a) The easiest way to write an equation for this line is to use intercept form:
.. x/x-intercept +y/y-intercept = 1
.. x/9 +y/3 = 1
b) The point (x, y) = (4, 2) does not satisfy the equation. (see the graph) These cannot be the dimensions of an inscribed rectangle.
c) The intersection with the line y=x is (x, y) = (2.25, 2.25). These are the dimensions of an inscribed square.
d) The equation can be rewritten as
.. x +3y = 9
.. y = 3 -(x/3)
Then the area can be written as
.. A = xy = x(3 -x/3)
e) For x=6, the area is
.. A = 6(3 -6/3) = 6*1 = 6 . . . . square units
f) The parabola described by A has zeros at x=0 and x=9. The vertex of that parabola is on the line of symmetry, at x = 4.5. That value of x gives the maximum area.