Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Analytic > Lines in a plane

..Orthonormal axes
..Equation and slope of a line
..Direction vector of a line
..Three points on one line.
..Equation of the line PQ with P(x1,y1) and Q(x2,y2)
..Orthogonality; Distances
..Area of the triangle P(x1,y1) Q(x2,y2) and R(x3,y3)


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

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).

Three points on one line.

Say that a line l has equation ax+by+c=0 .
Three points P(x1,y1) Q(x2,y2) and R(x3,y3) are on line l if and only if 
 
        a x1 + b y1 + c = 0
        a x2 + b y2 + c = 0
        a x3 + b y3 + c = 0
From the theory of homogenious systems of linear equations, we know that previous system has a solution for a,b and c if and only if 
 
        |x1    y1     1|
        |x2    y2     1| = 0
        |x3    y3     1|
Conclusion :
Three points P(x1,y1) Q(x2,y2) and R(x3,y3) are on line l 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|

Orthogonality; Distances

These topics are in the 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 formule 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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