Linear fns (1 Viewer)

mayngo · Aug 16, 2015

Sorry for asking so much >.< but i need help

InteGrand · Aug 16, 2015

mayngo said:
Sorry for asking so much >.< but i need help

$The slope of the line is $m=\tan \theta = \tan 30^\circ = \frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.$

$So in point-slope form, the line's equation is:$

$$y-(-2)=\frac{\sqrt{3}}{3} \left(x-\sqrt{3} \right)$.$

$This can be simplified to: $y+2=\frac{\sqrt{3}}{3}x - 1$$

$$\Longleftrightarrow y = \frac{\sqrt{3}}{3}x-3$.$

Linear fns (1 Viewer)

mayngo

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InteGrand

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