How do you find the slope of the line that passes through (6,-10), (6,14)?

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Answer 1 · 2016-11-23T14:17:58

The slope is undefined (infinitely steep).

For $(x_{1}; y_{1}) and (x_{2}; y_{2})$ $(x_1;y_1) and (x_2;y_2)$

$(6, - 10) and (6, 14)$ $(6,-10) and (6,14)$

The equation for a line is $(y_{2} - y_{1}) = m (x_{2} - x_{1})$ $(y_2-y_1) =m(x_2-x_1)$

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

So, substituting

$m = \frac{14 - (- 10)}{6 - 6}$ $m= [14-(-10)]/(6-6)$

$m = \frac{24}{0} = \infty$ $m = 24/0 = ∞$

The line is vertical, the slope is undefined, or infinity

That means the line belongs to family of line x=constant,
in this case, $x = 6$ $" " x=6$