# Standard Form Equation

## Equation

An equation is a statement which equates two expressions. There is a  wide application of equations in mathematics.

## Standard form

Standard form is the method of representing mathematical expressions in the simple form. It helps us to simplify complex equations into simpler equations

## Standard form of a linear equation

The standard form of a linear equation is given by Ax+ By = C where A, B and C are constants and x and y are variables.
Example: -
5x + 6y = 8 is an example of a straight line in the standard form.  Here A = 5,       B = 6 and C = 8.

The equation 4y - 7x = 9 is also an example of straight line in the standard form.  Here A = -7, B = 4 and C = 9.

Example 1: -
Find the equation of the line passing through the points (2, 12) and (5, 16) and express it in the standard form.
Solution: -
We know that the line passing through two points (x1, y1) and (x2, y2) is given by
y - y1 = m (x - x1) where m is the slope of the line, which is given by                      m =  (y2 - y1)/(x2 - x1)
Given that the line passes through (x1, y1)  = (2, 12) and  (x2, y2) = (5 , 16).
Lets first find the slope.
Slope m = (y2 - y1)/(x2 - x1) = (16 - 12 )/(5 - 2) = 4/3
Substituting in the equation of the line, we get
y - 12 = 4/3 (x - 2)
Cross multiplying we get,
3y - 36 = 4x - 8
Adding 8 on both sides, we get
3y - 28 = 4x
subtracting 3y on both sides we get
4x - 3y = -28, which is the required equation.  Here A = 4, B = -3and C = -28.

Example 2: -
Find the equation of the line passing through (5, 8) and having a slope 4.
Solution: -
We know that if a line passes through a point (x1, y1) and have slope m, then the equation is given by y - y1 = m (x - x1).  This form is known as point slope form.
Given that the line passing through (5, 8).  That is (x1, y1) = (5, 8).
Slope is 4.  That is m = 4
Therefore the required line is
y - 8 = 4 (x - 5)
y - 8 = 4x - 20
adding 20 both sides, we get
y + 12  = 4x
subtracting y on both sides, we get
4x - y = 12, which is the required equation.Here A = 4, B = -1 and C = 12.

## Standard form of a circle

I
f the center of the circle is at the point (h, k) and has radius of the circle is r, then the equation of the circle is given by (x - h)2+ (y - k)2= r2
This representation of the circle is called the standard form.

Example 1: -
Write the equation of the circle in the standard form, whose center is (-2, 7) and radius is 5.

Solution: -
Given center of the circle,(h, k) = (-2, 7) and radius of the circle, r = 5.
We know that the standard form of a circle is (x - h)2 + (y - k)2 = r2
Substituting for h, k and r, we get
(x - (- 2))2 + (y - 7)2 = 52
(x + 2)2 + (y - 7)2 = 52
Simplifying we get,
(x + 2)2 + (y - 7)2 = 25
So the required equation is (x + 2)2 + (y - 7)2 = 25.

Example 2: -
Write the equation of the circle in the standard form whose center is (1, 0) and radius is 6.
Solution: -
Given center of the circle, (h, k) = (1, 0) and radius of the circle, r = 6.
We know that the standard form of a circle is (x - h)2 + (y - k)2= r2
Substituting for h, k and r, we get
(x - 1)2 + (y - 0)2 = 62
Simplifying we get,
(x - 1)2 + (y)2 = 36
So the required equation is (x - 1)2 + y2 = 36.