# Q : 3     Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.              (i)   $x-\sqrt{3}y+8=0$

G Gautam harsolia

Given equation is
$x-\sqrt{3}y+8=0$
we can rewrite it as
$-x+\sqrt3y=8$
Coefficient of x is -1 and y is $\sqrt3$
Therefore, $\sqrt{(-1)^2+(\sqrt3)^2}= \sqrt{1+3}=\sqrt4=2$
Now, Divide both the sides by 2
we will get
$-\frac{x}{2}+\frac{\sqrt3y}{2}= 4$
we can rewrite it as
$x\cos 120\degree + y\sin 120\degree= 4 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Now, we know that the normal form of the line is
$x\cos \theta + y\sin \theta= p \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Where $\theta$ is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin
On comparing equation (i) and (ii)
we wiil get
$\theta = 120\degree \ \ and \ \ p = 4$
Therefore,  the angle between perpendicular and the positive x-axis and  perpendicular distance  from the origin is $120\degree \ and \ 4$  respectively

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