Question

The point a' is the reflection of the point a about the line $z\overline{b}+\overline{z}b+c=0$, then show that $a^{\prime }\overline{b}+\overline{a}b+c=0$.

Answer 1

The point Q $\equiv$ a' is the reflection of P $\equiv$ a about the given line. Then the given line is the right bisector of the Join PQ. Let R be any point z on the line.
Then we have $PR=QR$
$|z-a|=|z-a^{\prime }|$
$\Rightarrow |z-a{|}^2=|z-a^{\prime }{|}^2$
$\Rightarrow (z-a)(\overline{z}-\overline{a})=(z-a^{\prime })(\overline{z}-\overline{a^{\prime }})$
$\Rightarrow z\overline{z}-\overline{a}z-a\overline{z}+a\overline{a}=z\overline{z}-\overline{a^{\prime }}z-a^{\prime }\overline{z}+a^{\prime }\overline{a^{\prime }}$
$\Rightarrow z(\overline{a^{\prime }}-\overline{a})+\overline{z}(a^{\prime }-a)+a\overline{a}-a^{\prime }\overline{a^{\prime }}=0$ ........ (1)
Since R is any point on the line, the equation (1) may be regarded as the equation of the given line,
Now comparing (1) with the given line, we have
$\frac{\overline{a^{\prime }}-\overline{a}}{\overline{b}}=\frac{a^{\prime }-a}{b}=\frac{a\overline{a}-a^{\prime }\overline{a^{\prime }}}{-c}=k\left(say\right)$
so that ${\overline{a}}^{\prime }-\overline{a}=\overline{b}k,a^{\prime }-a=bk$
and $a\overline{a}-a^{\prime }{\overline{a}}^{\prime }=-ck$
Now $a^{\prime }\overline{b}+\overline{a}b=\frac{1}{k}\left\{a^{\prime }\right({\overline{a}}^{\prime }-\overline{a})-\overline{a}(a^{\prime }-a\left)\right\}$
$=\frac{1}{k}(-ck)$
or $a^{\prime }\overline{b}+\overline{a}b+c=0$ which is the required condition.
Ans: 1