Question

$P$ is the point $(-1,2)$, a variable line through $P$ cuts the co-ordinate axes in $A$ and $B$ respectively. $Q$ is a point on $AB$ such that $PA,PQ$ and $PB$ are in $H.P.$ Show that the locus of $Q$ is the line $y=2x$.

Answer 1

Any line through $P(-1,2)$ is
$\frac{x+1}{\cos \theta }=\frac{y-2}{\sin \theta }=r_1,r_2,r_3$
where $r_1,r_2,r_3$ are respectively the distance of the points $A,Q$ and $B$ from the given point $P(-1,2)$.
Point $A$ is $(r_1\cos \theta -1,r_1\sin \theta +2)$
where $r_1\sin \theta +2=0$
as the point $A$ lies on $x-axis$.
Point $B$ is $(r_3\cos \theta -1,r_3\sin \theta +2)$
where $r_3\cos \theta -1=0$
as the point $B$ lies on $y-axis$
Let the point $Q$ be $(h,k)$.
$\therefore h=r_2\cos \theta -1$
and $k=r_2\sin \theta +2$
We have to find the locus of the point $Q$.
Also it is given that $r_1,r_2,r_3$ are in $H.P.$
$\therefore \frac{2}{r_2}=\frac{1}{r_1}+\frac{1}{r_3}=-\frac{\sin \theta }{2}+\cos \theta$, by $\left(1\right)$and $\left(2\right)$
or $\frac{2}{r_2}=-\frac{1}{2}\frac{(k-2)}{r_2}+\frac{(h+1)}{r_2}$, by $\left(3\right)$
or $2=(-k/2)+1+h+1$
or $k=2h$ or $y=2x$