Trisection of the line segment joining …

Let P and Q be the points of trisection as shown below

Then, AP : PB = 1 : 2
and AQ : QB = 2 : 1

1. 
   Here, {tex}\frac{m_{1}}{m_{2}}=\frac{1}{2},{/tex} A(x₁, y₁) = (2, -3) and B(x₂, y₂) = (4, -1)
   For internally, {tex}P=\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right){/tex}
   {tex}\Rightarrow \quad P=\left(\frac{1 \times 4+2 \times 2}{1+2}, \frac{1 \times(-1)+2 \times(-3)}{1+2}\right){/tex}
   {tex}\Rightarrow \quad P=\left(\frac{4+4}{3}, \frac{-1-6}{3}\right) \Rightarrow P=\left(\frac{8}{3}, \frac{-7}{3}\right){/tex}
2. Here, {tex}\frac{m_{1}}{m_{2}}=\frac{2}{1}{/tex}, A(x₁, y₁) = (2, -3)
   and B(x₂, y₂) = (4, -1)
   For internally, {tex}Q=\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right){/tex}
   
   {tex}\Rightarrow \quad Q=\left(\frac{2 \times 4+1 \times 2}{2+1}, \frac{2 \times(-1)+1 \times(-3)}{2+1}\right){/tex}
   {tex}\Rightarrow \quad Q=\left(\frac{8+2}{3}, \frac{-2-3}{3}\right) \Rightarrow Q=\left(\frac{10}{3}, \frac{-5}{3}\right){/tex}