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.