Ahsome
  • Ahsome
Absolute Value question
Mathematics
jamiebookeater
  • jamiebookeater
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Ahsome
  • Ahsome
Assuming \(|x+iy-1| = |x+iy-2i|\), express \(y\) in terms of \(x\) Where \(x,y\in\mathbb{R}\)
Ahsome
  • Ahsome
Any idea, @ganeshie8?
anonymous
  • anonymous
|dw:1433247964097:dw|

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anonymous
  • anonymous
z is a complex number equidistant from (1,0) and (0,2) so it is a straight line perpendicular to the line joining both of them and passing through the mid point you will get the relation between y and x
ganeshie8
  • ganeshie8
Nice!
Ahsome
  • Ahsome
I am confused. What do you mean by \(\mathbf{equidistant }\)?
ganeshie8
  • ganeshie8
let \(z=x+iy\) then the given equation is same as \[|z-1| = |z-2i|\]
ganeshie8
  • ganeshie8
recall that |z-1| gives you the distance between \(z\) and \(1+0i\) and |z-2i| gives the distance between \(z\) and \(0+2i\)
ganeshie8
  • ganeshie8
if you don't like that geometric thingy, you may work it algebraically : \[|x+iy-1| = |x+iy-2i|\\~\\|(x-1)+iy|=|x+i(y-2)|\\~\\(x+1)^2+y^2 = x^2+(y-2)^2\] simplify
Ahsome
  • Ahsome
May I ask, why you squared everything?
Ahsome
  • Ahsome
NVM, REALISED
Ahsome
  • Ahsome
I get \[y=\dfrac{2x}{4}+\dfrac{3}{4}\]. Is that right?
Ahsome
  • Ahsome
Wait, no
Ahsome
  • Ahsome
Why did: \[|(x−1)+iy|=|x+i(y−2)| \]Become\[ (x+1)^2+y^2=x^2+(y−2)^2\], Instead of\[ (x-1)^2+y^2=x^2+(y−2)^2\]
ganeshie8
  • ganeshie8
Ahh thats just my mistake!
Ahsome
  • Ahsome
Ahh, phew ;)
Ahsome
  • Ahsome
The answer I get is \[\huge{y=\dfrac{x}{2}+\dfrac{3}{4}}\]
Ahsome
  • Ahsome
The answer in the book is\[\huge{y=\dfrac{x}{4}+\dfrac{3}{4}}\]
Ahsome
  • Ahsome
What do you think is the issue, @ganeshie8?
ganeshie8
  • ganeshie8
y = x/2 + 3/4 is right
ganeshie8
  • ganeshie8
your textbook must be having a typo
Ahsome
  • Ahsome
Ahh, ok then. Thanks so much!. Could you help me with another question?
ganeshie8
  • ganeshie8
wil try, pls post as a new q so that it gets bumped in the top and others see
Ahsome
  • Ahsome
Ok then :D

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