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
11 and 12 Grade Math
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