34. Divide the segment $A B$, given by the points $A(5 ; 1)$ and $B(-4 ;-14)$, into three equal parts.
Solution. Let $M$ and $N$ be the points of division (Fig. 40). We will form the ratios for these points. We have $|A M|:|M B|=1: 2$, i.e., $\lambda_{M}=1 / 2 ; \quad|A N|:|N B|=2: 1$, i.e., $\lambda_{N}=2$. Now, substituting these ratios and the coordinates of points $A$ and $B$ into formula (3), we find

$$
x_{M}=\frac{x_{A}+\lambda_{M} x_{B}}{1+\lambda_{M}}=\frac{5+\frac{1}{2}(-4)}{1+\frac{1}{2}}=2
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-124.jpg?height=291&width=191&top_left_y=1516&top_left_x=962)

Fig. 40

$$
\begin{gathered}
y_{M}=\frac{y_{A}+\lambda_{M} y_{B}}{1+\lambda_{M}}=\frac{1+\frac{1}{2}(-14)}{1+\frac{1}{2}}=-4, \text { i.e., } M(2 ;-4) \\
\qquad x_{N}=\frac{x_{A}+\lambda_{N} x_{B}}{1+\lambda_{N}}=\frac{5+2(-4)}{1+2}=-1 \\
y_{N}=\frac{y_{A}+\lambda_{N} y_{B}}{1+\lambda_{N}}=\frac{1+2(-14)}{1+2}=-9, \text { i.e., } N(-1 ;-9)
\end{gathered}
$$