Question

6) Without graphing, what is the equation of the line that passes through the points (-4, 3) and (6,-2)? ​

Answer 1

`(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-2}-\stackrel{y1}{3}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-4)}}} \implies \cfrac{-5}{6 +4} \implies \cfrac{ -5 }{ 10 } \implies - \cfrac{ 1 }{ 2 }`

`\begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{- \cfrac{ 1 }{ 2 }}(x-\stackrel{x_1}{(-4)}) \implies y -3 = - \cfrac{ 1 }{ 2 } ( x +4) \\\\\\ y-3=- \cfrac{ 1 }{ 2 }x-2\implies {\Large \begin{array}{llll} y=- \cfrac{ 1 }{ 2 }x+1 \end{array}}`