If $z = x + i y, | 3 z - 1 | = 3 | z - 2 |$ and then locus of z is?

x = 0

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x^{2} + y^{2} = 3

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x = y

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6 x = 7

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Solution

The correct option is D $6 x = 7$
Given:

$z = x + i y$

$| 3 z - 1 | = 3 | z - 2 |$ …… (1)

Substitute the value of z in equation (1).

$| 3 (x + i y) - 1 | = 3 | (x + i y) - 2 |$

Simplify the above equation.

$| 3 x + 3 i y - 1 | = 3 | (x + 2) + i y |$

$| (3 x - 1) + 3 i y | = 3 | (x + 2) + i y |$

$\sqrt{{(3 x - 1)}^{2} + {(3 i y)}^{2}} = 3 \sqrt{{(x + 2)}^{2} + {(i y)}^{2}}$

$6 x^{2} + 1 - 6 x - 9 y^{2} = 9 (x^{2} + 4 - 4 x - y^{2})$

$1 - 6 x = 36 - 36 x$

$6 x = 7$

Hence, the required result is found.

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