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
$(m^2 - 1) x - my + b (m^2 + 1) + am = 0$