Question 1027800
The desired line(s) should have the form 7y + x +c = 0, for some undetermined constant(s) c.

The formula fro the distance between a point and the line is given by {{{d = abs(7y[0]+x[0]+c)/sqrt(7^2+1^2) = abs(7*-2+4+c)/sqrt(50) = abs(c-10)/sqrt(50) = sqrt(2)}}}.

==> {{{abs(c-10)=sqrt(50)*sqrt(2) = sqrt(100) = 10}}}.

==> c - 10 = 10, or c - 10 = -10
==> c = 20, or c = 0.

Therefore there are two lines satisfying the given conditions, namely
7y + x + 20 = 0, or 7y + x = 0,

on both sides of the point (4,-2) and perpendicular to  y=7x+1.