If a variable line drawn through the intersection of the lines $$\frac{x}{3} + \frac{y}{4} = 1$$ and $$\frac{x}{4} + \frac{y}{3} = 1$$, meets the coordinate axes at A and B, $$(A \neq B)$$, then the locus of the midpoint of AB is:

To solve this problem, we need to find the locus of the midpoint of the segment AB, where AB is part of a variable line passing through the intersection of the given lines $$\frac{x}{3} + \frac{y}{4} = 1$$ and $$\frac{x}{4} + \frac{y}{3} = 1$$, and meeting the coordinate axes at points A and B (with A ≠ B).

First, we find the point of intersection of the two given lines. The equations are:

$$ \frac{x}{3} + \frac{y}{4} = 1 \quad \text{(Equation 1)} $$

$$ \frac{x}{4} + \frac{y}{3} = 1 \quad \text{(Equation 2)} $$

To eliminate fractions, multiply both equations by 12 (the LCM of 3 and 4):

For Equation 1: $$ 12 \times \frac{x}{3} + 12 \times \frac{y}{4} = 12 \times 1 $$ gives $$ 4x + 3y = 12 $$ \quad $$\text{(Equation 1a)}$$

For Equation 2: $$ 12 \times \frac{x}{4} + 12 \times \frac{y}{3} = 12 \times 1 $$ gives $$ 3x + 4y = 12 $$ \quad $$\text{(Equation 2a)}$$

Now solve the system:

$$ 4x + 3y = 12 \quad \text{(1a)} $$

$$ 3x + 4y = 12 \quad \text{(2a)} $$

Multiply Equation 1a by 4 and Equation 2a by 3:

$$ 4 \times (4x + 3y) = 4 \times 12 \implies 16x + 12y = 48 \quad \text{(Equation 1b)} $$

$$ 3 \times (3x + 4y) = 3 \times 12 \implies 9x + 12y = 36 \quad \text{(Equation 2b)} $$

Subtract Equation 2b from Equation 1b:

$$ (16x + 12y) - (9x + 12y) = 48 - 36 \implies 16x + 12y - 9x - 12y = 12 \implies 7x = 12 \implies x = \frac{12}{7} $$

Substitute $$ x = \frac{12}{7} $$ into Equation 1a:

$$ 4 \times \frac{12}{7} + 3y = 12 \implies \frac{48}{7} + 3y = 12 \implies 3y = 12 - \frac{48}{7} = \frac{84}{7} - \frac{48}{7} = \frac{36}{7} \implies y = \frac{36}{7} \times \frac{1}{3} = \frac{12}{7} $$

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

The variable line passes through this point and meets the coordinate axes at A(a, 0) and B(0, b), where a and b are the x-intercept and y-intercept, respectively. The equation of this line in intercept form is:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

Since the line passes through $$ \left( \frac{12}{7}, \frac{12}{7} \right) $$, substitute these coordinates:

$$ \frac{\frac{12}{7}}{a} + \frac{\frac{12}{7}}{b} = 1 \implies \frac{12}{7a} + \frac{12}{7b} = 1 $$

Factor out $$ \frac{12}{7} $$:

$$ \frac{12}{7} \left( \frac{1}{a} + \frac{1}{b} \right) = 1 \implies \frac{12}{7} \left( \frac{a + b}{ab} \right) = 1 \implies \frac{12(a + b)}{7ab} = 1 \implies 12(a + b) = 7ab \quad \text{(Equation 3)} $$

The midpoint of AB, where A is (a, 0) and B is (0, b), is $$ M \left( \frac{a}{2}, \frac{b}{2} \right) $$. Denote the midpoint as (h, k), so:

$$ h = \frac{a}{2}, \quad k = \frac{b}{2} \implies a = 2h, \quad b = 2k $$

Substitute these into Equation 3:

$$ 12(2h + 2k) = 7 \times (2h) \times (2k) \implies 12 \times 2(h + k) = 7 \times 4hk \implies 24(h + k) = 28hk $$

Divide both sides by 4:

$$ 6(h + k) = 7hk \implies 7hk - 6h - 6k = 0 $$

This equation relates h and k. For the locus, replace h and k with x and y:

$$ 7xy - 6x - 6y = 0 \implies 7xy = 6x + 6y \implies 7xy = 6(x + y) $$

Comparing with the options, this matches option A.

Hence, the correct answer is Option A.