we know that AC is 3/4 of the way from AB, meaning, if we cut AB in 4 equal pieces, AC takes 3 of those, and CB takes 1, meaning, AC and CB are on a 3:1 ratio.
[tex] \bf ~~~~~~~~~~~~\textit{internal division of a line segment}
\\\\\\
A(-10,-6)\qquad B(6,2)\qquad
\qquad \stackrel{\textit{ratio from A to B}}{3:1}
\\\\\\
\cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{1}\implies \cfrac{A}{B} = \cfrac{3}{1}\implies 1A=3B\implies 1(-10,-6)=3(6,2)\\\\[-0.35em]
~\dotfill\\\\
C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em]
~\dotfill [/tex]
[tex] \bf C=\left(\cfrac{(1\cdot -10)+(3\cdot 6)}{3+1}\quad ,\quad \cfrac{(1\cdot -6)+(3\cdot 2)}{3+1}\right)
\\\\\\
C=\left( \cfrac{-10+18}{4}~~,~~\cfrac{-6+6}{4} \right)\implies C=\left( \cfrac{8}{4}~,~\cfrac{0}{4} \right)\implies C=(2,0) [/tex]
