Question

Point A is located at (5, 10) and point B is located at (20, 25).

Answer 1

Let's say the point is C, so C partitions AB into two pieces, where AC is at a ratio of 3 and CB is at  a ratio of 7, thus 3:7,

$\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(5,10)\qquad B(20,25)\qquad \qquad 3:7 \\\\\\ \cfrac{AC}{CB} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(5,10)=3(20,25)\\\\ -------------------------------\\\\ { C=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)}\\\ -------------------------------$

$\bf C=\left(\cfrac{(7\cdot 5)+(3\cdot 20)}{3+7}\quad ,\quad \cfrac{(7\cdot 10)+(3\cdot 25)}{3+7}\right) \\\\\\ C=\left( \cfrac{35+60}{10}~~,~~\cfrac{70+75}{10} \right)\implies C=\left(\cfrac{95}{10}~~,~~\cfrac{145}{10} \right) \\\\\\ C=\left( \cfrac{19}{2}~~,~~\cfrac{29}{2} \right)\implies C=\left( 9\frac{1}{2}~~,~~14\frac{1}{2} \right)$