find distance of the point(1,2) from the line 4x + 7y +18= 0 along the line x-y=8.

equation of line pependicular to the 4x+7y+18=0

and passing through (1,2)

7x-4y=-1

perpendicular distance of (1,2) from the line 4x+7Y=8

36/√65

angle between x-y=8 and 7x-4y=-1

Φ=tan^-1[(7/4-1)/1+7/4]

  =tan^-1[3/11]

distance of (1,2) from the line 4x+7y+18=0 along x-y=8

=rcosΦ=36/√65

r=36/√65*11/√130

=36*11/65√2