Which ordered pairs could be points on a line parallel to the line that contains (3, 4) and (–2, 2)? check all that apply.

Step-by-step explanation:

As we go from (–2, 2) to  (3, 4), x increases by 5 and y increases by 4.  Thus, the slope of the line through (–2, 2) and (3, 4) is

m = rise / run = 4/5.

Use the slope-intercept form of the equation of a straight line:

y = mx + b becomes 4 = (4/5)(3) + b.  Multiplying all three terms by 5, we eliminate the fraction:  20 = 12 + b.  Thus, b = 8, and the equation of the line through (–2, 2) and (3, 4) is y = (4/5)x + 8.

A line parallel to this one would have the form  y = (4/5)x + b; note that the slopes of these two lines are the same, but the y-intercept, b, would be different if the two lines do not coincide.

Unfortunately, you have not shared the ordered pairs given in this problem statement.  

You could arbitrarily let b = 0.  Then the parallel line has equation

y = (4/5)x; if x = 3, then y = (4/5)(3) = 12/5, and so (3, 12/5) lies on the parallel line.

the answer would be c:   ax + b = ax + c

step-by-step explanation: