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Question

The point a' is the reflection of the point a about the line z¯¯b+¯¯¯zb+c=0, then show that a¯¯b+¯¯¯ab+c=0.

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Solution

The point Q a' is the reflection of P 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
|za|=|za|
|za|2=|za|2
(za)(¯¯¯z¯¯¯a)=(za)(¯¯¯z¯¯¯¯a)
z¯¯¯z¯¯¯aza¯¯¯z+a¯¯¯a=z¯¯¯z¯¯¯¯aza¯¯¯z+a¯¯¯¯a
z(¯¯¯¯a¯¯¯a)+¯¯¯z(aa)+a¯¯¯aa¯¯¯¯a=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
¯¯¯¯a¯¯¯a¯¯b=aab=a¯¯¯aa¯¯¯¯ac=k(say)
so that ¯¯¯a¯¯¯a=¯¯bk,aa=bk
and a¯¯¯aa¯¯¯a=ck
Now a¯¯b+¯¯¯ab=1k{a(¯¯¯a¯¯¯a)¯¯¯a(aa)}
=1k(ck)
or a¯¯b+¯¯¯ab+c=0 which is the required condition.
Ans: 1

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