## Reduction of Linear Equation in Different Forms

Subject: Optional Mathematics

#### Overview

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. This note gives the information about the standard forms of reducing general equation and to find a straight line from the perpendicular length.

#### 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, k= $\frac{1}{A^2 + B^2}$
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 ± ($\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$)

##### Things to remember
1. 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.
2. Ax + By + C = 0 is± ($\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$)
3. Ax + By + C = 0
4. $\frac{1}{2}$(x1y2 - x2y1 + x2y3 - x3y2+ x3y1 - x1y3) = 0
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Reduction of Linear Equation in Different Forms

Solution:

Here,
3x + 4y + 5 = 0
Reduce it into slope intercept from, 4y = -3x - 5
or, y = $\frac{-3}{4}$ x - $\frac{5}{4}$
which is in the form y = mx + c,
so, on comparing, we get,
Slope (m) = $\frac{-3}{4}$ and
y-intercept (c) = $\frac{-5}{4}$

Solution:

Here,
4x + 5y + 20 = 0
or, $\frac{4x}{-20}$ + $\frac{5y}{-20}$ = 1
or, $\frac{x}{-20/4}$ + $\frac{y}{-20/5}$
or, $\frac{x}{-5}$ + $\frac{y}{-4}$ = 1
which is in the form $\frac{x}{a}$ + $\frac{y}{b}$ = 1,
∴ x-intercept (a) = -5 and y-intercept (b) = -4

Solution:

x - $\sqrt{3}$ y - 6 = 0
or, x - $\sqrt{3}$ y = 6
Now,
= $\sqrt{(coefficient \: of \: x)^2 + (coefficient \: of \: y)^2}$
= $\sqrt{(1)^2 + (-\sqrt{3})^2}$
= $\sqrt{1 + 3}$
= $\sqrt{4}$
= 2.

Dividing both sides by 2, we get,
$\frac{1}{2}$ x - $\frac{\sqrt{3}}{2}$ y = $\frac{6}{2}$
or, $\frac{1}{2}$ x + ($\frac{\sqrt{3}}{2}$) y = 3
which is in form x cosα + y sinα = p
∴p = 3 units, cosα = $\frac{1}{2}$ and sinα = -$\frac{\sqrt{3}}{2}$
Since, cosα is positive and sinα = -$\frac{\sqrt{3}}{2}$, α lies in the 4th quardrant.
∴ cosα = $\frac{1}{2}$
cosα = cosα (360° - 60°)
α = 300°
∴ x cos 300° + y sin 300° = 3

Solution:

Here,
2x + 5y - 20 = 0
or, 2x + 5y = 20
or, $\frac{2x}{20}$ + $\frac{5y}{20}$ = 1
or, $\frac{x}{10}$ + $\frac{y}{4}$ = 1
Which is in the form $\frac{x}{a}$ + $\frac{y}{b}$ = 1
∴ X-intercept (a) = 10
y-intercept (b) = 4
∴ Area of ΔAOB = $\frac{1}{2}$ OA × OB
= $\frac{1}{2}$ × 10 × 4
= 20 sq. units

Solution:

The equation of line is 3x + 4y + 7 = 0
Perpendicular length (p) = |$\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}$|

we have, (x1,y1)= (-3,0)

So, p =  |$\frac{3\times(-3)+ 4\times 0}{\sqrt{3^2 + 4^2}}$|= $\frac{9}{5}$

Solution:

The equation of line is 3x + 4y + 7 = 0
Perpendicular length (p) =$\vert \frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\vert$
Where (x1 , y1) is the point from whhich the perpendicular is drawn on the line
or, p = $\vert\frac{3 . (-3) + 4 . 0 + 7}{\sqrt{(3)^2 + (4)^2}}\vert$
or, p = $\vert\frac{-9 + 7}{\sqrt{9 + 16}}\vert$
or, p = $\vert\frac{-2}{5}\vert$
∴ p = $\frac{2}{5}$ units

Solution:

Here,
Let $\frac{x}{a}$ + $\frac{y}{b}$ = 1 be the equation of a line peprendicular length frawn from origin (0, 0) to $\frac{x}{a}$ + $\frac{y}{b}$ = 1 and its equation is given by
or, p = $\vert\frac{0 . \frac{1}{a} + 0 . \frac{1}{b} - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}\vert$
or, p = $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$
Squaring on both sides we get,
or, p2 = $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$
or, $\frac{1}{a^2}$ + $\frac{1}{b^2}$ = $\frac{1}{p^2}$
∴ $\frac{1}{p^2}$ = $\frac{1}{a^2}$ + $\frac{1}{b^2}$ proved.