A variable straight line drawn through the point of intersection of the straight lines $\frac xa + \frac yb=1$ and $\frac xb + \frac ya=1$ meets the co ordinate axes at $A$ and $B$. Prove that the locus of the mid point of $AB$ is $$2xy(a+b)=ab(x+y)$$
I found the point of intersection but could not proceed any further. The point of intersection is $\left[\frac{ab}{(a+b)}, \frac{ab}{(a+b)}\right]$. So please solve from here onward
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 $$\color{red}{2xy(a+b)=ab(x+y)}$$