Question

The complex equation of straight line $L$ is $(4+3 i) z+(4-3 i) \bar{z}=24$. Find the area of the triangle formed by $L$, real and the imaginary axes.

Answer 1

$(4+3 i) z+(4-3 i) \bar{z}=24 \,\,\,\ldots (i) $
Let $P$ and $Q$ be points on real and imaginary axes respectively where line $L$ intersects.
Let $P$ and $Q$ represent the complex numbers $z_{0}$ and $z_{1}$. Then

$z_{0}=\bar{z}_{0}$ and $z_{1}=-\bar{z}_{1}$
(i) gives
$(4+3 i) z_{0}+(4-3 i) z_{0}=24$
$\Leftrightarrow 8 z_{0}=24 $
$\Rightarrow z_{0}=3$
Also
$(4+3 i) z_{1}+(4-3 i)\left(-z_{1}\right)=24$
$\Leftrightarrow z_{1}=\frac{24}{6 i}=-4 i$
$\Rightarrow|O P|=3,|O Q|=4$
$\Rightarrow $ Area of $\triangle P O Q=\frac{1}{2}(3)(4)=6$