Iforgot how to do this if you could with one that would be great !

If the relationship is parallel, you find the gradient/slope (m) in the equation y=mx+c. Then substitute the point and the gradient/slope into the equation y-y1=m(x-x1), after that you work it out algebraically making y the subject.

Question 1... Line: y=3x+5   Point: (1/3, 4)   Relationship: Parallel
m = 3      y-4=3(x-1/3)
               y-4=3x-1
                  y=3x+3

If the relationship is perpendicular then you find the gradient/slope (m) from the equation y=mx+c, after doing that you sub m into the equation m1*m2= -1 (solve algebraically to make m2 the subject). Then you substitute the answer of m2 and the points into the equation y-y1=m2(x-x1).

Question 5... Line y=-3x+5   Point: (3, 2)   Relationship: Perpendicular
m= -3
-3*m2= -1            y-2=1/3(x-3)
     m2= 1/3          y-2=1/3x-1
                               y=1/3x+1

I can send you the photo of the working if you wish, since it's a bit hard to read online.

Just a second zzz