(!) 11. Formulate and solve a similar problem when there are $n$ material points (P. 145).

(!*) 12. Let $\angle A_{1}^{\prime} O A_{2}^{\prime}$ be one of the angles formed by two intersecting lines $a_{1}$ and $a_{2}$ (Fig. 18). From an arbitrary point $P$ inside this angle, we drop perpendiculars $P B_{1}$ and $P B_{2}$ to the lines $a_{1}$ and $a_{2}$. Let $k_{1}$ and $k_{2}$ be some given positive numbers, and $d$ be a given segment. Find the geometric locus of such points $P$ (lying inside the given angle $\angle A_{1}^{\prime} O A_{2}^{\prime}$) for which

$$
k_{1} \cdot P B_{1} + k_{2} \cdot P B_{2} = d . \quad(\mathrm{Y} .121, \mathrm{P} .146 .)
$$
11. Let $n$ material points $A_{1} \equiv\left(A_{1}, m_{1}\right), \boldsymbol{A}_{2} \equiv$ $\equiv\left(A_{2}, m_{2}\right), \ldots, A_{n} \equiv\left(A_{n}, m_{n}\right)$ be located in one plane on the same side of some line $s$. Let their distances from the line $s$ be $y_{1}, y_{2}, \ldots, y_{n}$, respectively. We need to find the distance $z$ of the center of mass $Z$ of this system from the line $s$.

We introduce the following notations:

Let $\boldsymbol{B}_{2}$ denote the union of material points $A_{1}$ and $A^{2}$; let $B_{s}$ denote the union of material points $\boldsymbol{B}_{2}$ and $A_{8}$;

let $\boldsymbol{B}_{n-1}$ denote the union of material points

$$
\text { points } B_{n-2} \text { and } A_{n-1} \text {; }
$$

let $Z$ denote the union of material points

$$
\text { points } B_{n-1} \text { and } A_{n} \text {. }
$$

Then, by the solution of problem 9,

$$
\begin{aligned}
& \text { Mom }_{s} \boldsymbol{B}_{2}=\text { Mom }_{s} \boldsymbol{A}_{1}+\text { Mom }_{s} \boldsymbol{A}_{2}, \\
& \text { Mom }_{s} \boldsymbol{B}_{3}=\text { Mom }_{s} \boldsymbol{B}_{2}+\text { Mom }_{s} \boldsymbol{A}_{3}, \\
& \operatorname{Mom}_{s} \boldsymbol{B}_{n-1}=\operatorname{Mom}_{s} \boldsymbol{B}_{n-2}+\operatorname{Mom}_{s} \boldsymbol{A}_{n-1}, \\
& \operatorname{Mom}_{s} \boldsymbol{Z}=\text { Mom }_{s} \boldsymbol{B}_{n-1}+\text { Mom }_{s} \boldsymbol{A}_{n} \text {. }
\end{aligned}
$$

By adding these equations term by term, we obtain after simplifications

$$
\text { Mom }_{s} Z=\text { Mom }_{s} A_{1}+\text { Mom }_{s} A_{2}+\ldots+\text { Mom }_{s} A_{n}
$$

or

$$
\left(m_{1}+m_{2}+\ldots+m_{n}\right) z=m_{1} y_{1}+m_{2} y_{2}+\ldots+m_{n} y_{n} .
$$

Hence,

$$
z=\frac{m_{1} y_{1}+m_{2} y_{2}+\ldots+m_{n} y_{n}}{m_{1}+m_{2}+\ldots+m_{n}} .
$$

Formula (1) shows that the moment of a system of $n$ material points lying in one plane and on the same side of some line, relative to this line, is equal to the moment of their union relative to the same line.