so ... notice the picture below... the ratio is 3:1 from C to D, meaning the segment CE takes 3 units, whilst the ED segment takes 1
anyhow
[tex]\bf \textit{internal division of a line segment}
\\ \quad \\\\
C(3,2)\qquad D(-5,-2)\qquad
ratio1=3\qquad ratio2=1\qquad 3:1
\\\\\\
\cfrac{C\underline{ E }}{\underline{ E }D}=\cfrac{ratio1}{ratio2}\implies
ratio2\cdot C=ratio1\cdot D
\\\\\\
1(3,2)=3(-5,-2)
\\ \quad \\\\
{{ E=\left(\cfrac{\textit{sum of "x" values}}{ratio1+ratio2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{ratio1+ratio2}\right)}}[/tex]
[tex]\bf \qquad thus
\\\\\\
E=\left(\cfrac{(1\cdot 3)+(3\cdot -5)}{3+1}\quad ,\quad \cfrac{(1\cdot 2)+(3\cdot -2)}{3+1}\right)[/tex]
