Question

Prove that the equation of a line through two given points in the Argand plane can be put in the form $z\overline{\beta }+\overline{z}\beta +c=0$ where $\beta$ is non-zero complex and $c$ is real.

• A. $z\overline{p}-p\overline{z}+c=0.$
  
• B. $z\overline{p}+p\overline{z}-c=0.$
  
• C. $z\overline{p}-p\overline{z}-c=0.$
  
• D. None of these

Answer 1

Answer: (D)

None of these

Let $P\left(z\right)$ be any point on the line $AB$.
$arg(z-z_1)-arg(z-z_2)=0$ or $\pi$
according as $P$ is outside or inside $AB$.
or $arg\frac{(z-z_1)}{(z-z_2)}=0$ or$\pi$
or ${tan}^{-1}\frac{X}{Y}=0$ or $\pi$ or $\frac{X}{Y}=0$ $\therefore y=0$
Hence from $\left(1\right)$ we conclude that
$\frac{z-z_1}{z-z_2}=$ purely real
$\therefore \frac{z-z_1}{z-z_2}=\left(\frac{\overline{z-z_1}}{z-z_2}\right)=\frac{\overline{z}-\overline{z_1}}{\overline{z}-\overline{z_2}}$
or $(z-z_1)(\overline{z}-{\overline{z}}_2)=(z-z_2)(\overline{z}-\overline{z_1})$
or $z\overline{z}-\overline{z}{z}_1-z{\overline{z}}_2+z_1{\overline{z}}_2=z\overline{z}-\overline{z}{z}_2-z{\overline{z}}_1+z_2{\overline{z}}_1$
or $z({\overline{z}}_1-{\overline{z}}_2)+\overline{z}(z_2-z_1)+z_1{\overline{z}}_2-z_2{\overline{z}}_1=0$
or $z\overline{(z_1-z_2)}-\overline{z}\left(z_1{-z}_2\right)+(z_1{\overline{z}}_2-\overline{z_1{\overline{z}}_2})=0$
Now we know that for any complex number $z=a+ib,z-\overline{z}2iImz=2ib=p.i$
Let $z_1-z_2=\beta$, then $\overline{z_1-z_2}=\overline{\beta }$
Hence from $\left(2\right)$, $z\overline{\beta }-\overline{z}\beta +ic=0$
Now $\frac{1}{i}=-i,\overline{i}=-i$
Divide by $i$
$-zi\overline{\beta }+\overline{z}i\beta +c=0.$
If $i\beta =p\overline{i\beta }=\overline{p}$
or $\overline{i}\overline{\beta }=\overline{p}$ or $-i\overline{\beta }=\overline{p}$
$\therefore z\overline{p}+p\overline{z}+c=0.$