To find the value of \( k \) in the equation of the locus of a point equidistant from two given points, we start with the concept of the perpendicular bisector. The perpendicular bisector of a segment connecting two points is the locus of points that are equidistant from these two points.

The given points are \((a, b)\) and \((c, d)\). The equation of the locus is:

\((a-c)x + (b-d)y + k = 0\)

To find the value of \( k \), let's derive the equation of the perpendicular bisector. The midpoint of the line segment joining \((a, b)\) and \((c, d)\) is:

\(\left( \frac{a+c}{2}, \frac{b+d}{2} \right)\)

The slope of the line joining \((a, b)\) and \((c, d)\) is:

\(\frac{d-b}{c-a}\)

Thus, the slope of the perpendicular bisector is the negative reciprocal:

\(-\frac{c-a}{d-b}\)

Using the point-slope form of a line, the equation of the perpendicular bisector is:

\(y - \frac{b+d}{2} = -\frac{c-a}{d-b}(x - \frac{a+c}{2})\)

Let's rearrange this into the standard form:

\((d-b)(y - \frac{b+d}{2}) = -(c-a)(x - \frac{a+c}{2})\)

Simplify to get the general form:

\((d-b)y + (c-a)x = (d-b)\frac{b+d}{2} + (c-a)\frac{a+c}{2}\)

Reorganizing terms leads to:

\((a-c)x + (b-d)y = \frac{c^2 + d^2 - a^2 - b^2}{2}\)

Comparing this equation with the standard form \((a-c)x + (b-d)y + k = 0\), we find:

\(k = \frac{c^2 + d^2 - a^2 - b^2}{2}\)

Thus, the value of \( k \) is:

\((c^2 + d^2 - a^2 - b^2) / 2\)

This matches the given correct answer option.