Lines in a plane

Orthonormal axes

In all that follows we assume an X-axis orthogonal to a Y-axis fixed in o and so that the unities on both axes have the same magnitude.
We call this : 'orthonormal axes'.

Equation and slope of a line

We know that, in a plane, each line l has an equation of the form 
 
        ax+by+c=0
If b is not 0, the slope of that line is -a/b .
Remark: ax+by = 0 is the equation of a line parallel to l and containing the origin o.

Direction vector of a line and parallelism

Let line l : ax+by+c=0 and l' : ax+by = 0
Each point P on l' is the image point of a vector P defining the direction of l. Then, P is called a direction vector of l.
If r is a real number (not 0 ) , then r.P is a direction vector too.
A simple choice for P is P(b,-a).

Take any two lines 

 
  a x + b y + c = 0

  d x + e y + f = 0

    The lines are parallel
<=>
    A direction vector of the first line is a multiple of
    the direction vector of the second line.
<=>
    (b, -a) is a multiple of (e, -d)
<=>
    (a, b)  is a multiple of (d, e)
<=>
    a e = b d
<=>
      | a   b |
      | d   e |  = 0

 
   The lines a x + b y + c = 0  and d x + e y + f = 0 are parallel
<=>
     a e = b d
<=>
      | a   b |
      | d   e |  = 0

Three points on one line. (collinear points)

 
   Three points P(x1,y1) Q(x2,y2) and R(x3,y3) are collinear
<=>
   There is a line  a x + b y + c = 0  that contains the points P, Q and R.
<=>
   There is a line  a x + b y + c = 0  such that
        a x1 + b y1 + c = 0
        a x2 + b y2 + c = 0
        a x3 + b y3 + c = 0
<=>
   There is an a, b and c, not all zero, such that
        a x1 + b y1 + c = 0
        a x2 + b y2 + c = 0
        a x3 + b y3 + c = 0
<=>
   The following homogeneous system with unknowns a, b and c
   has a solution different from (0,0,0).
        a x1 + b y1 + c = 0
        a x2 + b y2 + c = 0
        a x3 + b y3 + c = 0
<=>
        |x1    y1     1|
        |x2    y2     1| = 0
        |x3    y3     1|
Conclusion :
Three points P(x1,y1) Q(x2,y2) and R(x3,y3) are collinear if and only if 
 
        |x1    y1     1|
        |x2    y2     1| = 0
        |x3    y3     1|

Equation of the line PQ with P(x1,y1) and Q(x2,y2)

From above we know that a variable point D(x,y) is on the line PQ if and only if 
 
        |x      y       1|
        |x1     y1      1| = 0
        |x2     y2      1|
So, this is the equation of the line PQ.
The line PQ with P(x1,y1) and Q(x2,y2) is 
 
        |x      y       1|
        |x1     y1      1| = 0
        |x2     y2      1|

Three lines through one point. (Concurrent lines)

In the same way as for collinear points you can show (exercise) that

Three lines 
 
        a x  + b y  + c  = 0
        a' x + b' y + c' = 0
        a" x + b" y + c" = 0
are concurrent if and only if 
 
      | a    b    c |
      | a'   b'   c'| = 0
      | a"   b"   c"|

Application related with concurrent and parallel lines

 
 a x + b y = 1         (1)
 a x +   y = b         (2)
   x + b y = a         (3)
are the equations of three lines. The parameters a and b are real and different.

Examine the relative position of the three lines for all values of a and b.


