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Question Number 66562 by Rio Michael last updated on 17/Aug/19
given that ∣z − i∣ = ∣z − 4 +3 i∣ sketch the locus of z find the catersian equation of this locus.
$${given}\:{that}\:\:\mid{z}\:−\:\mathrm{i}\mid\:=\:\mid{z}\:−\:\mathrm{4}\:+\mathrm{3}\:\mathrm{i}\mid \\ $$$${sketch}\:{the}\:{locus}\:{of}\:\:{z} \\ $$$${find}\:{the}\:{catersian}\:{equation}\:{of}\:{this}\:{locus}. \\ $$
Commented by mathmax by abdo last updated on 17/Aug/19
let z =x+iy we have ∣z−i∣=∣z−4+3i∣⇔∣x+iy−i∣=∣x+iy−4+3i∣ ⇔∣x+i(y−1)∣=∣x−4+i(y+3)∣ ⇔(√(x^2 +(y−1)^2 ))=(√((x−4)^2 +(y+3)^2 )) ⇔x^2 +y^2 −2y+1 =x^2 −8x +16 +y^2 +6y +9 ⇔ −2y+1 =−8x +6y +25 ⇔−2y+1+8x−6y−25 =0 ⇔ 8x−8y−24 =0 ⇔x−y−3 =0 so the locus is a line.
$${let}\:{z}\:={x}+{iy}\:\:\:\:{we}\:{have}\:\mid{z}−{i}\mid=\mid{z}−\mathrm{4}+\mathrm{3}{i}\mid\Leftrightarrow\mid{x}+{iy}−{i}\mid=\mid{x}+{iy}−\mathrm{4}+\mathrm{3}{i}\mid \\ $$$$\Leftrightarrow\mid{x}+{i}\left({y}−\mathrm{1}\right)\mid=\mid{x}−\mathrm{4}+{i}\left({y}+\mathrm{3}\right)\mid\:\Leftrightarrow\sqrt{{x}^{\mathrm{2}} \:+\left({y}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\:={x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{16}\:+{y}^{\mathrm{2}} +\mathrm{6}{y}\:+\mathrm{9}\:\Leftrightarrow \\ $$$$−\mathrm{2}{y}+\mathrm{1}\:=−\mathrm{8}{x}\:+\mathrm{6}{y}\:+\mathrm{25}\:\Leftrightarrow−\mathrm{2}{y}+\mathrm{1}+\mathrm{8}{x}−\mathrm{6}{y}−\mathrm{25}\:=\mathrm{0}\:\Leftrightarrow \\ $$$$\mathrm{8}{x}−\mathrm{8}{y}−\mathrm{24}\:=\mathrm{0}\:\Leftrightarrow{x}−{y}−\mathrm{3}\:=\mathrm{0}\:\:\:{so}\:{the}\:{locus}\:{is}\:{a}\:{line}. \\ $$
Answered by mr W last updated on 17/Aug/19
∣z − i∣ = ∣z − 4 +3 i∣ means the distance from the variable point P(x,y) to the point A(0,1) is equal to the distance from P to the point B(4,−3). the locus of point P is therefore the perpendicular bisector of AB, i.e. the locus is a line. z=x+yi ∣z−i∣=(√(x^2 +(y−1)^2 )) ∣z−4+3i∣=(√((x−4)^2 +(y+3)^2 )) ⇒x^2 +(y−1)^2 =(x−4)^2 +(y+3)^2 ⇒y=x−3 ← eqn. of locus
$$\mid{z}\:−\:\mathrm{i}\mid\:=\:\mid{z}\:−\:\mathrm{4}\:+\mathrm{3}\:\mathrm{i}\mid\:{means} \\ $$$${the}\:{distance}\:{from}\:{the}\:{variable}\:{point} \\ $$$${P}\left({x},{y}\right)\:{to}\:{the}\:{point}\:{A}\left(\mathrm{0},\mathrm{1}\right)\:{is}\:{equal} \\ $$$${to}\:{the}\:{distance}\:{from}\:{P}\:\:{to}\:{the}\:{point} \\ $$$${B}\left(\mathrm{4},−\mathrm{3}\right).\:{the}\:{locus}\:{of}\:{point}\:{P}\:{is} \\ $$$${therefore}\:{the}\:{perpendicular}\:{bisector} \\ $$$${of}\:{AB},\:{i}.{e}.\:{the}\:{locus}\:{is}\:{a}\:{line}. \\ $$$${z}={x}+{yi} \\ $$$$\mid{z}−{i}\mid=\sqrt{{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mid{z}−\mathrm{4}+\mathrm{3}{i}\mid=\sqrt{\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{y}={x}−\mathrm{3}\:\leftarrow\:{eqn}.\:{of}\:{locus} \\ $$
Commented by Rio Michael last updated on 17/Aug/19
thanks so much sir
$${thanks}\:{so}\:{much}\:{sir} \\ $$

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