8,9

On the plane, points $A(1 ; 2), B(2 ; 1), C(3 ;-3), D(0 ; 0)$ are given. They are the vertices of a convex quadrilateral $A B C D$. In what ratio does the point of intersection of its diagonals divide the diagonal $A C$?
Since the lines $A C$ and $B D$ are not parallel to the $O Y$ axis, their equations can be found in the form $y=a x+b$. Substituting the coordinates of points $A$ and $C$ into this equation, we get the system

$$
\left\{\begin{array}{l}
2=a+b \\
-3=3 a+b
\end{array}\right.
$$

from which we find that $a=-\frac{5}{2}, b=\frac{9}{2}$. Similarly, we obtain the equation of the line $B D: y=\frac{1}{2} x$. The coordinates of the point $M$ of intersection of the lines $A C$ and $B D$ are found from the system

$$
\left\{\begin{array} { l } 
{ y = - \frac { 5 } { 2 } x + \frac { 9 } { 2 } } \\
{ y = \frac { 1 } { 2 } x }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{3}{2} \\
y=\frac{3}{4}
\end{array}\right.\right.
$$

Consider the vectors $\overrightarrow{A M}$ and $\overrightarrow{M C}$. Since the coordinates of a vector are equal to the differences of the corresponding coordinates of the end and the start, then

$$
\overrightarrow{A M}=\left(\frac{3}{2}-1 ; \frac{3}{4}-2\right)=\left(\frac{1}{2} ;-\frac{5}{4}\right), \overrightarrow{M C}=\left(3-\frac{3}{2} ;-3-\frac{3}{4}\right)=\left(\frac{3}{2} ;-\frac{15}{4}\right)
$$

From the obtained equalities, it follows that $\overrightarrow{M C}=3 \overrightarrow{A M}$. Therefore, the point $M$ divides the diagonal $A C$ in the ratio $3: 1$, counting from point $A$.

## Answer

$1: 3$, counting from point $A$.