Find the equation of the line perpendicular to y=x and passing at a distance 3sqrt2 from (4,1).

The equation of the line perpendicular to y = x is y = -x+
constant  or x+y+k = 0....(1)

If ax+by+c = 0 is any line ,
then its distance d from a point (x1,y1) is given by:

d =
(ax1+by1+c)|/(a^2+b^2).

Therefore if the the point (4,1)
and the line x+y+k are at distance of 3sqrt2,
then:

3*2^(1/2) = (4+1+k)/(1^2+1^2)^(1/2)
.

3*2^(1/2) =
(5+k)/2^(1/2).

3* 2(1/2)*2^(1/2) =
5+k.

3*2 = 5+k.

k = 6-5 =
1.

So we substitute k = in (1) and get : x+y+1 =
0

Therefore the required equation of the line
is x+y+1 = 0.