Question

Two fixed points A and B are taken on the coordinates axes such that OA = a and OB = b. Two variable points A' and B' are taken on the same axes such that OA' + OB' = OA + OB. Find the locus of the point of intersection of AB' and A'B.

Answer 1

Let $A=(a,0)$ and $B=(0,b)$

and $A^{\prime }=(a^{\prime },0)$ and $B^{\prime }=(0,b^{\prime })$

Now $A{B}^{\prime }=\frac{x}{a}+\frac{y}{b^{\prime }}=1$ ………$\left(1\right)$

and $A^{\prime }B=\frac{x}{a^{\prime }}+\frac{y}{b}=1$ …………$\left(2\right)$

Subtract $\left(1\right)$ from $\left(2\right)$

$x(\frac{1}{a}-\frac{1}{a^{\prime }})+y(\frac{1}{b^{\prime }}-\frac{1}{b})=0$

$x\left(\frac{a^{\prime }-a}{a{a}^{\prime }}\right)+y\left(\frac{b-b^{\prime }}{bb}\right)=0$

Now, it is given

$0A^{\prime }+O{B}^{\prime }=OA+OB$

$a^{\prime }+b^{\prime }=a+b$

$b^{\prime }-b=a-a^{\prime }$

$b-b^{\prime }=a^{\prime }-a$

$\therefore (a^{\prime }-a)[\frac{x}{a{a}^{\prime }}+\frac{y}{b{b}^{\prime }}]=0$

$\Rightarrow \frac{x}{a{a}^{\prime }}+\frac{y}{b{b}^{\prime }}=0$.