In this post we will discuss about equations of a line in different form. Every equation is related to other equation and can be derived from one other. In the post I have derived the different forms of equations of a straight line. I have started from an equation which is called twopoint form and end with the normal form. normal form is also called perpendicular form. I have given graphs of four equations.
The equation of a line can be found if we know anyone of
 two points
 one point and slope of the line
 slope of the line and y intercept
 angle made by the perpendicular and its length to the line from the origin
The corresponding four types of equations are as follows
 two points form: (y−y_{1})/(x−x_{1}) = (y_{2}−y_{1})/(x_{2}−x_{1})
 pointslope form: (y−y_{1}) = m (x−x_{1})
 slopeintercept form: y = mx + c
 normal form: x cos α + y sin α = p
The four equations in the graph are
y = 3x + 4, y = x + 7 , y = 2x− 5 and y = −x + 3.
Let A(x _{1}, y _{1}) and B(x _{2}, y _{2}) be two points. Then the four types of equations can be framed as follows:
 As the slope of the line will be constant. So, if a variable point is (x,y) then
(y−y_{1})/(x−x_{1}) = (y_{2}−y_{1})/(x_{2}−x_{1})
 If the value of the slope is m then we can substitute,
(y_{2}−y_{1})/(x_{2}−x_{1}) = m
and get the equation of the line as
(y−y_{1})/(x−x_{1}) = m
(y−y_{1})= m (x−x_{1})
 Expanding (y−y_{1})=m(x−x_{1})
y−y_{1}= mx − mx_{1}
y = mx−(mx_{1}− y_{1})
As [−(mx_{1} − y_{1})] is a constant and can be substituted for c
y = mx + c
From the above equation we get y = c when x = 0. Hence, c is the y intercept.
 Normal form is found by considering the angle which the perpendicular from the origin to the line makes with the xaxis and its length.
In the figure the angle is α and the length of the perpendicular is p. Equation of line in two point form is
(x_{1}, y_{1}) ≡ (a,0) and (x_{2}, y_{2}) ≡ (0,b)
Using the two point form
(y−y_{1})/(x−x_{1}) = (y_{2}−y_{1})/(x_{2}−x_{1})
y/(x − a) = b/(− a)
−ay = bx − ab
bx + ay = ab
x/a + y/b = 1
a = p sec α
b = p cosec α
x/(p sec α) + y/(p cosec α) = 1
x cos α + y sin α = p
