Question

Line CD has a slope of -1/4. If point C is at (3t - 2, 4) & point D is at (t + 2, 15), find the value of t

Answer 1

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