check the picture below.
[tex]\bf \left. \qquad \right.\textit{internal division of a line segment}
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a(-1,2)\qquad b(7,14)\qquad
\qquad 1:3
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\cfrac{aP}{Pb} = \cfrac{1}{3}\implies \cfrac{a}{b} = \cfrac{1}{3}\implies 3a=1b\implies 3(-1,2)=1(7,14)
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{ P=\left(\cfrac{\textit{sum of "x" values}}{r1+r2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{r1+r2}\right)}[/tex]
[tex]\bf -------------------------------\\\\
P=\left(\cfrac{(3\cdot -1)+(1\cdot 7)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 14)}{1+3}\right)
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P=\left( \cfrac{-3+7}{4}~,~\cfrac{6+14}{4} \right)[/tex]
