Themes > Science > Mathematics > Algebra > Foci of a conic section > Topics and Problems > Lines in a plane - Orthogonality; Distances

..Orthogonal lines
..Equation and slope of a line
..Direction vector of a line
..Orthogonal lines formule
..Orthogonal lines and slope

..Distance from a point to a line

..Normal vector to a line
..Normal equation of a line
..Distance between a point and a line
..Bisecting lines


Orthogonal lines

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).
(b,-a) are the coordinates of a direction vector of the line ax + by + c = 0

Orthogonal lines formule

Take two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0
Direction vectors are P(b,-a) and Q(b',-a'). Now, 
 
        l and l' are orthogonal
                <=>
        P and Q are orthogonal
                <=>
                P.Q = 0
                <=>
        b.b' + a.a' = 0
                <=>
        a.a' + b.b' = 0
Two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0 are orthogonal if and only if a.a' + b.b' = 0 .

Orthogonal lines and slope

Take two lines l : ax + by + c = 0 and l' : a'x + b'y + c' = 0
If b and b' are not zero, the slope is m = -a/b and m' = -a'/b' .
Now, 
 
        l and l'are orthogonal
                <=>
        a.a' + b.b' = 0
                <=>
         aa'
        ---- +  1  = 0
         bb'

                <=>
          m.m' + 1 = 0
                <=>
             m.m' = -1
Line l with slope m and line l' with slope m' are orthogonal if and only if m.m' = -1.

Distance from a point to a line

Normal vector to a line

Let l : ax + by + c = 0
The vector P(b,-a) is a direction vector of l.
The vector Q(a,b) is orthogonal with P because P.Q = 0.
Hence, Q(a,b) is a vector orthogonal to l.
We call Q(a,b) a normal vector to l.

Normal equation of a line

Let l : ax + by + c = 0
Then also l : rax + rby + rc = 0 with r a real number not 0.
We calculate a suitable r such that the normal vector Q(ra,rb) is a unity vector. 
 
        Q(ra,rb) is a unity vector
                <=>
                Q.Q = 1
                <=>
           ra.ra + rb.rb = 1
                <=>

          r2 (a2  + b2  ) = 1
                <=>
               1                    - 1
        r = ------------  or  r = ------------
             ___________           ____________
            V a2  + b2            V a2  + b2
Substitute the positive value of r in the equation of l. 
 
               ax +  by + c
        l :  ----------------- = 0
                ____________
               V a2  + b2
This equation is called a normal equation of l. Hence, 
 
         lx + my + n = 0 is a normal equation of line l.
                        <=>
                  Q(l,m) is a unity normal vector to l
                        <=>

                      l2  + m2  = 1

 
        l : lx + my + n = 0 is a normal equation of l.

                        <=>

                      l2  + m2  = 1

Distance between a point and a line

Take a point P(a,b) and a line l with normal equation lx + my + n = 0 .
Then Q(l,m) is a unity normal vector to l.
Call S(c,d) the intersection of l with the perpendicular dropped on l through P.
Then the distance from P to l is |P,S|.
Now, PS = r.Q and |P,S| = |r| .
 

 
        PS = r.Q

=>      S - P = r. Q

=>      S.Q - P.Q = r.Q.Q

=>      cl + dm - (al + bm) = r. 1
                        since S on l holds lc  +  md = -n

=>      -n -al - bm = r

=>      la  + mb  + n = -r

=>      |la  + mb  + n| = |r|
The distance between a point P(a,b)
and a line l with normal equation lx + my + n = 0 is
| la + mb + n | .

Bisecting lines

Take lines l and l' with normal equations lx + my + n = 0 and l'x + m'y + n' = 0 .
A point P(a,b) is on a bisecting line if and only if 
 
        distance from P to l = distance from P to l'
                        <=>
        |la + mb + n| = |l'a + m'b + n'|
                        <=>
        la + mb + n = l'a + m'b + n' or la + mb + n = -(l'a + m'b + n')
The two bisecting lines of l and l' with normal equations lx + my + n = 0 and l'x + m'y + n' = 0 are
 
        lx + my + n = l'x + m'y + n'
and
        lx + my + n = -(l'x + m'y + n')


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