Slope of line Z.a¯+Z¯.a+b=0, b ∈ R is

z=x+iygiven ⇒(x+iy) a−+(x−iy)a+b=0(a−+a)x+i(a−−a)y+b=0m=−(a−+a)i(a−−a)=a−+ai(a−−a)=2Re(a)i 2i im(a)=−Re(a)Im(a)

Slope of line Z.a¯+Z¯.a+b=0, b ∈ R is

z=x+iygiven ⇒(x+iy) a−+(x−iy)a+b=0(a−+a)x+i(a−−a)y+b=0m=−(a−+a)i(a−−a)=a−+ai(a−−a)=2Re(a)i 2i im(a)=−Re(a)Im(a)

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Slope of line $Z . \bar{a} + \overset{}{\bar{Z} . a + b = 0, b \in R i s}$

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$\frac{- Re (a)}{Im (a)}$

$\frac{- Im (a)}{Re (a)}$

$\frac{\begin{array}{l} Re (a) \end{array}}{Im (a)}$

$\frac{Im (a)}{Re (a)}$

answer is A.

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Detailed Solution

$\begin{array}{l} z = x + i y \\ g i v e n \Rightarrow \\ (x + i y) \bar{a} + (x - i y) a + b = 0 \\ (\bar{a} + a) x + i (\bar{a} - a) y + b = 0 \\ m = \frac{- (\bar{a} + a)}{i (\bar{a} - a)} = \frac{\bar{a} + a}{i (\bar{a} - a)} \\ = \frac{2 Re (a)}{i 2 i i m (a)} \\ = \frac{- Re (a)}{Im (a)} \end{array}$