The line passing through (1, 0) and (-2, sqrt | vedclass

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(C) The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.Substituting the given points

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.Substituting the given points $(1, 0)$ and $(-2, \sqrt{3})$:$m = \frac{\sqrt{3} - 0}{-2 - 1} = \frac{\sqrt{3}}{-3} = -\frac{1}{\sqrt{3}}$.Since $m = 	an(	heta)$,we have $	an(	heta) = -\frac{1}{\sqrt{3}}$.Since the slope is negative,the angle $	heta$ lies in the second quadrant.$	heta = 180^o - 30^o = 150^o$.

The line passing through $(1, 0)$ and $(-2, \sqrt{3})$ makes an angle of ...... with the $x$-axis. (in $^o$)

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