Question

The equation of the straight line which makes an angle of ${15}^{\circ }$ with the positive direction of x-axis and cuts an intercept of length 4 on the negative direction of y-axis, is

• A. $(2-\sqrt{3})x-y-4=0$
• B. $(2-\sqrt{3})x+y-4=0$
• C. $(2+\sqrt{3})x-y-4=0$
• D. $(2+\sqrt{3})x-3y+7=0$

Answer 1

The correct option is A

$(2-\sqrt{3})x-y-4=0$

The slope of the line = m = tan ${15}^{\circ }$= tan (${45}^{\circ }–{30}^{\circ }$)

$=\frac{tan{45}^{\circ }-tan{30}^{\circ }}{1+tan{45}^{\circ }.tan{30}^{\circ }}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{{(\sqrt{3}-1)}^2}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$

Also, the intercept on y-axis = c = –4

Therefore, the equation of the line is

$y=(2-\sqrt{3})x-4or(2-\sqrt{3})x-y-4=0$