0
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .