Correct Answer - C
We have to find the locus of the point (h,k) whose image on the line 2x-y-1 =0 lies on the line y=x. Now, the image of (h,k) on the line 2x-y-1=0 is given by
`(x_(2)-h)/(2) = (y_(2)-h)/(-1)=-(2(2h-k-1))/(5)`
`"or "x_(2) = (-3h+4k+4)/(5)`
`"and "y_(2) = (4h+3k-2)/(5)`
The point lies on y=x. Then,
`(-3h+4k+4)/(5) = (4h+3k-2)/(5) " or "7h-k=6`