line through the point $A (2, 4)$ intersects the line $x + y = 9$ at the point $P.$ The minimum distance of $AP,$ is

The equation of a line passing through $A(2, 4)$ is $\frac{x-2}{\cos \theta }=\frac{y-4}{\sin \theta }$

Suppose it cuts the line $x + y = 9$ at point $P$ whose coordinates are given by

  $\frac{x-2}{\cos \theta }=\frac{y-4}{\sin \theta }=r$ i.e. $x=2+r\cos \theta ,y=4+r\sin \theta$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2+r\cos \theta +4+r\sin \theta =9\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}r(\cos \theta +\sin \theta )=3\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}r=\frac{3}{\cos \theta +\sin \theta }\end{array}$

$\Rightarrow r\ge \frac{3}{\sqrt{2}}\left[\because \cos \theta +\sin \theta \le \sqrt{2}\therefore \frac{1}{\cos \theta +\sin \theta }\ge \frac{1}{\sqrt{2}}\right]$

Hence, the minimum value of $AP$ is $\frac{3}{\sqrt{2}}$