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Find point P that divides segments AB into a 2:3 ratio.

The point should be closer to A. A(-3,1) and B(3,5) P=

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\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(-3,1)\qquad B(3,5)\qquad \qquad \stackrel{\textit{ratio from A to B}}{2:3} \\\\\\ \cfrac{A\underline{P}}{\underline{P} B} = \cfrac{2}{3}\implies \cfrac{A}{B} = \cfrac{2}{3}\implies 3A=2B\implies 3(-3,1)=2(3,5)\\\\ -------------------------------\\\\ P=\left(\cfrac{\textit{sum of


\bf -------------------------------\\\\ P=\left(\cfrac{(3\cdot -3)+(2\cdot 3)}{2+3}\quad ,\quad \cfrac{(3\cdot 1)+(2\cdot 5)}{2+3}\right) \\\\\\ P=\left(\cfrac{-9+6}{5}~~,~~\cfrac{3+10}{5} \right)\implies P=\left(-(3)/(5)~~,~~(13)/(5) \right)
User Hussain Rahimi
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