Suppose, I have three points: $A(2,5),B(7,9)$ & $C(5,\frac{37}{5})$. Now, I need to find the ratio in which $C$ divides $A$ & $B$. Now, let us use the internal section formula:
$$(x,y)=(\frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2})$$
where $(x,y)$ divides the line segment created by $(x_1,y_1)$ & $(x_2, y_2)$ internally in the ratio $m_1:m_2$. Now, let $C$ divides $A$ and $B$ in the ratio $m_1:m_2$. Now,
$$\frac{7m_1+2m_2}{m_1+m_2}=5$$
$$\implies7m_1+2m_2=5m_1+5m_2$$
$$\implies2m_1-3m_2=0...(i)$$
Again, $$\implies\frac{9m_1+5m_2}{m_1+m_2}=\frac{37}{5}$$
$$\implies45m_1+25m_2=37m_1+37m_2$$
$$\implies8m_1-12m_2=0$$
$$\implies2m_1-3m_2=0...(ii)$$
Here, the equations $(i)$ and $(ii)$ are the same. So, how am I supposed to find $m_1:m_2$? Is my process wrong?
Clearly both equations will be the same as the ratio that $C$ cuts $AB$ should be the same regardless of how the segment is projected.
Once you got $2m_1-3m_2=0$ you should have continued like $$2m_1=3m_2$$ $$\frac{m_1}{m_2}=\frac{3}{2}$$ $$m_1:m_2=\boxed{3:2}$$
The fact that there are infinite solutions to $2m_1-3m_2=0$ is because equivalent fractions/ratios can be made by multiplying the "numerator" and "denominator" both by a constant. Regardless, the value of the ratio between the two variable is always invariant despite the fact that there are infinite solutions.