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)
• 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|

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|