## Ahsome one year ago Absolute Value question

1. ahsome

Assuming $$|x+iy-1| = |x+iy-2i|$$, express $$y$$ in terms of $$x$$ Where $$x,y\in\mathbb{R}$$

2. ahsome

Any idea, @ganeshie8?

3. anonymous

|dw:1433247964097:dw|

4. 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

5. ganeshie8

Nice!

6. ahsome

I am confused. What do you mean by $$\mathbf{equidistant }$$?

7. ganeshie8

let $$z=x+iy$$ then the given equation is same as $|z-1| = |z-2i|$

8. ganeshie8

recall that |z-1| gives you the distance between $$z$$ and $$1+0i$$ and |z-2i| gives the distance between $$z$$ and $$0+2i$$

9. 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

10. ahsome

May I ask, why you squared everything?

11. ahsome

NVM, REALISED

12. ahsome

I get $y=\dfrac{2x}{4}+\dfrac{3}{4}$. Is that right?

13. ahsome

Wait, no

14. ahsome

Why did: $|(x−1)+iy|=|x+i(y−2)|$Become$(x+1)^2+y^2=x^2+(y−2)^2$, Instead of$(x-1)^2+y^2=x^2+(y−2)^2$

15. ganeshie8

Ahh thats just my mistake!

16. ahsome

Ahh, phew ;)

17. ahsome

The answer I get is $\huge{y=\dfrac{x}{2}+\dfrac{3}{4}}$

18. ahsome

The answer in the book is$\huge{y=\dfrac{x}{4}+\dfrac{3}{4}}$

19. ahsome

What do you think is the issue, @ganeshie8?

20. ganeshie8

y = x/2 + 3/4 is right

21. ganeshie8

your textbook must be having a typo

22. ahsome

Ahh, ok then. Thanks so much!. Could you help me with another question?

23. ganeshie8

wil try, pls post as a new q so that it gets bumped in the top and others see

24. ahsome

Ok then :D