A variable line, drawn through the point of intersection of the straight lines (x/a)+(y/b) = 1 and (x/b)+(y/a)=1, meets the coordinate axes in A & B . We have to Show that the locus of the mid point of AB is the curve 2xy(a + b) = ab(x + y)
I tried as
Equation of line can be written as L$_1$+kL$_2$= 0
From this i got A(a,0) and B(0,b).
Let $(h, k)$ be the coordinates of the mid-point of the line $AB$ then it will intersect the coordinate axes at the points $A(2h, 0)$ & $B(0, 2k)$ respectively hence line $AB$ has x-intercept $2h$ & y-intercept $2k$,
Now, the equation of the line $AB$ is given by intercept form as $$\frac{x}{2h}+\frac{y}{2k}=1\tag 1$$
Now, since the above line $AB$ passes through the intersection point of the lines: $\frac{x}{a}+\frac{y}{b}=1$ & $\frac{x}{b}+\frac{y}{a}=1$ hence setting the coordinates of the intersection point $\left(\frac{ab}{a+b}, \frac{ab}{a+b}\right)$ in (1),
$$\frac{\frac{ab}{a+b}}{2h}+\frac{\frac{ab}{a+b}}{2k}=1$$ $$\frac{1}{h}+\frac{1}{k}=\frac{2(a+b)}{ab}$$ or $$\frac{h+k}{hk}=\frac{2(a+b)}{ab}$$ or $$2hk(a+b)=ab(h+k)$$ Now, setting $h=x$ & $k=y$ in the above equation, the locus of the mid-point of line $AB$ is given as follows $$\color{red}{2xy(a+b)=ab(x+y)}$$
The equation of any straight line passing through the intersection of the given two lines is $$ bx+ay-ab+K(ax+by-ab)=0$$ where $K$ is arbitrary constant
$$ x\dfrac1{\dfrac{ab(K-1)}{(b+Ka)}}+y\cdot\dfrac1{\dfrac{ab(K-1)}{a+Kb}}=1$$
If $(p,q)$ is the midpoint, $$2p=\dfrac{ab(K-1)}{(b+Ka)}$$ and $$2q=\dfrac{ab(K-1)}{a+Kb}$$
Equate the values of $K$ to eliminate this.
You have proceeded rightly. The equation of the variables line $AB $ is $(bx+ay-ab) + k (ax+by-ab) =0$
Then we get $A (\frac {ab (1+k)}{b + ak},0)$ and $B (0,\frac {ab (1+k)}{a +bk}) $. Let $(h,k) $ be the mid point of $AB $.
Thus $h =\frac {ab (1+k)}{2 (b+ak)} $ and $k=\frac {ab (1+k)}{2 (a+bk)} $. Now, $$\frac{1}{2}[\frac {1}{h}+\frac {1}{k}] = \frac {1}{2}[\frac {2 (b+ak)+2 (a+bk)}{ab (1+k)}] = \frac {a+b}{ab} $$ and thus the locus of the midpoint is $$\boxed {2xy(a + b) = ab(x + y)} $$ Hope it helps .