Problem 9.7. On the coordinate plane, point $A$ is marked. On the $O x$ axis, point $B$ is marked, and on the $O y$ axis, point $C$ is marked.

It is known that the equations of the lines $A B, B C, A C$ in some order have the form $y=a x+4$, $y=2 x+b$ and $y=\frac{a}{2} x+8$ for some real numbers $a$ and $b$.

Find the sum of the coordinates of point $A$. Specify all possible values.
Answer: 13 or 20.

Solution. Let the graph of the function $y=a x+4$ be denoted by $l_{1}$, the function $y=2 x+b-$ by $l_{2}$, and the function $y=\frac{a}{2} x+8-$ by $l_{3}$.

From the condition, it follows that $l_{1}, l_{2}, l_{3}$ are pairwise intersecting lines (each of the points $A, B, C$ belongs to some two of them). Immediately note that the case $a=0$ is impossible, otherwise the lines $y=a x+4$ and $y=\frac{a}{2} x+8$ would be parallel.

Point $C$ lies on the $O y$ axis, as well as on two of the three given lines. This means that their equations, when substituting $x=0$, give the same values (equal to the ordinate of point $C$). Therefore, $C$ cannot be the intersection point of lines $l_{1}$ and $l_{3}$.

It is easy to see that the line $l_{1}$ intersects the $O y$ axis at the point $-\frac{4}{a}$, the line $l_{2}$ at the point $-\frac{b}{2}$, and the line $l_{3}$ at the point $\frac{-16}{a}$. Among these three numbers, there are two identical ones (equal to the abscissa of point B). Clearly, $-\frac{4}{a} \neq \frac{-16}{a}$, so there are two possible cases.

- Suppose $-\frac{4}{a}=-\frac{b}{2}$, i.e., $l_{1}$ and $l_{2}$ intersect at point $B$. Then $C$ is the intersection point of lines $l_{2}$ and $l_{3}$, i.e., their equations at $x=0$ coincide, and $b=8$. From the equality $-\frac{4}{a}=-\frac{b}{2}$, we get that $a=1$. Then $A$ is the intersection point of $l_{1}$ and $l_{3}$, which are the graphs of the functions $y=x+4$ and $y=\frac{1}{2} x+8$. It is easy to understand that these two lines intersect at the point $(8 ; 12)$, so the answer in this case is $8+12=20$.
- Suppose $-\frac{b}{2}=-\frac{16}{a}$, i.e., $l_{2}$ and $l_{3}$ intersect at point $B$. Then $C$ is the intersection point of lines $l_{1}$ and $l_{2}$, i.e., their equations at $x=0$ coincide, and $b=4$. From the equality $-\frac{b}{2}=-\frac{16}{a}$, we get that $a=8$. Then $A$ is the intersection point of $l_{1}$ and $l_{3}$, which are the graphs of the functions $y=8 x+4$ and $y=4 x+8$. It is easy to understand that these two lines intersect at the point $(1 ; 12)$, so the answer in this case is $1+12=13$.