Question

Find the equation to, and the length of, the perpendicular drawn from the point $(1,1)$ upon the straight line $3x+4y+5=0$, the angle between the axes being ${120}^o$.

Answer 1

Length of perpendicular $=p=\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2-2ab\cos \omega }}\sin \omega$

$p=\frac{3.1+4.1+5}{\sqrt{3^2+4^2-2\left(3\right)\left(4\right)\cos {120}^{\circ }}}\sin {120}^{\circ }p=\frac{12}{\sqrt{37}}\times \frac{\sqrt{3}}{2}p=\frac{6\sqrt{111}}{37}$

When the axes is inclined angle between two line is given by

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

Slope of given line $=m=-\frac{3}{4}$

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

$\theta =\frac{\pi }{2}\Rightarrow \tan \theta =\infty \frac{m-m^{\prime }\sin \omega }{1+(m+m^{\prime })\cos \omega +m{m}^{\prime }}=\infty \Rightarrow 1+(m+m^{\prime })\cos \omega +m{m}^{\prime }=01+(-\frac{3}{4}+m^{\prime })\cos {120}^{\circ }-\frac{3}{4}{m}^{\prime }=01+(-\frac{3}{4}+m^{\prime })\frac{-1}{2}-\frac{3}{4}{m}^{\prime }=0\Rightarrow m^{\prime }=\frac{11}{10}$

Equation of line passing through $(1,1)$ with slope $m^{\prime }$ is

$y-1=\frac{11}{10}(x-1)10y-10=11x-1110y-11x+1=0$