5. Given $A\left(\frac{1}{3}, \frac{1}{a}\right), B\left(\frac{1}{4}, \frac{1}{b}\right), C\left(\frac{1}{5}, \frac{1}{c}\right)$ satisfy $\frac{a}{b+c}=\frac{1}{3}, \frac{b}{a+c}=\frac{1}{2}$. Then the positions of points $A, B, C$ are suitable for
(A) on the same straight line
(B) forming an acute triangle
(C) forming a right triangle
(D) forming an obtuse triangle
5. A.

From $\frac{a}{b+c}=\frac{1}{3}, \frac{b}{a+c}=\frac{1}{2}$, we get the following by combining ratios:
$$
\begin{array}{c}
\frac{a}{a+b+c}=\frac{1}{4}=\frac{3}{12}, \\
\frac{b}{a+b+c}=\frac{1}{3}=\frac{4}{12} . \\
\text { Therefore, }\left\{\begin{array}{l}
a=3 t, \\
b=4 t, \\
a+b+c=12 t .(t \neq 0)
\end{array}\right.
\end{array}
$$

Thus, $\left\{\begin{array}{l}a=3 t, \\ b=4 t, \\ a+b+c=12 t .(t \neq 0)\end{array}\right.$
which means $a=3 t, b=4 t, c=5 t$.
Therefore, $A, B, C$ are all on the line $y=\frac{x}{t}$.