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Wednesday, August 20, 2014

Preliminaries in plane geometry (Part 2)

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
  1. two points
  2. one point and slope of the line
  3. slope of the line and y intercept
  4. angle made by the perpendicular and its length to the line from the origin
The corresponding four types of equations are as follows
  1. two points form: (y−y1)/(x−x1) = (y2−y1)/(x2−x1)
  2. point-slope form: (y−y1) = m (x−x1)
  3. slope-intercept form: y = mx + c
  4. 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:
  1. 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)

  2. 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)

  3. 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.

  4. 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

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