The equation $$ |z-i-2|=|z-3 i+2| $$ where $$ z=x+i y $$ describes the line $$ y=\frac{1}{5} x-\frac{1}{5} $$ $$ y=-2 x+2 $$ $$ y=2 x+2 $$ $$ y=\frac{5}{4} x+\frac{3}{8} $$ None of the others N N N

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Let $$ z = x + i y $$. We rewrite the expression inside each modulus:

1. For the first modulus,
   $$
   z - i - 2 = x + i y - i - 2 = (x - 2) + i (y - 1).
   $$
2. For the second modulus,
   $$
   z - 3i + 2 = x + i y - 3i + 2 = (x + 2) + i (y - 3).
   $$

The given equation is
$$
|z - i - 2| = |z - 3i + 2|,
$$
or equivalently,
$$
|(x-2)+i(y-1)| = |(x+2)+i(y-3)|.
$$

Recall that for a complex number $$ a+ib $$, the modulus is given by
$$
|a+ib| = \sqrt{a^2 + b^2}.
$$
Thus, the equation becomes
$$
\sqrt{(x-2)^2 + (y-1)^2} = \sqrt{(x+2)^2 + (y-3)^2}.
$$

Squaring both sides to eliminate the square roots gives
$$
(x-2)^2 + (y-1)^2 = (x+2)^2 + (y-3)^2.
$$

Now, expand both sides:

Left side:
$$
(x-2)^2 + (y-1)^2 = x^2 - 4x + 4 + y^2 - 2y + 1 = x^2 + y^2 - 4x - 2y + 5.
$$

Right side:
$$
(x+2)^2 + (y-3)^2 = x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 + y^2 + 4x - 6y + 13.
$$

Set the two expressions equal:
$$
x^2 + y^2 - 4x - 2y + 5 = x^2 + y^2 + 4x - 6y + 13.
$$

Cancel $$x^2 + y^2$$ from both sides:
$$
-4x - 2y + 5 = 4x - 6y + 13.
$$

Collect like terms by bringing all terms to one side:
$$
-4x - 2y + 5 - 4x + 6y - 13 = 0.
$$
$$
-8x + 4y - 8 = 0.
$$

Divide the entire equation by 4:
$$
-2x + y - 2 = 0.
$$

Solve for $$ y $$:
$$
y = 2x + 2.
$$

Thus, the locus is the line 
$$
y = 2x + 2.
$$
This corresponds to the third option.
The equation $$ |z - i - 2| = |z - 3i + 2| $$ represents the line $$ y = 2x + 2 $$.

The equation where describes the line

None of the others
N
N
N

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