Given `A-=(1,1)` and `A B` is any line through it cutting the x-axis at `Bdot` If `A C` is perpendicular to `A B` and meets the y-axis in `C` , then the equation of the locus of midpoint `P` of `B C` is (a) `x+y=1` (b) `x+y=2` (c) `x+y=2x y` (d) `2x+2y=1`

Options

A. x+y=1
B. x+y=2
C. x+y=2xy
D. 2x+2y=1

Explanation

To solve the problem, we need to find the equation of the locus of the midpoint \( P \) of the line segment \( BC \), where \( A = (1, 1) \), \( B \) is a point on the x-axis, and \( C \) is a point on the y-axis such that \( AC \) is perpendicular to \( AB \).

### Step-by-Step Solution:

1. **Identify Points**:
- Let \( A = (1, 1) \).
- Let \( B = (x_1, 0) \) be a point on the x-axis.
- Let \( C = (0, y_1) \) be a point on the y-axis.

2. **Find the Slopes**:
- The slope of line \( AB \) can be calculated as:
\[ \text{slope of } AB = \frac{0 - 1}{x_1 - 1} = \frac{-1}{x_1 - 1} \]
- The slope of line \( AC \) can be calculated as:
\[ \text{slope of } AC = \frac{y_1 - 1}{0 - 1} = \frac{y_1 - 1}{-1} = 1 - y_1 \]

3. **Use the Perpendicularity Condition**:
- Since \( AC \) is perpendicular to \( AB \), the product of their slopes should equal \(-1\):
\[ \left(\frac{-1}{x_1 - 1}\right) \cdot (1 - y_1) = -1 \]
- Simplifying this gives:
\[ \frac{(1 - y_1)}{(x_1 - 1)} = 1 \]
- Rearranging leads to:
\[ 1 - y_1 = x_1 - 1 \implies y_1 = 2 - x_1 \]

4. **Find the Midpoint \( P \)**:
- The midpoint \( P \) of \( BC \) is given by:
\[ P = \left(\frac{x_1 + 0}{2}, \frac{0 + y_1}{2}\right) = \left(\frac{x_1}{2}, \frac{y_1}{2}\right) \]
- Substituting \( y_1 = 2 - x_1 \):
\[ P = \left(\frac{x_1}{2}, \frac{2 - x_1}{2}\right) \]

5. **Express Coordinates of \( P \)**:
- Let \( h = \frac{x_1}{2} \) and \( k = \frac{2 - x_1}{2} \).
- From \( h \), we have \( x_1 = 2h \).
- Substituting \( x_1 \) in the expression for \( k \):
\[ k = \frac{2 - 2h}{2} = 1 - h \]

6. **Find the Locus Equation**:
- Rearranging gives:
\[ h + k = 1 \]
- Substituting back \( h \) and \( k \) with \( x \) and \( y \):
\[ x + y = 1 \]

### Conclusion:
The equation of the locus of the midpoint \( P \) of \( BC \) is:
\[ \boxed{x + y = 1} \]