If a variable line drawn through the intersection of the lines $\frac{x}{3}+\frac{y}{4}=1$ and $\frac{x}{4}+\frac{y}{3}=1\text{, }$meets the 

coordinate axes at A and B $(A\ne B)$ then the locus of the midpoint of AB is

Given equations of lines can be written as 

$4x+3y-12=0$ and $3x+4y-12=0$

Equation of line passing through the intersection of
these two lines is given by

$\begin{array}{l}(4x+3y-12)+\lambda (3x+4y-12)=0\\ \Rightarrow x(4+3\lambda )+y(3+4\lambda )-12(1+\lambda )=0\end{array}$

Above line meets the coordinate axes at points A and B .

Now, coordinates of point A are $\left(\frac{12(1+\lambda )}{4+3\lambda },0\right)$ and 

coordinates of point B are $\left(0,\frac{12(1+\lambda )}{3+4\lambda }\right)$

$\therefore$ Coordinates of mid-point of AB are given by

$h=\frac{6(1+\lambda )}{4+3\lambda }$     …(i),                       $k=\frac{6(1+\lambda )}{3+4\lambda }$               …(ii)

Eliminating $\lambda$  from (i) and (ii), we get $6(h+k)=7hk$

$\therefore$ Locus of the mid-point of AB is $6(x+y)=7xy$