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 equation of the line perpendicular to $x-\sqrt{3}y+4=0$ is$\sqrt{3}x+y+\lambda =0$.
This line passes through (1, 2).

$$\begin{gathered} \therefore \sqrt{3}+2+\lambda =0 \\ \Rightarrow \lambda =-\sqrt{3}-2 \end{gathered}$$

Substituting the value of $\lambda$, we get $\sqrt{3}x+y-\sqrt{3}-2=0$

Let d be the perpendicular distance from the origin to the line $\sqrt{3}x+y-\sqrt{3}-2=0$

$d=\left|\frac{0-0-\sqrt{3}-2}{\sqrt{1+{\displaystyle 3}}}\right|=\frac{\sqrt{3}+2}{2}$

Hence, the required perpendicular distance is $\frac{\sqrt{3}+2}{2}$