If a variable line drawn through the intersection of the lines `x/3+y/4=1`and `x/4+y/3=1` meets the coordinate axes at `A` and `B, (A !=B),` then the locus of the midpoint of `AB` is (A) `6xy = 7 (x+y)` (B) `4 (x+y)^2 - 28(x+y) + 49 =0` (C) `7xy = 6(x+y)` (D) `14 (x+y)^2 - 97(x+y) + 168 = 0`

Options

A. `7xy=6(x+y)`
B. `4(x+y)^(2)-28(x+y)+49=0`
C. `6xy=7(x+y)`
D. `14(x+y)^(2)-97(x+y)+168=0`

Answer

To find the locus of the midpoint of the line segment \( AB \) formed by the intersection of the lines \( \frac{x}{3} + \frac{y}{4} = 1 \) and \( \frac{x}{4} + \frac{y}{3} = 1 \), we will follow these steps:

### Step 1: Find the intersection point of the two lines.
We start with the equations of the lines:
1. \( \frac{x}{3} + \frac{y}{4} = 1 \) (Equation 1)
2. \( \frac{x}{4} + \frac{y}{3} = 1 \) (Equation 2)

To eliminate the fractions, we can multiply both equations by 12:
- From Equation 1:
\[ 12 \left(\frac{x}{3} + \frac{y}{4}\right) = 12 \implies 4x + 3y = 12 \]
- From Equation 2:
\[ 12 \left(\frac{x}{4} + \frac{y}{3}\right) = 12 \implies 3x + 4y = 12 \]

### Step 2: Solve the system of equations.
Now we have:
1. \( 4x + 3y = 12 \) (Equation 3)
2. \( 3x + 4y = 12 \) (Equation 4)

Next, we can multiply Equation 3 by 3 and Equation 4 by 4:
- \( 12x + 9y = 36 \) (Equation 5)
- \( 12x + 16y = 48 \) (Equation 6)

Now, subtract Equation 5 from Equation 6:
\[ (12x + 16y) - (12x + 9y) = 48 - 36 \implies 7y = 12 \implies y = \frac{12}{7} \]

### Step 3: Substitute \( y \) back to find \( x \).
Substituting \( y = \frac{12}{7} \) into Equation 3:
\[ 4x + 3\left(\frac{12}{7}\right) = 12 \implies 4x + \frac{36}{7} = 12 \]
Multiply through by 7 to eliminate the fraction:
\[ 28x + 36 = 84 \implies 28x = 48 \implies x = \frac{12}{7} \]

Thus, the intersection point is \( \left(\frac{12}{7}, \frac{12}{7}\right) \).

### Step 4: Equation of the variable line through the intersection point.
Let the slope of the variable line be \( m \). The equation of the line in point-slope form is:
\[ y - \frac{12}{7} = m\left(x - \frac{12}{7}\right) \]

### Step 5: Find points \( A \) and \( B \) where the line intersects the axes.
1. **Finding point \( A \) (where \( y = 0 \))**:
\[ 0 - \frac{12}{7} = m\left(x - \frac{12}{7}\right) \implies -\frac{12}{7} = m\left(x - \frac{12}{7}\right) \]
Rearranging gives:
\[ x = \frac{12}{7} - \frac{12}{7m} \]
So, \( A = \left(\frac{12}{7} - \frac{12}{7m}, 0\right) \).

2. **Finding point \( B \) (where \( x = 0 \))**:
\[ y - \frac{12}{7} = m(0 - \frac{12}{7}) \implies y - \frac{12}{7} = -\frac{12m}{7} \]
Rearranging gives:
\[ y = \frac{12}{7} - \frac{12m}{7} \]
So, \( B = \left(0, \frac{12}{7} - \frac{12m}{7}\right) \).

### Step 6: Find the midpoint \( M \) of \( AB \).
The midpoint \( M \) is given by:
\[ M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = \left(\frac{\left(\frac{12}{7} - \frac{12}{7m}\right) + 0}{2}, \frac{0 + \left(\frac{12}{7} - \frac{12m}{7}\right)}{2}\right) \]
Calculating gives:
\[ M = \left(\frac{12}{14} - \frac{6}{7m}, \frac{12}{14} - \frac{6m}{7}\right) \]

### Step 7: Find the locus of the midpoint.
Let \( x = \frac{12}{14} - \frac{6}{7m} \) and \( y = \frac{12}{14} - \frac{6m}{7} \).

From the relationship \( m = -\frac{y}{x} \), substituting this into the equations will yield the locus equation. After simplification, we find that the locus of the midpoint is:
\[ 7xy = 6(x + y) \]

### Final Answer:
Thus, the locus of the midpoint of \( AB \) is given by:
\[ \boxed{7xy = 6(x+y)} \]