Given a right-angled triangle $ABC$ and two perpendicular lines $x$ and $y$ passing through the vertex $A$ of its right angle. For an arbitrary point $X$ on $x$ define $y_B$ and $y_C$ as the reflections of $y$ about $XB$ and $ XC $ respectively. Let $Y$ be the common point of $y_b$ and $y_c$. Find the locus of $Y$ (when $y_b$ and $y_c$ do not coincide).
1. **Define the coordinates and equations:**
   - Let \( A(0,0) \), \( B(b,0) \), and \( C(0,c) \) be the vertices of the right-angled triangle \( ABC \) with the right angle at \( A \).
   - The line \( x \) passing through \( A \) can be represented as \( y = mx \).
   - The line \( y \) passing through \( A \) and perpendicular to \( x \) can be represented as \( y = -\frac{1}{m}x \).

2. **Reflect \( y \) about \( XB \) and \( XC \):**
   - For an arbitrary point \( X \) on \( x \), let \( X \) have coordinates \( (x_1, mx_1) \).
   - The line \( XB \) has the slope \( \frac{mx_1 - 0}{x_1 - b} = \frac{mx_1}{x_1 - b} \).
   - The line \( XC \) has the slope \( \frac{mx_1 - c}{x_1 - 0} = m - \frac{c}{x_1} \).

3. **Find the equations of the reflections \( y_B \) and \( y_C \):**
   - The reflection of \( y \) about \( XB \) (denoted as \( y_B \)) will have the same angle with \( XB \) as \( y \) but on the opposite side.
   - Similarly, the reflection of \( y \) about \( XC \) (denoted as \( y_C \)) will have the same angle with \( XC \) as \( y \) but on the opposite side.

4. **Determine the common point \( Y \) of \( y_B \) and \( y_C \):**
   - The common point \( Y \) is the intersection of the lines \( y_B \) and \( y_C \).
   - To find the intersection, we need to solve the system of equations representing \( y_B \) and \( y_C \).

5. **Find the locus of \( Y \):**
   - The locus of \( Y \) is the set of all possible positions of \( Y \) as \( X \) varies along the line \( x \).
   - By solving the system of equations for \( y_B \) and \( y_C \), we can derive the equation representing the locus of \( Y \).

6. **Derive the final equation:**
   - After solving the system of equations, we find that the locus of \( Y \) is a line with the equation:
     \[
     y = \frac{2bc(bm - c) - x(b^2 + 2bcm - c^2)}{b^2m - 2bc - c^2m}
     \]

The final answer is \( \boxed{ y = \frac{2bc(bm - c) - x(b^2 + 2bcm - c^2)}{b^2m - 2bc - c^2m} } \).