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

Options

A. `(5)/(sqrt(2))`
B. `(7)/(sqrt(2))`
C. `(3)/(sqrt(2))`
D. `(1)/(sqrt(2))`

Answer

C

Explanation

The equation of a line passing through A(2,4) is `(x-2)/(cos theta)=(y-4)/(sin theta)`
Suppose it cuts the line `x+y=9` at point P whose coordinates are given by
`(x-2)/(cos theta)=(y-4)/(sin theta)=r` I.e. `x=2+r cos theta, y=4+r sin theta`
`therefore 2+rcos theta+4+rsin theta=9`
`implies r(cos theta+sin theta)=3`
`implies r=(3)/(cos theta+sin theta)`
`implies r ge (3)/(sqrt(2)) [because cos theta + sin theta le sqrt(2) therefore (1)/(cos theta + sin theta ge (1)/(sqrt(2)))]`
Hence , the minimum value of AP is `(3)/(sqrt(2))`