Question

Determine the equation of the line joining the points whose affixes are $z_{1}and{z}_2$ in argand plane. Hence prove that the equation of the line joining the points $\alpha andi\beta where\alpha ,\beta ϵRand\alpha ,\beta \ne 0$ is
$\frac{z}{2}(\frac{1}{\alpha }-\frac{i}{\beta })+\frac{\overline{z}}{2}(\frac{1}{\alpha }+\frac{i}{\beta })$

Answer 1

1st method : If z be any point on the line joining $z_{1}and{z}_2$ then

$arg\left(\frac{z-z_1}{z-z_2}\right)=0$

$\therefore \frac{z-z_1}{z-z_2}=e^{i,0}or{e}^{i,\pi }$

$\therefore \frac{z-z_1}{z-z_2}=\overline{\left(\frac{z-z_1}{z-z_2}\right)}=\frac{\overline{z}-\overline{z_1}}{\overline{z}-\overline{z_2}}$

Cross-multiply and cancel $z\overline{z}$ from both sides and we have

$z(\overline{z_1}-\overline{z_2})-\overline{z}(z_1-z_2)+z_1\overline{z_2}-\overline{z_1}{z}_2=0$

or $\left|\begin{array}{ccc}z& \overline{z}& 1\\ z_1& \overline{z_1}& 1\\ z_2& \overline{z_2}& 1\end{array}\right|=0$

2nd method : If $z_1=x_1+i{y}_{1}and{z}_2=x_2+i{y}_2$ then equation of line is

$\left|\begin{array}{ccc}x& y& 1\\ x_1& y_1& 1\\ x_2& y_2& 1\end{array}\right|=0$

Apply $C_1+i{C}_{2}andthenmultiply{C}_{2}by2i$.

$\left|\begin{array}{ccc}z& 2iy& 1\\ z_1& 2i{y}_1& 1\\ z_2& 2i{y}_2& 1\end{array}\right|=0or\left|\begin{array}{ccc}z& z-\overline{z}& 1\\ z_1& z_1-\overline{z_1}& 1\\ z_2& z_2-\overline{z_2}& 1\end{array}\right|=0$

Apply $C_2-C_1$ and take -1 common.

$\left|\begin{array}{ccc}z& \overline{z}& 1\\ z_{!}& \overline{z_1}& 1\\ z_2& \overline{z_2}& 1\end{array}\right|=0$

Deduction :

Here $z_1=\alpha ,\overline{z_1}=\alpha ,z_2=i\beta \therefore \overline{z_2}=-i\beta$

$\therefore Lineis$ $\left|\begin{array}{ccc}z& \overline{z}& 1\\ \alpha & \alpha & 1\\ i\beta & -i\beta & 1\end{array}\right|=0$

or $z(\alpha +i\beta )-\overline{z}(\alpha -i\beta )-2i\alpha \beta =0$

Dividing throughout by $2i\alpha \beta$ and putting $\frac{1}{i}=-i$

$\frac{z}{2}(\frac{1}{\alpha }-\frac{i}{\beta })+\frac{\overline{z}}{2}(\frac{1}{\alpha }+\frac{i}{\beta })=1$