Write the equation of the line that passes through (–1, 5) and has a slope of 3 in point-slope form.
a: x + 1 = 3(y – 5)
b: x – 5 = 3(y + 1)
c: y – 5 = 3(x + 1)
d: y + 1 = 3(x – 5)

Well first we eliminate the the ones that does not have a slope of 3.
In this case is A and B
A: x + 1 = 3(y – 5)
    x=3(y-5)-1
    x/3= y-5-1
    x/3 = y-6
  y= 1/3x+6
B: x – 5 = 3(y + 1)
     x=3(y+1)+5
    x/3= y+1+5
    x/3 = y+11
  y= 1/3x-11

So we are left with C or D
C: y – 5 = 3(x + 1)
    y= 3(x+1)+5
     y= 3x+3+5
     y= 3x+8
D: y + 1 = 3(x – 5)
    y= 3(x-5)-1
     y= 3x-15-1
     y= 3x-16

To check wich one have the line that passes though (-1,5) we put -1 as x and y as 5 so lets check
  C.   5= 3(-1)+8
         5= -3+8
         5= 5
D.     5= 3(-1)-16
        5= -3-16
        5= -19
Well since 5 is not equal to 19 we can assume that you answer is C were 5 is equal to 5.

There is no association.