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BITSAT 2005 Solved Question Paper

Section: Mathematics

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Question : 24 of 150

Marks: +1, -0

The number of solutions for the equations

|z-1|=|z-2|=|z-i|

one solution
3 solutions
2 solutions
no solution

Solution:

Let

z=x+iy

\therefore |z-1| = |z-2| = |z-i|

\Rightarrow |(x-1)+iy| = |(x-2)+iy|

= |(x+i(y-1)|

\Rightarrow x^{2}-2x+1+y^{2} = x^{2}+4-4x+y^{2}

= x^{2}+y^{2}+1-2y

Taking Ist and IInd term

⇒

-2x+1 = 4-4x

⇒

2x = 3

......(i)

Taking IInd and IIIrd term

\Rightarrow 4-4x = 1-2y

⇒

4x-2y = 3

.............(ii)

Taking Ist and IIIrd term

-2x+1 = 1-2y

\Rightarrow 2x-2y = 0

\Rightarrow x = y

..........(iii)

From (i)

x = \frac{3}{2}

On putting value of

x

in Eq. (iii), we get

y = \frac{3}{2}

On putting the value of

x

and

y

in Eq. (ii), we get

4\left(\frac{3}{2}\right) - 2\left(\frac{3}{2}\right) = 3

\Rightarrow 3=3

\therefore

One solution exist.

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