[tex]\bf C(\stackrel{x_1}{3t-2}~,~\stackrel{y_1}{4})\qquad
D(\stackrel{x_2}{t+2}~,~\stackrel{y_2}{15})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{15-4}{(t+2)-(3t-2)}=\stackrel{slope}{-\cfrac{1}{4}}
\\\\\\
\cfrac{11}{t+2-3t+2}=-\cfrac{1}{4}\implies \cfrac{11}{4-2t}=-\cfrac{1}{4}\implies 44=2t-4
\\\\\\
48=2t\implies \cfrac{48}{4}=t\implies 12=t[/tex]
