# Reason behind alternate answers to a locus problem

The problem :Locus of mid point of a segment (I don't have the required reputation to ask my question regarding an answer here in the comments)

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 $$2xy(a + b) = ab(x + y)$$.

The solution :

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 using the intercept form as $$\frac{x}{2h}+\frac{y}{2k}=1\tag 1$$

Now, since the line $$AB$$ passes through the intersection of the lines: $$\frac{x}{a}+\frac{y}{b}=1$$ & $$\frac{x}{b}+\frac{y}{a}=1$$ hence the coordinates of the intersection point are $$\left(\frac{ab}{a+b}, \frac{ab}{a+b}\right)$$ which can be substituted into (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, substitute $$h=x$$ & $$k=y$$ in the above equation, the locus of the mid-point of line $$AB$$

is given as follows $$\color{red}{2xy(a+b)=ab(x+y)}$$

My Question :

If the problem is attempted using $$(h/2,k/2)$$ as the coordinates of the mid-point of $$AB$$ then the result is as follows :$$\color{red}{xy(a+b)=ab(x+y)}$$ (if I use $$h=x$$ and $$y=k$$).

I am unable to get my head around why $$x=2h$$ and $$y=2k$$ in this scenario?

The locus is formed by the midpoint, which in your case is $$(h/2,k/2)$$. Hence you must set $$x=h/2$$ and $$y=k/2$$, giving the same equation as before.
With $$x=h$$ and $$y=k$$ you get the equation of another locus.