$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{6-1}\implies \cfrac{-5}{5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-1(x-1) \\\\\\ y-4=-x+1\implies y=-x+5$

$\bf (\stackrel{x_1}{-12}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-14}{6-(-12)}\implies \cfrac{-15}{6+12}\implies \cfrac{-15}{18}\implies -\cfrac{5}{6}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-14=-\cfrac{5}{6}[x-(-12)] \\\\\\ y-14=-\cfrac{5}{6}(x+12)\implies y-14=-\cfrac{5}{6}x-10\implies y=-\cfrac{5}{6}x+4$

7 0

2 years ago