### Equation of Line

 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 two-point 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−y1)/(x−x1) = (y2−y1)/(x2−x1) point-slope form: (y−y1) = m (x−x1) slope-intercept 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(x1, y1) and B(x2, y2) 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−y1)/(x−x1) = (y2−y1)/(x2−x1) If the value of the slope is m then we can substitute, (y2−y1)/(x2−x1) = m and get the equation of the line as (y−y1)/(x−x1) = m (y−y1)= m (x−x1) Expanding (y−y1)=m(x−x1) y−y1= mx − mx1 y = mx−(mx1− y1) As [−(mx1 − y1)] 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 x-axis 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 (x1, y1) ≡ (a,0) and (x2, y2) ≡ (0,b) Using the two point form (y−y1)/(x−x1) = (y2−y1)/(x2−x1) 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