Question

Which of the following represent a straight line through $z_1$ and $z_2$?

• A. $z=z_1+t(z_2-z_1),t\in R$
• B. $Im\left(\frac{z-z_1}{z_2-z_1}\right)=0$
• C. $\left|\begin{array}{ccc}z& \overline{z}& 1\\ z_1& \overline{z_1}& 1\\ z_1& \overline{z_2}& 1\end{array}\right|=0$
• D. $\frac{z-z_1}{z-z_2}=e^{i\theta }.\theta \in R$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $z=z_1+t(z_2-z_1),t\in R$
B $Im\left(\frac{z-z_1}{z_2-z_1}\right)=0$
C $\left|\begin{array}{ccc}z& \overline{z}& 1\\ z_1& \overline{z_1}& 1\\ z_1& \overline{z_2}& 1\end{array}\right|=0$

(a) This represents the equation in parametric form:
$z=z_1+t(z_2-z_1)=z_1-t\left(z_1\right)+t\left(z_2\right)=z_1(1-t)+(1-(1-t\left)\right)z_2=t^{\prime }\left(z_1\right)+(1-t^{\prime })\left(z_2\right)$ where $t^{\prime }=1-t$
(b) Since $z,z_{1}and{z}_2$ are collinear, $arg\left(\frac{z-z_1}{z_2-z_1}\right)=0=>Im\left(\frac{z-z_1}{z_2-z_1}\right)=0$
(c) Since $z,z_{1}and{z}_2$ are collinear, the area of the triangle formed by these is zero.
Thus, $\frac{1}{4}\left|\begin{array}{ccc}1& \overline{z}& z\\ 1& \overline{z_1}& z_1\\ 1& \overline{z_2}& z_2\end{array}\right|=0$ and hence the result follows.
(d) This represents the equation of a line which is the perpendicular bisector of the segment joining $z_{1}and{z}_2$.
Hence, (a), (b), (c) are correct.