How do you find the slope of (-2, 5) and (-2, 8)?

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Answer 1 · 2016-04-10T10:08:25

$s l o p e = \infty$ $slope=oo$

$P_{1} = (- 2, 5) the point of P_{1}$ $P_1=(-2,5)" the point of " P_1$
$P_{2} = (- 2, 8) the point of P_{2}$ $P_2=(-2,8)" the point of "P_2$

$s l o p e = \frac{P_{2y} - P_{1y}}{P_{2x} - P_{1x}}$ $slope=(P_"2y"-P_"1y")/(P_"2x"-P_"1x")$

$s l o p e = \frac{8 - 5}{- 2 - (- 2)}$ $slope=(8-5)/(-2-(-2))$

$s l o p e = \frac{3}{0}$ $slope=3/0$

$s l o p e = \infty$ $slope=oo$

Answer 2 · 2016-04-10T10:10:34

Slope of line joining $(- 2, 5)$ $(-2,5)$ and $(- 2, 8)$ $(-2,8)$ is $\infty$ $oo$ i.e. it is perpendicular to $x$ $x$ -axis.

Slope of line joining $(x_{1}, y_{1})$ $(x_1,y_1)$ and $(x_{2}, y_{2})$ $(x_2,y_2)$ is given by

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

Hence slope of line joining $(- 2, 5)$ $(-2,5)$ and $(- 2, 8)$ $(-2,8)$ is

$\frac{8 - 5}{(- 2) - (- 2)} = \frac{3}{- 2 + 2} = \frac{3}{0} = \infty$ $(8-5)/((-2)-(-2))=3/(-2+2)=3/0=oo$

Hence, slope of line joining $(- 2, 5)$ $(-2,5)$ and $(- 2, 8)$ $(-2,8)$ is $\infty$ $oo$ i.e. it is perpendicular to $x$ $x$ -axis.