Question

Reduce the equation into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis.
$x+\sqrt{3}y+8=0$

Answer 1

Solution -

$x+\sqrt{3}y+8=0$

$\frac{x}{-8}+\frac{\sqrt{3}y}{-8}-1=0\to \frac{x}{-8}+\frac{y}{\frac{-8}{\sqrt{3}}}=1$--- Normal Form

$\perp$ distance from origin $=\left|\frac{8}{\sqrt{1+3}}\right|=4$

Angle with positive x-axis $=ta{n}^{-1}\left(\frac{-8/\sqrt{3}}{-8}\right)={30}^{\circ }$

Angle with $\perp$ positive x-axis $={90}^{\circ }-{30}^{\circ }={60}^{\circ }$