drop a perpendicular from P on the line (let point of intersection be A).

(this is the shortest distance between P and line)

now from P draw a line parallel to the given line and take an arbitrary point B on it.(take initial point of line as P)

now angle APB=90 degrees.

if this drawn line want to touch the other line then angle APB must be smaller then 90 degrees.

suppose we draw a line at the angle which is just smaller than 90 degrees.

let this angle be x degrees.

but since an angle greater than x but smaller than 90 can still be obtained which may be given by

(x+90)/2

hence a point of intersection (T) belonging to l can always be obtained such that PT>PT(o)

or

no point T(o) exist such that PT(o)>PT.

2) let the equation of line be

ax+by+c=0

let coordinates of P be (r,t)

now let the coordinates of perpendicular from P on line be (m,n)

suppose there is a point T(o) whose distance from T is maximum.

Let its coordinates be (M,N)

but another point T can be obtained on l whose coordinates can be given by

(m+M+c/a,n+N) or (m+M,n+N+c/b) whose distance from P i.e PT will be greater than PT(o)

so

there is no point T(o) belonging to l such that PT(o)>PT