Question

the normal form of the line x√3+y+4=0​

Answer 1

Question :-

• Reduce to normal form : √3x +y + 4 = 0

Answer

Given :

• A general equation of line : √3x + y + 4 = 0

To find :-

• Normal form of equation.

Method:-

To reduce the general equation Ax + By + C = 0 in to normal form (x cos α + y sin α = p):

We have the general equation Ax + By + C = 0.

Step I: Transfer the constant term to the right hand side and make it positive.

Step II: Divide both sides by √(Coefficient of x)²+(Coefficient of y)²

The obtained equation will be in the normal form.

Solution :-

General equation is √3x + y + 4 = 0

⇛ - √3x - y = 4

$\bf \:Divide \: both \: sides \: by = \sqrt{ {( \sqrt{3}) }^{2} + {(1)}^{2} }$

$\bf\implies \: \sqrt{3 + 1} = 2$

So, we get

$\bf\implies \: - \dfrac{ \sqrt{3} }{2} x - \dfrac{1}{2} = 2$

which is the required normal form, where

$\bf\implies \:cos \alpha = - \dfrac{ \sqrt{3} }{2} \: and \: sin \alpha = - \dfrac{1}{2} \: and \: p = 2$

As cosα and sinα are negative

⇛ α lies in 3rd quadrant.

⇛ α =  π +  π/6

⇛ α = 7π/6.

So, required normal form √3x + y + 4 = 0 is

x cos (7π/6) + y sin (7π/6) = 2.

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This normal form indicates that the line √3x + y + 4 = 0 is at a distance of 2 units from the origin and normal makes an angle of 7π/6 with positive direction of x - axis.