Question

Find the perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line $x=\sqrt{3}y+4=0$.

Answer 1

The perpendicular of (1, 2) on the straight line $x-\sqrt{3}y=-4$

Then, the equation is

$y-y_1=m^{{}^{\prime }}(x-x_1)$

$x_1=1,y_1=2,m=\frac{1}{\sqrt{3},m^{{}^{\prime }}=-\sqrt{3}}$

$y-2=-\sqrt{3}(x-1)$

$y+\sqrt{3}x-(2+\sqrt{3})=0\dots \dots \left(i\right)$

The perpendicular distance from (0, 0) to (i) is

$\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$a=\sqrt{3},b=1,c=-(2+\sqrt{3})$

$x_1=0,y_1=0$

$=\frac{|\sqrt{3}\left(0\right)+1\left(0\right)+(-2-\sqrt{3})|}{(\sqrt{3}{)}^2+(1{)}^2}=\frac{2+\sqrt{3}}{2}$