The problem requires finding the equation that represents all points (x, y) that are the same distance away from two fixed points: A(6, 2) and B(4, 6). This means we need to use the distance formula and set the distances equal.

Distance Calculation

Let P be the point (x, y). The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

The distance PA from P(x, y) to A(6, 2) is:

$ PA = \sqrt{(x - 6)^2 + (y - 2)^2} $

The distance PB from P(x, y) to B(4, 6) is:

$ PB = \sqrt{(x - 4)^2 + (y - 6)^2} $

Equidistant Condition

Since the point (x, y) is equidistant from A and B, we have PA = PB.

$ \sqrt{(x - 6)^2 + (y - 2)^2} = \sqrt{(x - 4)^2 + (y - 6)^2} $

Equation Simplification

To eliminate the square roots, square both sides of the equation:

$ (x - 6)^2 + (y - 2)^2 = (x - 4)^2 + (y - 6)^2 $

Expand the squared terms:

$ (x^2 - 12x + 36) + (y^2 - 4y + 4) = (x^2 - 8x + 16) + (y^2 - 12y + 36) $

Cancel out the $x^2$ and $y^2$ terms from both sides:

$ -12x + 36 - 4y + 4 = -8x + 16 - 12y + 36 $

Combine constants and simplify:

$ -12x - 4y + 40 = -8x - 12y + 52 $

Rearrange the terms to group x, y, and constants:

$ (-8x + 12x) + (-12y + 4y) + (52 - 40) = 0 $

$ 4x - 8y + 12 = 0 $

Divide the entire equation by 4:

$ x - 2y + 3 = 0 $

Isolate the x and y terms to match the option format:

$ x - 2y = -3 $

Final Relation

The relation between x and y such that the point (x, y) is equidistant from (6, 2) and (4, 6) is $x - 2y = -3$.