General Form into Normal Form

We will learn the transformation of general form into normal form.

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

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

Let the normal form of the given equation ax + by + c = 0……………. (i) be 

x cos α + y sin α - p = 0, where p > 0. ……………. (ii)

Then, the equations (i) and (ii) are the same straight line i.e., identical.

⇒ \(\frac{A}{cos α}\) = \(\frac{B}{sin α}\) = \(\frac{C}{-p}\)

⇒ \(\frac{C}{P}\) = \(\frac{-A}{cos α}\) = \(\frac{-B}{sin α}\) = \(\frac{+\sqrt{a^{2} + b^{2}}}{\sqrt{cos^{2} α + sin^{2} α}}\) = +  \(\sqrt{A^{2} + B^{2}}\)

Therefore, p = \(\frac{C}{\sqrt{A^{2} + B^{2}}}\), cos α = - \(\frac{A}{\sqrt{A^{2} + B^{2}}}\) and sin α = - \(\frac{B}{\sqrt{A^{2} + B^{2}}}\)

So, putting the values of cos α, sin α and p in the equation (ii) we get the form,

⇒ - \(\frac{A}{\sqrt{A^{2} + B^{2}}}\) x - \(\frac{B}{\sqrt{A^{2} + B^{2}}}\) y - \(\frac{C}{\sqrt{A^{2} + B^{2}}}\) =  0, when c > 0

⇒ \(\frac{A}{\sqrt{A^{2} + B^{2}}}\) x +  \(\frac{B}{\sqrt{A^{2} + B^{2}}}\) y = - \(\frac{C}{\sqrt{A^{2} + B^{2}}}\), when c < 0

Which is the required normal form of the general form of equation Ax + By + C = 0.

 

Algorithm to Transform the General Equation to Normal Form

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

Step II: Divide both sides by \(\sqrt{(\textrm{Coefficient of x})^{2} + (\textrm{Coefficient of y})^{2}}\).

The obtained equation will be in the normal form.


Solved examples on transformation of general equation into normal form:

1. Reduce the line 4x + 3y - 19 = 0 to the normal form.

Solution:

The given equation is 4x + 3y - 19 = 0

First shift the constant term (-19) on the RHS and make it positive.

4x + 3y = 19 ………….. (i)

Now determine \(\sqrt{(\textrm{Coefficient of x})^{2} + (\textrm{Coefficient of y})^{2}}\)

= \(\sqrt{(4)^{2} + (3)^{2}}\)

= \(\sqrt{16 + 9}\)

= √25

= 5

Now dividing both sides of the equation (i) by 5, we get

\(\frac{4}{5}\)x + \(\frac{3}{5}\)y = \(\frac{19}{5}\)

Which is the normal form of the given equation 4x + 3y - 19 = 0.

 

2. Transform the equation 3x + 4y = 5√2 to normal form and find the perpendicular distance from the origin of the straight line; also find the angle that the perpendicular makes with the positive direction of the x-axis.

Solution:    

The given equation is 3x + 4y = 5√2 ……..….. (i)

Dividing both sides of equation (1) by + \(\sqrt{(3)^{2} + (4)^{2}}\) = + 5 we get,

⇒ \(\frac{3}{5}\)x + \(\frac{4}{5}\)y = \(\frac{5√2}{5}\)

⇒ \(\frac{3}{5}\)x + \(\frac{4}{5}\)y = √2

Which is the normal form of the given equation 3x + 4y = 5√2.

Therefore, the required, perpendicular distance from the origin of the straight line (i) is √2 units.

If the perpendicular makes an angle α with the positive direction of the x-axis then,

cos α = \(\frac{3}{4}\) and sin α = \(\frac{4}{5}\)

Therefore, tan α = \(\frac{sin α}{cos α }\) = \(\frac{\frac{4}{5}}{\frac{3}{5}}\) = \(\frac{4}{3}\)

⇒ α = tan\(^{-1}\)\(\frac{4}{3}\).




11 and 12 Grade Math 

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