A geometry problem by Rahil Sehgal

Put the equation of the line in slope-intercept form to find for the slope of the line.

$x-\sqrt{3}y+4=0$ $\implies$ $y=\dfrac{1}{\sqrt{3}}x+\dfrac{4}{\sqrt{3}}$ $\implies$ The slope is $\dfrac{1}{\sqrt{3}}$ . Slopes of perpendicular lines are negative reciprocals. Therefore, the slope of the perpendicular line to point $(1,2)$ is $-\sqrt{3}$ . From point-slope form of an equation of a line, we have

$y-y_1=m(x-x_1)$

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

$\sqrt{3}x+y-2-\sqrt{3}$ $\implies$ Equation of the perpendicular line.

The distance form the origin is,

$d=\huge\lvert$ $\dfrac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}$ $\huge\rvert$ $=$ $\huge\lvert$ $\dfrac{0+0-2-\sqrt{3}}{\sqrt{3+1}}$ $\huge\rvert$ $=\large\boxed{1.866}$ $\color{#D61F06}\large\boxed{\text{answer}}$