Question

The slope of a line, which makes an angle of $30^{\circ }$ with the positive direction of $Y$-axis measured anti-clockwise, is

• A. $\sqrt{3}$
• B. $\frac{-1}{\sqrt{3}}$
• C. $-\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

Given, $\angle YPQ=30^{\circ }$
To find slope of line $AQ$.

Here, $\angle YPQ=\angle OPA$ (vertically opposite angles)
$\because \angle OPA+\angle POA+\angle PAO=180^{\circ }$
$(\because$ sum of all angles of a triangle is $180^{\circ })$
$\Rightarrow 30^{\circ }+90^{\circ }+\angle PAO=180^{\circ }$
$\Rightarrow \angle PAO=180^{\circ }-120^{\circ }=60^{\circ }$
$\Rightarrow \angle PAX=180^{\circ }-60^{\circ }=120^{\circ }$
$\therefore$ Slope of line $AQ=m=\tan 120^{\circ }$
$\left[\because m=\tan \theta \tan \left(180^{\circ }-\theta \right)=-\tan \theta \right]$
$=\tan \left(180^{\circ }-60^{\circ }\right)$
$=-\tan 60^{\circ }=-\sqrt{3}$