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

Answer

To find the locus of the midpoint of the line segment AB, where A and B are the points where the line through (1, 1) intersects the x-axis and y-axis respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Points A and B**:
Let the point A be (a, 0) on the x-axis and the point B be (0, b) on the y-axis.

2. **Equation of the Line**:
The equation of the line passing through the point (1, 1) can be expressed in the intercept form:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]

3. **Substituting the Point (1, 1)**:
Since the line passes through the point (1, 1), we substitute x = 1 and y = 1 into the equation:
\[ \frac{1}{a} + \frac{1}{b} = 1 \]

4. **Finding a Relationship Between a and b**:
To solve for a and b, we can rearrange the equation:
\[ \frac{b + a}{ab} = 1 \implies a + b = ab \]

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

6. **Substituting a and b Back**:
Substitute \( a \) and \( b \) into the equation \( a + b = ab \):
\[ 2h + 2k = (2h)(2k) \]
Simplifying gives:
\[ 2(h + k) = 4hk \]

7. **Final Equation**:
Dividing through by 2:
\[ h + k = 2hk \]
Rearranging gives:
\[ h + k - 2hk = 0 \]

8. **Replacing h and k with x and y**:
Since \( h \) and \( k \) represent the coordinates of the midpoint M, we can replace them with x and y:
\[ x + y - 2xy = 0 \]

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