Introduction to the Problem

The problem presented here involves finding the locus of a point (P(x, y)) such that the sum of the squares of its distances to two given lines, (L_1: x - 2y 7) and (L_2: 2x - y 3), is constant, equal to 7. This geometric problem is approached through algebraic methods and can be quite insightful for understanding the relationship between points and lines in a coordinate plane.

Step-by-Step Solution

Step 1: Calculating Distances from a Point to Each Line

The first step is to calculate the distance from a point (P(x, y)) to each of the given lines. The formula for the distance from a point ((x_0, y_0)) to a line (Ax By C 0) is:

d Ax By C A 2 B 2

Distance from (P(x, y)) to (L_1: x - 2y - 7 0)

The coefficients for (L_1) are (A 1), (B -2), and (C -7). Substituting these into the distance formula:

d_1 1 x-2y-7 1 2 - 2 2 x-2y-7 5

Distance from (P(x, y)) to (L_2: 2x y - 3 0)

The coefficients for (L_2) are (A 2), (B 1), and (C -3). Using the distance formula again:

d_2 2 2x y-3 2 2 1 2 2x y-3 5

Step 2: Setting Up the Equation

According to the problem, the sum of the squares of these distances equals 7:

d_1 1 2 d_2 2 2 7

Substituting the distances we calculated:

x-2y-7 5 2 2x y-3 5 2 7

Multiplying through by 5 to eliminate the denominator:

x-2y-7 2 2x y-3 2 35

Step 3: Removing the Absolute Values

To find the locus, we need to consider the squares of the distances, which already implicitly take care of the signs. Therefore, we can directly proceed to:

x-2y-7 2 2x y-3 2 35

Step 4: Final Equation of the Locus

The final equation representing the locus is:

x-2y-7 2 2x y-3 2 35

This describes the locus of points (P(x, y)) such that the sum of the squares of the distances from the two lines is 7.

Conclusion

The algebraic approach used in this problem demonstrates the interplay between lines and points in a coordinate plane. By using the distance formula and understanding the geometric interpretation of distances, we can derive the locus of a point that satisfies the given conditions. This method is a quintessential example of utilizing basic algebra and geometry to solve advanced geometric problems.