Question

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

Answer 1

Finding the slope of all sides of triangle


Given,
$ABC$ is an equilateral triangle,
$\Rightarrow \angle A=\angle B=\angle C={60}^{\circ }$

$\therefore \text{Slope of}AC=\tan {60}^{\circ }=\sqrt{3}$

Now, drawing a ray extended from $AB$, forming a linear pair, $\angle ABC\text{and}\angle DBC$.
$\Rightarrow \angle DBC={120}^{\circ }$

$\therefore \text{Slope of}BC=\tan {120}^{\circ }$
$=\tan ({90}^{\circ }+{30}^{\circ })$
$=-\cot {30}^{\circ }$ $[\because \tan (90°+\theta )=-\cot \theta ]$
$=-\sqrt{3}$

Also, we know that, the slope of any line parallel to $x-\text{axis}$ is $0$.

$\therefore \text{Slope of}AB=0$

Hence, the slope of $AB,AC\text{and}BC$ is $0,\sqrt{3}\text{and}-\sqrt{3}$ respectively.