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$$^{-1}$$$$\frac{4}{3}$$.

The Straight Line

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