The equation of a line in explicit form is:

#y=mx+q#, where #m# is the slope and #q# the y-intercept.

It is easier to show the procedure with some example:

#y=2#: this line is parallel to the x-axis and it passes from the point #P(0,2)#.

#x=3#: this line is parallel to the y-axis and it passes from the point #P(2,0)#.

#y=x+1#: this line is parallel to the bisector of the I and III quadrants and it passes from the point #P(0,1)#.

graph{x+1 [-10, 10, -5, 5]}

#y=-x-1#: this line is parallel to the bisector of the II and IV quadrants and it passes from the point #P(0,-1)#.

graph{-x-1 [-10, 10, -5, 5]}

#y=2/3x+1#: we have to find the point #P(0,1)#, from this point we have to "count" 3 units to the right and then 2 units to the up, so we can find the point #Q(3,3), then we have to join the two point found.

graph{2/3x+1 [-10, 10, -5, 5]}

#y=-1/2x-1#: we have to find the point #P(0,-1)#, from this point we have to "count" 2 units to the left and then 2 units to the up, so we can find the point #Q(-2,0), then we have to join the two point found.

graph{-1/2x-1 [-10, 10, -5, 5]}

The difference in these two last examples is the "choice" of the "right" and the "left". Right, if the #m# is positive; left, if the #m# is negative.