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Step-by-Step Solution
Step 1: Write down the given lines in standard form
The two given lines are:
1) $ \frac{x}{3} + \frac{y}{4} = 1 $ which can be rewritten as $4x + 3y - 12 = 0$.
2) $ \frac{x}{4} + \frac{y}{3} = 1 $ which can be rewritten as $3x + 4y - 12 = 0$.
Step 2: Form the general equation of the variable line through the intersection
A line passing through the intersection of these two lines can be written as:
$ L_{1} + \lambda \, L_{2} = 0 $
Substituting $L_{1} : 4x + 3y - 12$ and $L_{2} : 3x + 4y - 12$, we get:
$ (4x + 3y - 12) + \lambda \,(3x + 4y - 12) = 0. $
Simplify to:
$ x(4 + 3\lambda) + y(3 + 4\lambda) - 12(1 + \lambda) = 0. $
Step 3: Find the intercepts on the coordinate axes
(a) To find the $x$-intercept (point A), set $y = 0$ in the equation
$ x(4 + 3\lambda) - 12(1 + \lambda) = 0. $
Solving, $ x = \frac{12(1 + \lambda)}{4 + 3\lambda}. $
So,
$$
A = \left( \frac{12(1 + \lambda)}{4 + 3\lambda}, \, 0 \right).
$$
(b) To find the $y$-intercept (point B), set $x = 0$ in the equation
$ y(3 + 4\lambda) - 12(1 + \lambda) = 0. $
Solving, $ y = \frac{12(1 + \lambda)}{3 + 4\lambda}. $
So,
$$
B = \left( 0, \, \frac{12(1 + \lambda)}{3 + 4\lambda} \right).
$$
Step 4: Determine the midpoint of AB
Let $(h,k)$ be the midpoint of AB. Then
$$
h = \frac{1}{2}\left(\frac{12(1 + \lambda)}{4 + 3\lambda} + 0\right) = \frac{12(1 + \lambda)}{2(4 + 3\lambda)} = \frac{6(1 + \lambda)}{4 + 3\lambda},
$$
$$
k = \frac{1}{2}\left(0 + \frac{12(1 + \lambda)}{3 + 4\lambda}\right) = \frac{12(1 + \lambda)}{2(3 + 4\lambda)} = \frac{6(1 + \lambda)}{3 + 4\lambda}.
$$
Step 5: Eliminate $\lambda$ to find the locus in terms of $x$ and $y$
We have
$$
h = \frac{6(1 + \lambda)}{4 + 3\lambda}, \quad k = \frac{6(1 + \lambda)}{3 + 4\lambda}.
$$
We aim to eliminate $\lambda$ and express the relation in $h$ and $k$. Upon elimination, one obtains:
$$
6(h + k) = 7 \, h k.
$$
Replacing $h$ with $x$ and $k$ with $y$ (as the midpoint locus is given by $(x,y)$), we get the locus:
$$
7xy = 6(x + y).
$$
Final Answer
Hence, the locus of the midpoint of AB is
$$
\boxed{7xy = 6(x + y).}
$$
