Find the point C such that AC and BC form a 2:3 ratio(-1, 1.2)(-0.6, 3)(0, 2.4)(0.5, 2)

[tex]\bf \left. \qquad \right.\textit{internal division of a line segment} \\\\\\ A(-3,5)\qquad B(3,0)\qquad \qquad 2:3 \\\\\\ \cfrac{AC}{CB} = \cfrac{2}{3}\implies \cfrac{A}{B} = \cfrac{2}{3}\implies 3A=2B\implies 3(-3,5)=2(3,0)\\\\ -------------------------------\\\\ { 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 -3)+(2\cdot 3)}{2+3}\quad ,\quad \cfrac{(3\cdot 5)+(2\cdot 0)}{2+3}\right) \\\\\\ C=\left( \cfrac{-9+6}{5}~~,~~\cfrac{15+0}{5} \right)[/tex]

and surely you know how much that is.