Question

A rectangle PQRS has its side PQ parallel to the line `y= mx` and vertices `P, Q`, 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 at `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`
`implies(alpha-a)/(b+h)=m`.........`(i)`
Again `RQbotPQ`

Slope of `RQ=-(1)/(m)implies(k-alpha)/(h-b)=-(1)/(m)`.........`(ii)`
From Eq. `(i)`, we get
`alpha-a=m(b+h)`
`impliesalpha=a+m(b+h)`..........`(iii)`
and from Eq. `(ii)`, we get
`k-alpha=-(1)/(m)(h-b)`
`impliesalpha=k+(1)/(m)(h-b)`.......`(iv)`
From Eqs. `(iii)` and `(iv)` , we get
`a+m(b+h)=k+(1)/(m)(h-b)`
`impliesam+m^(2)(b+h)=km+(h-b)`
`implies(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`