The line segment joining the points $\mathrm{A}(3,-4)$  and $\mathrm{B}(1,2)$  is trisected at the points $\mathrm{P}$ and $\mathrm{Q}$ , where $\mathrm{P}$ is nearer to $\mathrm{A}$ . Find the coordinates of the point $\mathrm{P}$ .

Given: The line segment joining the points $\mathrm{A}(3,-4)$  and $\mathrm{B}(1,2)$  is trisected at the point P and Q, where P is nearer to A. 
We have to find the coordinates of the point P by using the section formula.
Given that the point P is the point of trisection of the line segment AB, so P divides AB in the ratio is 1:2 That means $\mathrm{m}=1$  and $\mathrm{n}=2$. 

The points are $\mathrm{A}(3,-4)$  and $\mathrm{B}(1,2)$.

So, ${\mathrm{x}}_1=3, {\mathrm{y}}_1=-4$  and ${\mathrm{x}}_2=1, {\mathrm{y}}_2=2$  

Now, first, we find the value of $\mathrm{x}$ using the section formula,

$\mathrm{x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}$

$\mathrm{x}=\frac{1\left(1\right)+2\left(3\right)}{1+2}=\frac{7}{3}$
   
Now, we find the value of $\mathrm{y}$,

$\mathrm{y}=\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}$

$\mathrm{y}=\frac{1\left(2\right)+2(-4)}{1+2}=-2$ 
Therefore, the coordinates of the point P is, $\mathrm{P}(\mathrm{x},\mathrm{y})=(\frac{7}{3},-2)$ .