Question

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

Answer 1

Here,$m=$Slope of the line $=\tan {15}^{\circ }=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=2-\sqrt{3}$

and $c=$ intercept on $y-$axis $=–4$

Therefore, equation of the required line will be $y=(2-\sqrt{3})x-4$

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