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
        

Answers (1)
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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