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Showing posts with label two point form. Show all posts
Showing posts with label two point form. Show all posts

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