Question 855083
Your line does not make a nice graph.
I like to include the origin, and pieces of the x-axis and y-axis to both sides.
In this case, including the y-intercept (0,-33/2) would be difficult.
Even including point (9,-3) may be tricky.
You only need two points (or one point and the slope) to draw a line, and you could use any point(s) for your line.
in line {{{y=(3/2)x-33/2}}} , for {{{x=9}}} --> {{{y=-3}}} (point (9,-3) as required),
and a slope of {{{3/2}}} means that
as {{{x}}} increases by {{{2}}} , {{{y}}} increases by {{{3}}} .
That gives us points (11,0), (13,3), (15,6).
For line {{{y=(-2/3)x+2}}} we could use point (0,2), and a slope of {{{-2/3}}} would take us to (3,0), (6,-2), and (9,-5)
{{{drawing(300,300,-2,18,-6,14,
grid(1),blue(line(-3,4,12,-6)),
blue(circle(0,2,0.3)),blue(circle(6,-2,0.3)),
green(line(5,-9,19,12)),
green(circle(9,-3,0.4)),green(circle(11,0,0.4)),
green(circle(13,3,0.4))
)}}}