Determine the equation of the line with the points (–13,10) and (16, 15).

1 answer:

3 0

$\bf (\stackrel{x_1}{-13}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{15}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{15-10}{16-(-13)}\implies \cfrac{5}{16+13}\implies \cfrac{5}{29}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-10=\cfrac{5}{29}[x-(-13)] \\\\\\ y-10=\cfrac{5}{29}(x+13)\implies y-10=\cfrac{5}{29}x+\cfrac{65}{29}\implies y=\cfrac{5}{29}x+\cfrac{65}{29}+10 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=\cfrac{5}{29}+\cfrac{355}{29}~\hfill$

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