Question

A line passing through the point $P(-2,3)$ intersects the $x$ and $y$ axes at points $A$ and $B$ respectively. Let $Q$ be a point on $AB$ such that $PA,PQ,$ and $PB$ are in H.P, then the locus of point $Q$

• A. passes through the origin
• B. has a slope of $\frac{3}{2}$
• C. passes through the point $(2,3)$
• D. has a slope of $\frac{-3}{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A passes through the origin
B has a slope of $\frac{3}{2}$
C passes through the point $(2,3)$

The equation of line passing through $(-2,3)$ is $\frac{x+2}{\cos \theta }=\frac{y-3}{\sin \theta }$

The co-ordinates of $A,Q$ and $B$ are $(r_1\cos \theta -2,r_1\sin \theta +3)$ $(r_2\cos \theta -2,r_2\sin \theta +3)$ and $(r_3\cos \theta -2,r_3\sin \theta +3)$ respectively.

As $A$ lies on the $x-$axis and $B$ lies on the $y-$axis,
$\therefore r_1\sin \theta +3=0\text{and}{r}_3\cos \theta -2=0$
$\Rightarrow \sin \theta =-\frac{3}{r_1}\text{and}\cos \theta =\frac{2}{r_3}$

Let $(h,k)$ be the co-ordinates of $Q.$
$\Rightarrow h=r_2\cos \theta -2$ and $k=r_2\sin \theta +3$
$\Rightarrow h=\frac{2r_2}{r_3}-2$ and $k=\frac{-3r_2}{r_1}+3$
$\Rightarrow \frac{h+2}{2}=\frac{r_2}{r_3}$ and $\frac{-(k-3)}{3}=\frac{r_2}{r_1}$

As $r_1,r_2,r_3$ are in H.P.,
$\frac{2}{r_2}=\frac{1}{r_1}+\frac{1}{r_3}$

$\Rightarrow 2=\frac{r_2}{r_1}+\frac{r_2}{r_3}$

$\Rightarrow 2=\frac{-k+3}{3}+\frac{h+2}{2}$

$\Rightarrow 12=-2k+6+3h+6$
$\Rightarrow 3h-2k=0$
$\therefore$ The locus of $Q\text{is}3x-2y=0$