The distance of the point (1,2) from the line `2x-3y-4=0` in the direction of the line `x+y=1` , is

Options

A. `sqrt(2)`
B. `5sqrt(2)`
C. `(1)/(sqrt(2))`
D. none of these

Answer

A

Explanation

The equation of a line passing through P(1,1) and parallel to `x+y=1`, is
`(x-1)/("cos"(3pi)/(4))=(y-1)/("sin"(3pi)/(4))`
Let PM=r. Then , the coordinates of M are given by

`(x-1)/(cos(3pi)/(4))=(y-1)/(sin(3pi)/(4))=r` or , `(1-(r)/(sqrt(2)),1+(r)/(sqrt(2)))`
`because` M lies on `2x-3y-4=0`
`therefore 2(1-(r)/(sqrt(2)))-3(1+(r)/(sqrt(2)))-4=0`
`implies 2-(2r)/(sqrt(2))-3-(3r)/(sqrt(2))-4=0 implies-5-(5r)/(sqrt(2))=0impliesr=-sqrt(2)`
Hence , PM =`sqrt(2)`