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*FIND THE SLOPE OF THE LINE THROUGH EACH PAIR OF POINTS*

d. (-4, 3) and (-6, -8)

e. (-7, -1) and (-7, 2)

f. (9, 4) and (-6, 4)

slope = y2 - y1
______
x2 - x1

User Talhature
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1 Answer

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(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-8}-\stackrel{y1}{3}}}{\underset{\textit{\large run}} {\underset{x_2}{-6}-\underset{x_1}{(-4)}}} \implies \cfrac{-11}{-6 +4} \implies \cfrac{ -11 }{ -2 } \implies \cfrac{11 }{ 2 } \\\\[-0.35em] ~\dotfill


(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-7}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{\textit{\large run}} {\underset{x_2}{-7}-\underset{x_1}{(-7)}}} \implies \cfrac{2 +1}{-7 +7} \implies \cfrac{ 3 }{ 0 } \implies \stackrel{\textit{vertical line}}{{\Large \begin{array}{llll} unde fined \end{array}}} \\\\[-0.35em] ~\dotfill


(\stackrel{x_1}{9}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{-6}-\underset{x_1}{9}}} \implies \cfrac{ 0 }{ -15 } \implies \stackrel{\textit{horizontal line}}{\text{\LARGE 0}}

keep in mind that whenever you notice the x-coordinates are the same, it's a vertical line, and whenever the y-coordinates are the same, is a horizontal line.

User Foxinni
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