The side AB of an equilateral triangle ABC is parallel to the x -axis. Find the slopes of all its sides.

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Solution

Slope of $B C = tan (120)$
$= tan (90 + 30)$
$= - cot 30 = - \sqrt{3}$
Slope of $A C = \sqrt{3}$

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The side AB of an equilateral triangle ABC is parallel to the x -axis. Find the slopes of all its sides.

Slope of BC=tan(120) =tan(90+30) =−cot30=−√3 Slope of AC=√3

Slope of $B C = tan (120)$
$= tan (90 + 30)$
$= - cot 30 = - \sqrt{3}$
Slope of $A C = \sqrt{3}$