\(1=a+q\\ q=1-a\)
\(\frac{-1}{3}=ab^{-1}+q\\ \frac{-1}{3}=\frac{a}{b}+q \qquad \qquad b\ne0 \\ \frac{-1}{3}=\frac{a}{b}+1-a\\ 3b(\frac{-1}{3})=3b(\frac{a}{b}+1-a)\\ -b=3a+3b-3ab\\ 0=3a+4b-3ab \qquad \text{a and b both equal 0 or}\\ -3a(1-b)=4b\\ 3a(b-1)=4b\\ a=\frac{4b}{3(b-1)} \qquad \qquad b\ne1\\ \)
If b =1 then
\(g(x)=a+q\\ g(x)=a+1-a\\ g(x)=1 \)
but then point G is not on the curve so b cannot equal 1
So we have
.\(q=1-a\\ a=\frac{4b}{3(b-1)}\\ where \;\; b\ne1 \;\;\;\; b\ne0\)
PLUS, I am not so sure that b can be negative........
The graph would be very broken if b was negative, I am not sure how much that matters.
Maybe another mathematician would like to comment here.
Anyway ,...There is no single solution for this.
Here is the graph.
https://www.desmos.com/calculator/bqqgr7eyms
Here is the graph.