Question

Find c such that (-5, -3), (10, 1), and (c, -9) lie on a line.

Answer 1

well, we know all three points are colinear, so the slope from the first two points is the same slope from either to the (c , -9) point, let's check the slope of the first two points given

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{1})~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-5)}}} \implies \cfrac{1 +3}{10 +5}\implies \cfrac{4}{15}`

so then, the slope from say hmmmm let's use (-5 , -3), the slope from (-5 , -3) and (c , -9) must also be the same 4/15

`(\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{c}~,~\stackrel{y_2}{-9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-9}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{c}-\underset{x_1}{(-5)}}} \implies \hspace{5em}\cfrac{-9 +3}{c +5}~~ = ~~\stackrel{slope}{\cfrac{4}{15}}\implies \cfrac{-6}{c+5}=\cfrac{4}{15} \\\\\\ -90=4c+20\implies -110=4c\implies \cfrac{-110}{4}=c\implies -\cfrac{55}{2}=c`