
$\therefore $ Area $=\frac{1}{2} \begin{vmatrix}x&y&1\\ -y&x&1\\ x-y&x+y&1\end{vmatrix} $
$ = \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 $= \frac{i}{4}\begin{vmatrix}z_{1}&\bar{z}_{1}&1\\ z_{2}&\bar{z}_{2}&1\\ z_{2}&\bar{z}_{3}&1\end{vmatrix}$
Replacing $z_{1} = z, z_{2} = iz, z_{3} = z+ iz$, we get
$ \frac{i}{4} \begin{vmatrix}z&\bar{z}&1\\ iz&-i\bar{z}&1\\ z+iz&\bar{z} -i\bar{z}&1\end{vmatrix} $
Using $R_{2 }\rightarrow R_{2} - R_{1}, R_{3} \rightarrow R_{3} - R_{1}$, we get
$=\frac{i}{4}\begin{vmatrix}z&\bar{z}&1\\ z\left(i-1\right)&-\bar{z}\left(1+i\right)&0\\ iz&-i\bar{z}&0\end{vmatrix} $
$ = \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).