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Find an equation of the line containing the given points. Use function notation to write the equation. (2/7, 3/7) and (-1/7, 13/14)

Respuesta :

[tex]\bf (\stackrel{x_1}{\frac{2}{7}}~,~\stackrel{y_1}{\frac{3}{7}})\qquad (\stackrel{x_2}{-\frac{1}{7}}~,~\stackrel{y_2}{\frac{13}{14}}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\frac{13}{14}-\frac{3}{7}}{~~-\frac{1}{7}-\frac{2}{7}~~}\implies \cfrac{~~\frac{13-6}{14}~~}{-\frac{3}{7}} \\\\\\ \cfrac{~~\frac{7}{14}~~}{-\frac{3}{7}}\implies \cfrac{7}{14}\cdot \cfrac{-7}{3}\implies \cfrac{1}{2}\cdot \cfrac{-7}{3}\implies -\cfrac{7}{6}[/tex]


[tex]\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-\cfrac{3}{7}=-\cfrac{7}{6}\left(x-\cfrac{2}{7} \right) \\\\\\ y-\cfrac{3}{7}=-\cfrac{7}{6}x+\cfrac{1}{3}\implies y=-\cfrac{7}{6}x+\cfrac{1}{3}+\cfrac{3}{7}\implies y=-\cfrac{7}{6}x+\cfrac{7+9}{21} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{7}{6}x+\cfrac{16}{21}~\hfill[/tex]


as for the 1/2 and the 1/3 parts


[tex]\bf \cfrac{7}{14}\implies \cfrac{7\cdot 1}{7\cdot 2}\implies \cfrac{7}{7}\cdot \cfrac{1}{2}\implies 1\cdot \cfrac{1}{2}\implies \boxed{\cfrac{1}{2}}\impliedby \textit{simplified} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{now, doing some distributing}}{-\cfrac{7}{6}\left( x-\cfrac{2}{7} \right)}\implies \left( -\cfrac{7}{6}\cdot x \right)+\left( -\cfrac{7}{6}\cdot -\cfrac{2}{7} \right)[/tex]


[tex]\bf \left( -\cfrac{7}{6}x \right)+\left( \cfrac{-7}{-7}\cdot \cfrac{2}{6} \right)\implies \left( -\cfrac{7}{6}x \right)+\left( 1\cdot \cfrac{2}{6} \right)\implies \left( -\cfrac{7}{6}x \right)+\cfrac{2}{6} \\\\\\ \left( -\cfrac{7}{6}x \right)+\cfrac{2\cdot 1}{2\cdot 3}\implies \left( -\cfrac{7}{6}x \right)+\cfrac{2}{2}\cdot \cfrac{1}{3} \\\\\\ \left( -\cfrac{7}{6}x \right)+1\cdot \cfrac{1}{3}\implies \boxed{-\cfrac{7}{6}x+\cfrac{1}{3}}[/tex]

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