Find the slope and the inclination of the line AB if :

(i) A = (-3, -2) and B = (1, 2)

(ii) A = (0, -$\sqrt{3}$) and B = (3, 0)

(iii) A = (-1, $2\sqrt{3}$) and B = (-2, $\sqrt{3}$)

By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

(i) A = (-3, -2) and B = (1, 2)

$\text{Slope of AB} = \dfrac{2 - (-2)}{1 - (-3)} \\[1em] = \dfrac{2 + 2}{1 + 3} \\[1em] = \dfrac{4}{4} = 1. \\[1em]$

Let inclination be θ,

∴ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°.

Hence, slope = 1 and inclination = 45°.

(ii) A = (0, -$\sqrt{3}$) and B = (3, 0)

$\text{Slope of AB} = \dfrac{0 - (-\sqrt{3})}{3 - 0} \\[1em] = \dfrac{\sqrt{3}}{3} \\[1em] = \dfrac{1}{\sqrt{3}}. \\[1em]$

Let inclination be θ,

∴ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, slope = $\dfrac{\sqrt{3}}{3}$ and inclination = 30°.

(iii) A = (-1, $2\sqrt{3}$) and B = (-2, $\sqrt{3}$)

$\text{Slope of AB} = \dfrac{\sqrt{3} - 2\sqrt{3}}{-2 - (-1)} \\[1em] = \dfrac{-\sqrt{3}}{-2 + 1} \\[1em] = \dfrac{-\sqrt{3}}{-1} = \sqrt{3}. \\[1em]$

Let inclination be θ,

∴ tan θ = $\sqrt3{}$

⇒ tan θ = tan 60°

⇒ θ = 60°.

Hence, slope = $\sqrt{3}$ and inclination = 60°.