Given line segment AB with endpoints A(-9, 2) and B(12, 8) what are the coordinated of point c that is partitioned one third from A to B. PLEASE HELP.

check the picture below.

so, from A to B cut at 1/3, simply means, splitting the AB segment into 1+3 equal pieces, and from those four, AB takes 1 piece, and CB takes the other 3 pieces.

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(-9,2)\qquad B(12,8)\qquad \qquad 1:3 \\\\\\ \cfrac{AC}{CB} = \cfrac{1}{3}\implies \cfrac{A}{B} = \cfrac{1}{3}\implies 3A=1B\implies 3(-9,2)=1(12,8)\\\\ -------------------------------\\\\ { 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)}[/tex]

[tex]\bf -------------------------------\\\\ C=\left(\cfrac{(3\cdot -9)+(1\cdot 12)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 8)}{1+3}\right) \\\\\\ C=\left(\cfrac{-27+12}{4}~~,~~\cfrac{6+8}{4} \right)\implies C=\left(-\frac{15}{4}~~,~~\frac{7}{2} \right) \\\\\\ C=\left(-3\frac{3}{4}~~,~~3\frac{1}{2} \right)[/tex]