How do you find slope of the line that contains (1,6) and (10, -9)?

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How do you find slope of the line that contains (1,6) and (10, -9)?

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Answer 1 · 2018-03-18T09:54:45

Slope $= - \frac{5}{3}$ $= -5/3$

Explanation:

Recall the slope formula.

The given points are (1,6) and (10, -9)

comparing with $(x_{1}, y_{1}) and (x_{2}, y_{2})$ $(x_1, y_1) and (x_2,y_2)$
$x_{1} = 1$ $x_1=1$ $d d d d d d d d d d d d d d$ $color(white)(dddddddddddddd$ $x_{2} = 10$ $x_2=10$
$y_{1} = 6$ $y_1=6$ $d d d d d d d d d d d d d d$ $color(white)(dddddddddddddd$ $y_{2} = - 9$ $y_2=-9$

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m = (y_2 - y_1) / (x_2 - x_1)$

$m = \frac{- 9 - 6}{10 - 1}$ $m = (-9 - 6) / (10 - 1)$

$m = \frac{- 15}{9}$ $m = (-15) / (9)$

Slope= $- \frac{15}{9} = - \frac{5}{3}$ $-15/9 = -5/3$

Answer 2 · 2018-03-18T10:12:03

$m = - \frac{5}{3}$ $m=-5/3$

Explanation:

$Slope (m) = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $"Slope "(m) = (y_2-y_1)/(x_2-x_1)$

Here , $x_{1} = 1$ $x_1=1$

$y_{1} = 6$ $y_1=6$

$x_{2} = 10$ $x_2=10$

$y_{2} = - 9$ $y_2=-9$

$\Rightarrow m = \frac{- 9 - 6}{10 - 1}$ $=> m = (-9-6)/(10-1)$

$\Rightarrow m = - \frac{15}{9}$ $=> m = -15/9$

$\Rightarrow m = - \frac{5}{3}$ $=> m = -5/3$

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How do you find slope of the line that contains (1,6) and (10, -9)?

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