in the equation 2x-3y=9 , 2x/3-3=y the slope is 2/3 and the poin (a)parallel to the given is the same to the slope parallel to this line so we can now subs. form the eqn y-y1=m(x-x1) so the eqn becomes y+1=2/3(x-4) or 3y+3=2x-8 or in gen.form 2x-3y-11=0 and (b) perpendicular to the line is -1 divided by the slope of the line which is 2/3 so the slope of the perpendicular line is -3/2 then the eqn perpendicular to the line is y+1=-3/2(x-4) or 2y+2=-3x+12 or in gen.form 3x-2y+10=0
