Question

A rectangle $PQRS$ has its side $PQ$ parallel to the line $y=mx$ and vertices $PQ$ and $S$ on the lines $y=a,x=b$ and $x=-b$ , respectively. Find the locus of the vertex $R$.

Answer 1

Let the coordinates of $Q$ be $(b,\alpha )$and that of $S$ be $(-b,\beta ).$ Suppose, $PR$ and $SQ$ meet in $G$. Since, $G$ is mid point of $SQ$, its $x$-coordinate must be $0$. Let the coordinates of $R$ be $(h,k)$.
Since, $G$ is mid point of $PR$, the $x$-coordinate of $P$ must be $-h$ and as $P$ lies on the line $y=a$, the coordinates of $P$ are $(-h,a)$. Since, $PQ$ is parallel to $y=mx$, slope of $PQ=m$
$\Rightarrow \frac{\alpha -a}{b+h}=m...(i)$
Again, $RQ\perp PQ$

Slope of $RQ=-\frac{1}{m}$
$\Rightarrow \frac{k-\alpha }{h-b}=-\frac{1}{m}...(ii)$
From Eq. (i), we get
$\alpha -a=m(b+h)$
$\Rightarrow \alpha =a+m(b+h)...(iii)$
and from Eq. (ii), we get
$k-\alpha =-\frac{1}{m}(h-b)$
$\Rightarrow \alpha =k+\frac{1}{m}(h-b)...(iv)$
From Eqs. (iii) and (iv), we get
$a+m(b+h)=k+\frac{1}{m}(h-b)$
$\Rightarrow am+m^2(b+h)=km+(h-b)$
$\Rightarrow (m^2-1)h-mk+b(m^2+1)+am=0$
Hence, the locus of vertex is
$(m^2-1)x-my+b(m^2+1)+am=0$