Question

Line inclination with
$\begin{array}{cc}Positivex-direction& Slopeofline\\ 1.0^o& p.0\\ 2.{90}^o& q.-\sqrt{3}\\ 3.{120}^o& r.-\frac{1}{\sqrt{3}}\\ 4.{150}^o& s.Notdefined\end{array}$

• A. 1 - p 2 - s 3 - q 4 - r
• B. 1 - p 2 - s 3 - r 4 - q
• C. 1 - p 2 - q 3 - r 4 - s
• D. 1 - s 2 - p 3 - q 4 - r

Answer 1

The correct option is A

1 - p 2 - s 3 - q 4 - r

If $\theta$ be the angle at which a straight line is inclined to the positive direction of x-axis, the slope of the line is defined by $m=tan\theta$ where $0^o\le \theta <{180}^o\theta \ne {90}^o$
1. when $\theta =0^o$
Slope =$tan\theta =tan{0}^o=0$

2. when $\theta ={90}^o$
at $\theta$ = ${90}^o$ slope of the line is NOT defined

3. when $\theta ={120}^o$
Slope of line =$tan{120}^o=tan({180}^o-{60}^o)=-tan{60}^o=-\sqrt{3}$

4. when $\theta ={150}^o$
Slope of line =$tan{150}^o=tan({180}^o-{30}^o)=-tan{30}^o=-\frac{1}{\sqrt{3}}$