Introduction:
The problem asks to find the locus of a point that is equidistant from the lines x y 4 = 0 and 7x y 20 = 0. In other words, we need to determine the set of all points that are at the same distance from both lines.

Understanding the Problem:
To solve this problem, we need to recall the distance formula between a point and a line. The distance between a point (x1, y1) and a line Ax + By + C = 0 is given by:

d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

Solving the Problem:
We will solve this problem step-by-step:

Step 1: Finding the distance from the first line:
The equation of the first line is x y 4 = 0. Comparing this with the standard form Ax + By + C = 0, we have A = 1, B = 1, and C = 4.
Let (x, y) be a point on the locus. The distance between this point and the first line is given by:

d1 = |x + y + 4| / sqrt(1^2 + 1^2)

Squaring both sides of the equation, we have:

d1^2 = (x + y + 4)^2 / 2

Step 2: Finding the distance from the second line:
The equation of the second line is 7x y 20 = 0. Comparing this with the standard form Ax + By + C = 0, we have A = 7, B = 1, and C = 20.
Let (x, y) be a point on the locus. The distance between this point and the second line is given by:

d2 = |7x + y + 20| / sqrt(7^2 + 1^2)

Squaring both sides of the equation, we have:

d2^2 = (7x + y + 20)^2 / 50

Step 3: Equating the distances:
Since the point (x, y) is equidistant from both lines, we can equate the distances:

d1^2 = d2^2

Substituting the expressions for d1^2 and d2^2 from the previous steps, we have:

(x + y + 4)^2 / 2 = (7x + y + 20)^2 / 50

Step 4: Simplifying the equation:
Expanding both sides of the equation and simplifying, we get:

25x^2 + 16xy + 25y^2 + 12x - 3y - 36 = 0

This is the equation of the locus of the point that is equidistant from the given lines.

Conclusion:
The locus of the point that is equidistant from the lines x y 4 = 0 and 7x y 20 = 0 is given by the equation 25x^2 + 16xy + 25y^2 + 12x - 3y - 36 = 0