Question:
Find the slope of a line whose inclination is
(i) $30^{\circ}$
(ii) $120^{\circ}$
(iii) $135^{\circ}$
(iv) $90^{\circ}$
Solution:
We know that the slope of a given line is given by
Slope = tanθ Where θ = angle of inclination
(i) Given that $\theta=30^{\circ}$
Slope $=\tan \left(30^{\circ}\right)=\frac{1}{\sqrt{3}}$
(ii) Given that $\theta=120^{\circ}$
Slope $=\tan \left(120^{\circ}\right)=\tan \left(90^{\circ}+30^{\circ}\right)=-\cot \left(30^{\circ}\right)=-\sqrt{3}$
(iii) Given that $\theta=135^{\circ}$
Slope $=\tan \left(135^{\circ}\right)=\tan \left(90^{\circ}+45^{\circ}\right)=-\cot \left(45^{\circ}\right)-1$
(iv) Given that $\theta=90^{\circ}$
Slope $=\tan \left(90^{\circ}\right)=\infty$