Question

Given $n$ straight lines and a fixed point $O$. Through $O$ is drawn a straight lines meeting these lines in the points $A_1,A_2,......A_n$ and on it is taken point $A$ such that $\frac{n}{OA}=\frac{1}{{OA}_1}+\frac{1}{{OA}_2}+\frac{1}{{OA}_3}+.....+\frac{1}{{OA}_n}$.
Prove that the locus of the point $A$ is a straight line.

Answer 1

Let the fixed point $O$ be chosen as origin and any straight line through $O$ be
$\frac{x-0}{\cos \theta }=\frac{y-0}{\sin \theta }=\underset{A_i}{r_i},\underset{A}{r}$
$\therefore O{A}_i=r_i,OA=r$
Again let the given $n$ straight lines be
$p_{i}x+q_{i}y=1(i=1,2,3,......n)$
The point $A_i$ is $(r_i\cos \theta ,r_i\sin \theta )$ and it lies on the above line
$\therefore (p_i\cos \theta +q_i\sin \theta )r_i=1$
$\therefore \frac{1}{O{A}_i}=\frac{1}{r_i}=p_i\cos \theta +q_i\sin \theta$
$\therefore \sum _{i=1}^n\frac{1}{O{A}_i}=\sum _{i=1}^n{p}_i\cos \theta +\sum _{i=1}^n{q}_i\sin \theta =\frac{n}{r}$
$\therefore \sum _{i=1}^n\frac{p_i}{n}\left(r\cos \theta \right)+\sum _{i=1}^n\frac{q_i}{n}\left(r\sin \theta \right)=1$
Therefore locus of the point $A(r\cos \theta ,r\sin \theta )$ is
$\frac{\sum p_i}{n}x+\frac{\sum q_i}{n}y=1$
Above equation represents a straight line.