Question

The equation of the line which makes an angle of ${15}^0$ with the positive X-axis and cuts an intercept of length $4$ on the negative Y-axis is

• A. $\left(\sqrt{3}-1\right)x-\left(\sqrt{3}+1\right)y-4=0$
• B. $\left(2-\sqrt{3}\right)x-y-4=0$
• C. $\left(2+\sqrt{3}\right)x-y-4=0$
• D. $\left(2+\sqrt{3}\right)x+y+4=0$

Answer 1

The correct option is A $\left(\sqrt{3}-1\right)x-\left(\sqrt{3}+1\right)y-4=0$

Given data :

$\theta =15$ and Intercept $=-4$

$C=-4$

Slope of line $=\tan \theta$ $\Rightarrow m=\tan {15}^o$

$m=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Equation of line in slope

Intercept form $=y_1=mx+c$

$y=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)x-4$

$(\sqrt{3}+1)y=(\sqrt{3}-1)x-4$

$(\sqrt{3}-1)x-(\sqrt{3}+1)y-4=0$

$\therefore$ The equation of the line is $(\sqrt{3}-1)x-(\sqrt{3}+1)y-1=0$