Question

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

Answer 1

Answer: 6

$(4+3i)z+(4-3i)\bar{z}=24\dots (i)$
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
$(4+3i)z_0+(4-3i)z_0=24$
$\iff 8z_0=24$
$\Rightarrow z_0=3$
Also
$(4+3i)z_1+(4-3i)\left(-z_1\right)=24$
$\iff z_1=\frac{24}{6i}=-4i$
$\Rightarrow |OP|=3,|OQ|=4$
$\Rightarrow$ Area of $△POQ=\frac{1}{2}(3)(4)=6$