Question

The axes being inclined at an angle of ${30}^o$, find the equation to the straight line which passes through the point $(-2,3)$ and is perpendicular to the straight line $y+3x=6$.

Answer 1

When axes are inclined angle between two lines is given by

$\tan \theta =\frac{m-m^{\prime }\sin \omega }{1+(m+m^{\prime })\cos \omega +m{m}^{\prime }}$

Slope of given line $m=-3$

Let slope of line perpendicular to given line be $m^{\prime }$

$\theta =\frac{\pi }{2}\Rightarrow \tan \theta =\infty \frac{m-m^{\prime }\sin \omega }{1+(m+m^{\prime })\cos \omega +m{m}^{\prime }}=\infty \Rightarrow 1+(m+m^{\prime })\cos \omega +m{m}^{\prime }=01+(-3+m^{\prime })\cos {30}^{\circ }-3m^{\prime }=01+(-3+m^{\prime })\frac{\sqrt{3}}{2}-3m^{\prime }=0\Rightarrow m^{\prime }=\frac{3\sqrt{3}-2}{\sqrt{3}-6}$

Equation of line passing through $(-2,3)$ and slope $m^{\prime }$ is

$(3\sqrt{3}-2)x-(\sqrt{3}-6)y+6\sqrt{3}-4+3\sqrt{3}-18=0(3\sqrt{3}-2)x-(\sqrt{3}-6)y+9\sqrt{3}-22=0$