A straight line through the point (1,1) meets the X-axis at A and Y-axis at B. The locus of the mid-point of AB is

Options

A. 2xy+x+y=0
B. x+y-2y=0
C. x+y+2=0
D. x+y-2=0

Explanation

To find the locus of the midpoint of the line segment AB, where A is the intersection of the line with the x-axis and B is the intersection with the y-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Points A and B**:
Let the coordinates of point A (where the line meets the x-axis) be \( A(a, 0) \) and the coordinates of point B (where the line meets the y-axis) be \( B(0, b) \).

2. **Equation of the Line**:
The equation of the line in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
Since the line passes through the point (1, 1), we can substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[ \frac{1}{a} + \frac{1}{b} = 1 \]

3. **Rearranging the Equation**:
Multiplying through by \( ab \) to eliminate the denominators gives:
\[ b + a = ab \]
Rearranging this, we have:
\[ ab - a - b = 0 \]

4. **Midpoint of AB**:
The coordinates of the midpoint M of segment AB can be calculated as:
\[ M\left(\frac{a + 0}{2}, \frac{0 + b}{2}\right) = M\left(\frac{a}{2}, \frac{b}{2}\right) \]
Let \( h = \frac{a}{2} \) and \( k = \frac{b}{2} \). Therefore, we have:
\[ a = 2h \quad \text{and} \quad b = 2k \]

5. **Substituting Back into the Equation**:
Substitute \( a \) and \( b \) back into the rearranged equation:
\[ (2h)(2k) - 2h - 2k = 0 \]
Simplifying this gives:
\[ 4hk - 2h - 2k = 0 \]

6. **Factoring Out Common Terms**:
Dividing the entire equation by 2:
\[ 2hk - h - k = 0 \]
Rearranging gives:
\[ h + k - 2hk = 0 \]

7. **Final Locus Equation**:
The locus of the midpoint M is given by the equation:
\[ h + k - 2hk = 0 \]
This can also be expressed as:
\[ x + y - 2xy = 0 \]
where \( x \) and \( y \) are the coordinates of the midpoint M.

### Conclusion:
The locus of the midpoint of AB is described by the equation:
\[ x + y - 2xy = 0 \]