Question Number 209707 by alcohol last updated on 19/Jul/24
![a,b ∈C : ab^− + b = 0 f : z′ = az^− + b such that f(M) = M′ 1. let z_A = z and z_(A′) = z′ and f(A) = A show that 2Re(b^− z) = bb^− (A is the set of invariant points and describes a line (△) ) 2. Deduce that (△) is a line with gradient u^( →) with affix z_u^→ = ib 3. show that (z_(MM ′) /z_u ) = ((bb^− − 2Re(bz^− ))/(ibb^− )) 4. show that 2Re(b^− z_0 ) = bb^_ where z_0 = ((z + z ′)/2) 5. Deduce that for M ∉ (△) , M is a perpendicular bisector of [MM ′]](https://www.tinkutara.com/question/Q209707.png)
$${a},{b}\:\in\mathbb{C}\::\:{a}\overset{−} {{b}}\:+\:{b}\:=\:\mathrm{0}\:{f}\::\:{z}'\:=\:{a}\overset{−} {{z}}\:+\:{b}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:{f}\left({M}\right)\:=\:{M}' \\ $$$$\mathrm{1}.\:{let}\:{z}_{{A}} \:=\:{z}\:{and}\:{z}_{{A}'} \:=\:{z}'\:{and}\:{f}\left({A}\right)\:=\:{A} \\ $$$${show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}\right)\:=\:{b}\overset{−} {{b}} \\ $$$$\left({A}\:{is}\:{the}\:{set}\:{of}\:{invariant}\:{points}\:{and}\right. \\ $$$$\left.\:{describes}\:{a}\:{line}\:\left(\bigtriangleup\right)\:\right) \\ $$$$\mathrm{2}.\:{Deduce}\:{that}\:\left(\bigtriangleup\right)\:{is}\:{a}\:{line}\:{with}\: \\ $$$${gradient}\:\overset{\:\rightarrow} {{u}}\:{with}\:{affix}\:{z}_{\overset{\rightarrow} {{u}}} \:=\:{ib} \\ $$$$\mathrm{3}.\:{show}\:{that}\:\frac{{z}_{{MM}\:'} }{{z}_{{u}} }\:=\:\frac{{b}\overset{−} {{b}}\:−\:\mathrm{2}{Re}\left({b}\overset{−} {{z}}\right)}{{ib}\overset{−} {{b}}} \\ $$$$\mathrm{4}.\:{show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}_{\mathrm{0}} \right)\:=\:{b}\overset{\_} {{b}}\:{where} \\ $$$$\:{z}_{\mathrm{0}} \:=\:\frac{{z}\:+\:{z}\:'}{\mathrm{2}} \\ $$$$\mathrm{5}.\:{Deduce}\:{that}\:{for}\:{M}\:\notin\:\left(\bigtriangleup\right)\:,\:{M}\:{is}\: \\ $$$${a}\:{perpendicular}\:{bisector}\:{of}\:\left[{MM}\:'\right] \\ $$