Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 19733 by Tinkutara last updated on 15/Aug/17

If ∣z + 1∣ = (√2)∣z − 1∣, then the locus  described by the point z in the argand  diagram is a

$$\mathrm{If}\:\mid{z}\:+\:\mathrm{1}\mid\:=\:\sqrt{\mathrm{2}}\mid{z}\:−\:\mathrm{1}\mid,\:\mathrm{then}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{described}\:\mathrm{by}\:\mathrm{the}\:\mathrm{point}\:{z}\:\mathrm{in}\:\mathrm{the}\:\mathrm{argand} \\ $$$$\mathrm{diagram}\:\mathrm{is}\:\mathrm{a} \\ $$

Answered by ajfour last updated on 15/Aug/17

let z=x+iy  ∣(x+1)+iy∣^2 =2∣(x−1)+iy∣^2   ⇒ x^2 +2x+1+y^2 =2x^2 −4x+2+2y^2   or  x^2 −6x+y^2 +1=0             (x−3)^2 +y^2 =(2(√2))^2         or        ∣z−3∣=2(√2) .

$$\mathrm{let}\:\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\mid\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{iy}\mid^{\mathrm{2}} =\mathrm{2}\mid\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{iy}\mid^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}+\mathrm{y}^{\mathrm{2}} =\mathrm{2x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2}+\mathrm{2y}^{\mathrm{2}} \\ $$$$\mathrm{or}\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{or}\:\:\:\:\:\:\:\:\mid\mathrm{z}−\mathrm{3}\mid=\mathrm{2}\sqrt{\mathrm{2}}\:. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com