Question

The equation of the line which makes an angle of ${15}^{\circ }$ with positive $x$-axis and cuts-off an intercept of $3$ unit on the negative $y$-axis is

• A. $y-(2-\sqrt{3})x+3=0$
• B. $y+(2-\sqrt{3})x+3=0$
• C. $y-(2-\sqrt{3})-3=0$
• D. $y-(2+\sqrt{3})x+3=0$

Answer 1

The correct option is A $y-(2-\sqrt{3})x+3=0$

Slope of the line $\left(m\right)=\tan {15}^{\circ }$
$=\tan ({45}^{\circ }-{30}^{\circ })=\frac{\tan {45}^{\circ }-\tan {30}^{\circ }}{1+\tan {45}^{\circ }\tan {30}^{\circ }}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$
and $y$ intercept$\left(c\right)=-3$
Hence the required equation is$y=(2-\sqrt{3})x-3$

$\Rightarrow y-(2-\sqrt{3})x+3=0$