Correct Answer - A
Slope of given line x+y-1=0 is
`"tan" theta = -1`
`therefore theta = 135^(@)`, which is inclination of line with x-axis.
Point P(2,1) is on the line.
Coordinates of point Q on the line at distance `3sqrt(2)` from it with decreasing ordinate are
`(2-3sqrt(2) "cos" 135^(@), 1-3sqrt(2)"sin"135^(@))`
`-=(2-3sqrt(2)(-(1)/(sqrt(2))), 1-3sqrt(2)((1)/(sqrt(2))))`
`-=(5,-2)`
If (h,k) is image of pont Q in the line x+y-1 =0, then
`(h-5)/(1) = (k+2)/(1) = (-2(5-2-1))/(2)`
`therefore h=3, k=-4`