The general equation is y-2=a(x-2) because (2,2) satisfies this equation, so (2,2) is on the line. We can find the x- and y-intercepts and hence find a.
When x=0, y-intercept=2-2a. When y=0, -2=ax-2a, so the x-intercept is (2a-2)/a.
The sum of these intercepts is 2-2a+(2a-2)/a=9.
Multiply through by a: 2a-2a2+2a-2=9a, 2a2+5a+2=0=(2a+1)(a+2). Therefore a=-½ or -2. So there appear to be two line equations: y-2=-½(x-2) and y-2=-2(x-2), that is, 2y-4=-x+2, 2y=6-x, y=3-½x; and y-2=-2x+4, y=6-2x. (These lines intersect at (2,2) and the sum of their intercepts is 6+3=9.)
[The question details and this solution do not correspond to the title of the question, which contains no details of the parallelogram.]