Let a given line `L_1` intersect the X and Y axes at P and Q respectively. Let another line `L_2` perpendicular to `L_1` cut the X and Y-axes at Rand S, respectively. Show that the locus of the point of intersection of the line PS and QR is a circle passing through the origin

Explanation

Let the equation of `L_)1)` be `xcosalpha+ysinalpha=P_(1)`
Then, any line perpendicular to `L_(1)` is

`xsinalpha-ycosalpha=p_(2)`
where, `p_(2)` is a variable.
Then, `L_(1)` meets X-axis at P `(p_(1) sec alpha , 0)` and Y-axis at `Q (0, p_(1)cosecalpha)`.
Similarly, `L_(1)` meets X-axis at `R(p_(2) cosec alpha, 0)` and Y-axis at S `(0,-p_(2)secalpha)`.
Now, equation of PS is,
`(x)/(p_(1)secalpha)+(y)/(-p_(2)secalpha)=1rArr (x)/p_(1)-(y)/(p_(2))=secalpha" "...(i)`
Similarly, equation of QR is
`(x)/(p_(2)cosecalpha)+(y)/(p_(1)cosecalpha)=1rArr (x)/p_(2)-(y)/(p_(1))=cosecalpha" "...(ii)`
Locus of point of intersection of PS and QR can be obtained by eliminating the variable `p_(2)` from Eqs, (i) and (ii).
`therefore
((x)/(p_(1))-secalpha)(x)/(y)+(y)/(p_(1))=cosecalpha`
`rArr (x-p_(1)secalpha)x+y^(2)=p_(1)y" cosec "alpha`
`rArr x^(2)+y^(2)-p_(1)xsecalpha-p_(1)y " cosec "alpha=0`
which is a circle through origin.