Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 9

The line has equation $2x-y+2=0$, with normal vector $(2,-1)$. The distance from a point to a line is the absolute value of the normal component of the displacement:The perpendicular foot is $(4,-6)-(16/5)(2,-1)=(-12/5,-14/5)$, which lies on the line and realizes this shortest distance.