  1. At least two lines are parallel.

    (1) and (2) are parallel if and only if ( a = 0 of b = 1 )
    (1) and (3) are parallel if and only if ( b = 0 of a = 1 )
    (2) and (3) are parallel if and only if ( a b = 1 )

    We now treat these cases separately

    1. a = 0 (then b is not 0) 
       
The equations of the lines are

       b y = 1         (1)
         y = b         (2)
   x + b y = 0         (3)
      The lines (1) and (2) are parallel and the third one intersects (1) and (2).
      (1) and (2) coincides if and only if ( b = 1 or b = -1)

    2. b = 0 (then a is not 0) 
       
The equations of the lines are

   a x       = 1         (1)
   a x +   y = 0         (2)
     x       = a         (3)
      The lines (1) and (3) are parallel and the second one intersects (1) and (3).
      (1) and (3) coincides if and only if ( a = 1 or a = -1)

    3. a = 1 (then b is not 1) 
       
The equations of the lines are

    x + b y = 1         (1)
    x +   y = b         (2)
    x + b y = 1         (3)
      The lines (1) and (3) coincides and the second one intersects (1) and (3).

    4. b = 1 (then a is not 1) 
       
The equations of the lines are

   a x +   y = 1         (1)
   a x +   y = 1         (2)
     x +   y = a         (3)
      The lines (1) and (2) coincides and the third one intersects (1) and (2).

    5. a.b = 1 (then a and b are different from 1) 

       
The equations of the lines are

   a x + (1/a) y = 1             (1)
   a x +       y = (1/a)         (2)
     x + (1/a) y = a             (3)
      The lines (2) and (3) are parallel but they don't coincide.
      The first line intersects the two parallel lines.

  2. No line is parallel to another one.

    a and b are different from 1 and from 0 and a.b is not 1.

    1. The three lines are concurrent if and only if 

       
  | a  b  1 |
  | a  1  b | = 0
  | 1  b  a |

<=>

  (a-1)(b-1)(a+b+1) = 0
      Since a and b are not equal to 1, we have :
      The three lines are concurrent if and only if a+b+1 = 0.

    2. a+b+1 is not 0

      The three lines form a triangle.

Orthogonality; Distances

These topics are covered in Lines - orthogonality and distances

Area of the triangle P(x1,y1) Q(x2,y2) and R(x3,y3)

The distance |Q,R| = 
 
              ___________________________
             |
            \| (x2 - x3)2  + (y2 - y3)2
From above, the equation of the line QR is 
 
        |x      y       1|
        |x2     y2      1| = 0
        |x3     y3      1|
If we calculate this determinant emanating from the first row, we find 
 
        x(y2 - y3) - y(x2-x3) + x2 y3 - x3 y2 = 0
To calculate the distance from P to the line QR, we write first the normal equation of a line QR 
 
            x(y2 - y3) - y(x2 - x3) + x2 y3 - x3 y2
            --------------------------------------- = 0
                    _________________________
                   |
                  \| (x2 - x3)2  + (y2 - y3)2

<=>
                    |x      y       1|
                    |x2     y2      1|
                    |x3     y3      1|
            --------------------------------------- = 0
                    ___________________________
                   |
                  \| (x2 - x3)2  + (y2 - y3)2
Now, to find the distance, we have to take the absolute value of the left side and we must replace x and y by the coordinates of P. The distance from P to QR is 
 
                    |x1     y1      1|
                    |x2     y2      1|
                    |x3     y3      1|
            | -------------------------------- |
                    _________________________
                   |
                  \| (x2 - x3)2 + (y2 - y3)2
The area of the triangle P(x1,y1) Q(x2,y2) and R(x3,y3) is 
 



                                         |x1     y1      1|
      _________________________          |x2     y2      1|
1    |                                   |x3     y3      1|
- . \| (x2 - x3)2  + (y2 - y3)2     | -------------------------------- |
2                                        _________________________
                                        |
                                       \| (x2 - x3)2  + (y2 - y3)2

<=>

                1    |x1     y1      1|
                -.|  |x2     y2      1| |
                2    |x3     y3      1|
This is a very simple formula to calculate the area of a triangle.
The area of the triangle P(x1,y1) Q(x2,y2) and R(x3,y3) is 
 

                1    |x1     y1      1|
                -.|  |x2     y2      1| |
                2    |x3     y3      1|



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