If a variable line passes through the point of intersectionof the lines `x + 2y - 1 = 0` and `2x - y-1 = 0` and meets the coordinate axes in `A` and `B`, then the locus of the midpoint of `AB` is : (A) ` x + 3y = 0` (B) `x + 3y = 10` (C) `x + 3y = 10xy` (D) None of these

Options

A. `x+3y=0`
B. `x+3y=10`
C. `x+3y=10xy`
D. None of these

Answer

To solve the problem, we need to find the locus of the midpoint of the line segment \(AB\) where the line passes through the point of intersection of the given lines and meets the coordinate axes.

### Step 1: Find the point of intersection of the lines
We have the equations of the lines:
1. \(x + 2y - 1 = 0\)
2. \(2x - y - 1 = 0\)

To find the intersection, we can solve these equations simultaneously.

From the first equation, we can express \(x\) in terms of \(y\):
\[ x = 1 - 2y \quad \text{(1)} \]

Substituting equation (1) into the second equation:
\[ 2(1 - 2y) - y - 1 = 0 \]
\[ 2 - 4y - y - 1 = 0 \]
\[ 1 - 5y = 0 \implies y = \frac{1}{5} \]

Now substituting \(y = \frac{1}{5}\) back into equation (1) to find \(x\):
\[ x = 1 - 2\left(\frac{1}{5}\right) = 1 - \frac{2}{5} = \frac{3}{5} \]

Thus, the point of intersection \(P\) is:
\[ P\left(\frac{3}{5}, \frac{1}{5}\right) \]

### Step 2: Equation of the variable line
Let the variable line passing through point \(P\) be represented in the intercept form:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
where \(A(a, 0)\) and \(B(0, b)\) are the intercepts on the x-axis and y-axis respectively.

### Step 3: Finding the coordinates of the midpoint \(M\)
The coordinates of the midpoint \(M\) of segment \(AB\) are given by:
\[ M\left(\frac{a}{2}, \frac{b}{2}\right) \]

### Step 4: Substitute the point \(P\) into the line equation
Since the line passes through point \(P\), we substitute \(P\) into the line equation:
\[ \frac{\frac{3}{5}}{a} + \frac{\frac{1}{5}}{b} = 1 \]
Multiplying through by \(5ab\) to eliminate the fractions:
\[ 3b + a = 5ab \quad \text{(2)} \]

### Step 5: Express \(a\) and \(b\) in terms of \(h\) and \(k\)
From the midpoint coordinates:
\[ a = 2h \quad \text{and} \quad b = 2k \]

Substituting \(a\) and \(b\) into equation (2):
\[ 3(2k) + 2h = 5(2h)(2k) \]
\[ 6k + 2h = 20hk \]

### Step 6: Rearranging to find the locus
Rearranging gives:
\[ 20hk - 2h - 6k = 0 \]
Factoring out common terms:
\[ 2h(10k - 1) - 6k = 0 \]
Rearranging:
\[ 2h(10k - 1) = 6k \]
Dividing by \(2\):
\[ h(10k - 1) = 3k \]
Now, substituting \(h\) with \(x\) and \(k\) with \(y\):
\[ x(10y - 1) = 3y \]
Expanding:
\[ 10xy - x = 3y \]
Rearranging gives:
\[ 10xy - 3y - x = 0 \]
This can be rewritten as:
\[ x + 3y = 10xy \]

### Final Answer
Thus, the locus of the midpoint of \(AB\) is:
\[ \boxed{x + 3y = 10xy} \]