given that ∣z − i∣ = ∣z − 4 +3 i∣ sketch the locus of z find the catersian equation of this locus.


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Question Number 66562 by Rio Michael last updated on 17/Aug/19

$g i v e n t h a t ∣ z - i ∣ = ∣ z - 4 + 3 i ∣$ $s k e t c h t h e l o c u s o f z$ $f i n d t h e c a t e r s i a n e q u a t i o n o f t h i s l o c u s .$
Commented by mathmax by abdo last updated on 17/Aug/19

$l e t z = x + i y w e h a v e ∣ z - i ∣=∣ z - 4 + 3 i ∣\Leftrightarrow∣ x + i y - i ∣=∣ x + i y - 4 + 3 i ∣$ $\Leftrightarrow∣ x + i (y - 1) ∣=∣ x - 4 + i (y + 3) ∣ \Leftrightarrow \sqrt{x^{2} + {(y - 1)}^{2}} = \sqrt{{(x - 4)}^{2} + {(y + 3)}^{2}}$ $\Leftrightarrow x^{2} + y^{2} - 2 y + 1 = x^{2} - 8 x + 16 + y^{2} + 6 y + 9 \Leftrightarrow$ $- 2 y + 1 = - 8 x + 6 y + 25 \Leftrightarrow - 2 y + 1 + 8 x - 6 y - 25 = 0 \Leftrightarrow$ $8 x - 8 y - 24 = 0 \Leftrightarrow x - y - 3 = 0 s o t h e l o c u s i s a l i n e .$
	Answered by mr W last updated on 17/Aug/19

	$∣ z - i ∣ = ∣ z - 4 + 3 i ∣ m e a n s$ $t h e d i s t a n c e f r o m t h e v a r i a b l e p o i n t$ $P (x, y) t o t h e p o i n t A (0, 1) i s e q u a l$ $t o t h e d i s t a n c e f r o m P t o t h e p o i n t$ $B (4, - 3) . t h e l o c u s o f p o i n t P i s$ $t h e r e f o r e t h e p e r p e n d i c u l a r b i s e c t o r$ $o f A B, i . e . t h e l o c u s i s a l i n e .$ $z = x + y i$ $∣ z - i ∣= \sqrt{x^{2} + {(y - 1)}^{2}}$ $∣ z - 4 + 3 i ∣= \sqrt{{(x - 4)}^{2} + {(y + 3)}^{2}}$ $\Rightarrow x^{2} + {(y - 1)}^{2} = {(x - 4)}^{2} + {(y + 3)}^{2}$ $\Rightarrow y = x - 3 \leftarrow e q n . o f l o c u s$
	Commented by Rio Michael last updated on 17/Aug/19

	$t h a n k s s o m u c h s i r$
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