Reduction of Linear Equation in Different Forms

Linear Equation Ax + By + C = 0

An equation between any two variables which straight line on a graph is known as a linear equation. In this equation, Ax + By + C = 0, where A, B, and C are neutrals and also A and B will not be together zero.
To prove this statement, let P(x₁, y₁), Q(x₂, y₂) and R(x₃, y₃) be nay three points on the locus represent by the equation Ax + By + C = 0. The coordinates of the points must satisfy the equation.
Hence,
Ax₁ + By₁ + C = 0 . . . . . . . . . . . . . . (i)
Ax₂ + By₂ + C = 0 . . . . . . . . . . . . . . (ii)
Ax₃ + By₃ + C = 0 . . . . . . . . . . . . . . (iii)
Solving first two equations by the rule of cross multiplication, we get
\(\frac{A}{y_1 - y_2}\) = \(\frac{B}{x_1 - x_2}\) = \(\frac{C}{x_1 y_2 - x_2 y_1}\) = k (say)
∴ A = k(y₁ - y₂), B = k(x₂ - x₁) and C = k(x₁y₂ - x₂y₁)
Substituting the values of A, B and C in the third equation, we get
k(y₁ - y₂) x₃ +k(x₂ - x₁) y₃ + k(x₁y₂ - x₂y₁) = 0
or, k(x₃y₁ - x₃y₂ + x₂y₃ - x₁y₃ + x₁y₂ - x₂y₁₎ = 0
or, x₁y₂ - x₂y₁ + x₂y₃ - x₃y₂+ x₃y₁ - x₁y₃ = 0
Multiplying both sides by \(\frac{1}{2}\), we get
\(\frac{1}{2}\)(x₁y₂ - x₂y₁ + x₂y₃ - x₃y₂+ x₃y₁ - x₁y₃) = 0
i.e. Area of ΔPQR = 0
This result sgows us that the points P, Q and R are collinear.
Thus, the general equation of first degree in x and y always represents a straight line.

Reduction of general equation of first degree to the three standard forms

There is three standard form to reduce the linear equation. They are given below:

Reduction to the slope intercept form
The first-degree general equation in x and y is
Ax + By + C = 0
This equation can be writtenas
By = -Ax - C
or, y = (-\(\frac{A}{B}\))x + (-\(\frac{C}{B}\)) which is of the form y = mx + c where
slope (m) = -\(\frac{A}{B}\) = -\(\frac{coefficient \: of \: x}{coefficient \: of \: y}\) and
Y-intercept (c) = -\(\frac{C}{B}\) = -\(\frac{content \: term}{coefficient \: of \: B}\)

Reduction of the double intercept form
The first-degree general equation in x and y is
Ax + By + C = 0
This equation can be writtenas
Ax + By = -C
Dividing both sides by -C, we get
\(\frac{Ax}{-C}\) + \(\frac{By }{-C}\) = 1
or, \(\frac{x}{\frac{-C}{A}}\) + \(\frac{y}{\frac{-C}{B}}\)= 1 which is of the form \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 where
X-interccept (a) = -\(\frac{C}{A}\) = -\(\frac{constant \: term}{coefficient \: of \: x}\)
Y-intercept (b) = -\(\frac{C}{B}\) = -\(\frac{constant \: term}{coefficient \: of \: y}\)

Reduction to the normal form
The equation Ax + By + C = 0 and x cosα = p will represent one and ssame straight line if their corresponding coefficients are proportional.
∴ \(\frac{cosα}{A}\) = \(\frac{sinα}{B}\) = \(\frac{-p}{C}\) = k (say)
Then, cosα = Ak, sinα = Bk and -p = Ck
Now,
(Ak)² + (Bk)² = cos²α + sin²α = 1
or, k² (A² + B²) = 1
or, k² = \(\frac{1}{A^2 + B^2}\)
or, k = ± \(\frac{1}{\sqrt{A^2 + B^2}}\)
∴ cosα = \(\frac{A}{±\sqrt{A^2 + B^2}}\), sinα = \(\frac{B}{±\sqrt{A^2 + B^2}}\) and p = \(\frac{-C}{±\sqrt{A^2 + B^2}}\)
Hence, the normal form is \(\frac{A}{±\sqrt{A^2 + B^2}}\)x + \(\frac{B}{±\sqrt{A^2 + B^2}}\)y = \(\frac{-C}{±\sqrt{A^2 + B^2}}\)
The + or - sign in the RHS being so chosen as to make the RHS positive.

Length of the perpendicular from a point on a straight line

To find the length of the perpendicular from a point on the line x cosα + y sinα = p

Let the equation of a line AB be x cosα + y sinα = p.
Then the length of a perpendicular from the origin on the line is p.
i.e. ON = p and∠AON =α
Let P(x₁, y₁) be any point and draw perpendicular PM from P to the line AB.
Through the point P, draw a line CD parallel to the given line AB. Let ON' be the perpendicular drawn from the origin to the CD such that ON' = p'.
Then,
PM = ON' - ON' = p' - p when p' > p and PM = ON - ON' = p - p' when p> p'.
Thus, PM = ± (p' - p). Here, the proper sign is taken so as make PM positive.
Now equation of CD is x cosα + y sinα = p'
But this line passes through the point (x₁, x₁)
So, x₁ cosα + y₁ sinα = p'
Hence, PM = ± (x₁ cosα + y₁ sinα - p) which is the length of the perpendicular drawn from (x₁, y₁) on the line x cosα + y sinα = p.

To find the length of the perpendicular from a point on the line Ax + By + C = 0
Here, the general equation of the first degree in x and y in Ax + By + C = 0.
Changing the equation into perpendicular form we get,
\(\frac{A}{\sqrt{A^2 + B^2}}\)x + \(\frac{B}{\sqrt{A^2 + B^2}}\)y + \(\frac{C}{\sqrt{A^2 + B^2}}\) = 0
Comparing this equation with x cosα + y sinα - p = 0 we get,
cosα = \(\frac{A}{\sqrt{A^2 + B^2}}\), sinα =\(\frac{B}{\sqrt{A^2 + B^2}}\) and p = -\(\frac{C}{\sqrt{A^2 + B^2}}\)
Now, length of the perpendicular drawn from the point (x₁, y₁) to the line x cosα + y sinα - p is

L = ± (x₁ cosα +y₁ sinα - p)

= ± (x₁\(\frac{A}{\sqrt{A^2 + B^2}}\) + y₁\(\frac{B}{\sqrt{A^2 + B^2}}\) -\(\frac{C}{\sqrt{A^2 + B^2}}\))

= ± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))

Hence, length of the perpendicular drawn from the point (x₁, y₁) to the line
Ax + By + C = 0 is ± (\(\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\))